Chemistry 102 Chapter 17 THERMODYNAMICS

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Chemistry 102 Chapter 17 THERMODYNAMICS"

Transcription

1 THERMODYNAMICS Thermodynamics is concerned with the energy changes that accompany chemical and physical processes. Two conditions must be fulfilled in order to observe a chemical or physical change: The change must be possible (determined by Thermodynamics) The change must occur at a reasonable speed (determined by Kinetics) Important Terms in Thermodynamics: 1. Nature of Energy: Energy is the capacity to do work, and work is defined as force acting through a distance. There are many types of energy. Among them are kinetic (moving), potential (stored), thermal energy and chemical energy. Energy can be transformed from one type to another through work. 2. System and Surroundings: System : The part of the universe we happen to be studying Surroundings: Everything outside the system Three types of systems can be present: Open: exchange of energy and matter Closed: exchange of energy but not matter Isolated: No exchange of energy or matter 1

2 3. State Function THERMODYNAMICS A quantity whose value depends only on the current state of the system, and does not depend on the prior history of the system. Two example of state functions are temperature, internal energy 2

3 INTERNAL ENERGY Internal Energy (E) is the sum of the kinetic and potential energies of all of the particles that compose the system. Internal energy is a state function. Internal Energy Kinetic Energy of the particles making the system Potential Energy of the particles making the system Results from the energy of motion of Results from electrons atoms molecules chemical bonding between atoms attraction between molecules The 1 st Law of Thermodynamics is the Law of Conservation of Energy applied to thermodynamic systems. When a system changes from one state to another (ex: warming), its internal energy changes from one definite energy to another. Change in Internal Energy : E = E products E reactants where: E products = final value of the internal energy E reactants = initial value of the internal energy 3

4 CHANGES IN INTERNAL ENERGY C (s) + O 2 (g) CO 2 (g) CO 2 (g) C (s) + O 2 (g) SUMMARY: If the reactants have higher internal energy than products, E sys is negative and energy flows out of the system and into the surroundings. If the reactants have lower internal energy than products, E sys is positive and energy flows into the system from the surroundings. 4

5 1 ST LAW OF THERMODYNAMICS The changes in internal energy ( E) are more important in thermodynamics than the absolute values of Internal Energy The changes in IE are measured by noting the exchanges of energy between the system and the surroundings. These exchanges occur as one of two forms: Heat and Work Heat (q) = the energy that moves into or out of the system because of a temperature difference between the system and its surroundings. When heat is lost, internal energy (E) decreases and q is therefore assigned a negative sign. When heat is absorbed, internal energy (E) increases and q is therefore assigned a positive sign. Work (w) = the energy exchange that results when a force (F) moves an object through a distance (d). work = force x distance If work is done by the system, internal energy (E) decreases and w is therefore assigned a negative sign. If work is done on the system, internal energy (E) increases and w is therefore assigned a positive sign. According to the 1 st law of thermodynamics, the changes in the internal energy of the system ( E) are the sum of the heat transferred (q) and the work done (w). E = q + w Examples: Determine the sign of the work for each of the changes shown below (carried out at constant P): 1) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) w = 2) CaCO 3 (s) CaO (s) + CO 2 (g) w = 3) CO 2 (g) + H 2 O (l) + CaCO 3 (s) Ca 2+ (aq) + 2 HCO 3 (aq) w = 5

6 1 ST LAW OF THERMODYNAMICS (ANALOGY) A rolling billiard ball has energy due to its motion. When it collides with the second ball, it does work, transferring energy to the second ball. The second ball now has now energy to roll away from the collision. The system can be defined as the first ball rolling across the table. The surroundings are the second ball and the pool table. Assume that when the first ball collides with the second ball, it loses all of its energy and remains still at the point of collision. After collision, the first ball loses all of its energy and therefore has E= 5.0 J. All this energy is gained by the surroundings. ( E sys = E surr ) In an actual situation, due to friction, some of the energy of the first ball is lost to the table as heat. As a result, energy change of the ball ( E sys ) is the sum of heat and work. On a table with a rough surface, more energy is lost to heat due to friction, leaving less available for work to transfer energy to the second ball. However, the sum of the heat and work still equals the change in energy of the first ball. Note that the amount of energy converted to heat and work depends on the details of the pool table and the path taken, while the change in internal energy of the rolling ball does not. This is because internal energy is a state function and only depends on the initial and final states of the ball. Heat and work are not state functions, and therefore are affected by the details of the ball s journey across the table. 6

7 Chemistry 102 HEAT AND WORK IN A PHYSICAL SYSTEM Case 1: A gas is enclosed in a vessel equipped with a movable piston, momentarily fixed in position. Heat passes from the surroundings to the vessel and the temperature is increased (q = J). In lifting the weight, the system does work equal to the energy gained by piston and the weight (w= -92 J). Heat Piston Gas E = q + w = (+165 J) + ( 92 J) = + 73 J Case 2: A gas is enclosed in a vessel equipped with a movable piston, momentarily fixed in position. Work is done on the system by pushing down the piston and thus compressing the gas (w= + 92 J). The gas pressure increases and causes the gas to warm up. This will cause heat to flow out of the system (q = 165 J). E = q + w = ( 165 J) + (+92 J) = 73 J Examples: 1. A cylinder and piston assembly is warmed by an external flame. As a result, the system absorbs 559 J of heat. The contents of the cylinder expand and perform 488 J of work during the expansion. What is the change in internal energy of the system? 2. A system releases 622 kj of heat and does 105 kj of work on the surroundings. What is the change in the internal energy of the system? 7

8 PRESSURE-VOLUME WORK To calculate the work associated with a chemical reaction, we limit discussion to those associated with a volume change and therefore called pressure-volume work. Pressure-volume work occurs when the force is caused by a volume change against an external pressure. Assume a cylinder, where the pressure exerted by the atmosphere is replaced with a piston, whose downward force of gravity, F, creates a pressure on the gas equivalent to that of the atmosphere. When the volume of the cylinder increases, it pushes against the external force. The work done by the system in expanding = (Force of Gravity) x (Distance the piston moves) w = F x h since V = A x h V F w = F x = x V A A Atmospheric pressure (P) Since the volume of cylinder is increasing, work is done by the system and therefore assigned a negative sign. w = P V 8

9 Examples: 1. In the reaction shown below, 1.00mol of Zn reacts with excess HCl to produce 1.00 mol of H 2 gas. At 25C and 1.00 atm, this amount of H 2 occupies 24.5 L. Calculate the change in Internal Energy ( U) for this reaction. (1 Latm = kj) Zn (s) + 2H 3 O + (aq) Zn 2+ (aq) + 2 H 2 O (l) + H 2 (g) q = kj/mol E = q + w q = kj w =? w (done by system) = PV = (1.00 atm)(24.5 L)(0.101 kj/1 Latm) = 2.47 kj E = q + w = ( kj ) + ( 2.47 kj) = kj 2. What is E when 1.00 mol of liquid water vaporizes at 100C? (H vap = kj/mol at 100 C) 3. Calculate the amount of work done in each of the following reactions. In each case, is the work done by or on the system? a) oxidation of HCl at 200 C 4 HCl (g) + O 2 (g) 2 Cl 2 (g) + 2 H 2 O (g) b) The decomposition of NO 2 at 300 C 2 NO (g) N 2 (g) + O 2 (g) 9

10 Examples: 4. A cylinder equipped with a piston expands against an external pressure of 1.58 atm. If the initial volume is L and the final volume is L, how much work (in J) is done? 5. When fuel is burned in a cylinder equipped with a piston, the volume expands from L to 1.45 L against an external pressure of 1.02 atm. In addition, 875 J is emitted as heat. What is E for the burning of the fuel? 6. Which statement is true of the internal energy of the system and surroundings following a process in which E sys = + 65 kj? a) The system and surroundings both lose 65 kj of heat. b) The system and surroundings both gain 65 kj of heat. c) The system loses 65 kj of energy and the surroundings gain 65 kj of energy. d) The system gains 65 kj of energy and the surroundings lose 65 kj of energy. 10

11 ENTHALPY AND ENTHALPY CHANGE Recall: Enthalpy = The Heat of Reaction, H, at constant pressure (q p ) Actually: Enthalpy = H = E + PV Since internal energy (E), pressure (P), and volume (V) are state functions, it follows that enthalpy (H) is also a state function. Therefore, the change in enthalpy (H) for any process occurring at constant pressure is given by the expression: H = E + P V H = (q p + w) + P V = q p + w w H = q p The enthalpy or reaction equals the Heat of Reaction at constant pressure. In practice, certain heats of reaction are measured and tabulated as heats of formation (H f ). H f = Standard Enthalpy of Formation of a Substance Enthalpy change for the formation of one mole of substance in its standard state from its elements in their reference form ( most stable form at 25C and 1.00 atm) and in their standard states (available in Appendix II in your textbook). The heat absorbed (endothermic) or released (exothermic) during a chemical reaction is defined as standard enthalpy change (H). H for a reaction can be calculated from known Hf values for reactants and products. H = n H f (products) n H f (reactants) 11

12 Examples: 1. Given the H f values below, calculate the heat absorbed or evolved in the reaction between NH 3 gas and CO 2 gas that yields aqueous urea (NH 2 CONH 2 ) and liquid water. 2 NH 3 (g) + CO 2 (g) CO(NH 2 ) 2 (aq) + H 2 O (l) H =??? H f (kj/mol) H = n H f (products) n H f (reactants) H = [( 319.2) + ( kj)] [2 x ( 45.9 ) + ( 393.5)] = kj Since H is negative, we conclude that the reaction is exothermic. 2. Use data in appendix II in your text to find H for the reactions shown below: a) NH 4 NO 3 (s) NO 2 (g) + 2 H 2 O (l) b) SiO 2 (s) + 3 C (graphite) SiC (s) + 2 CO (g) 12

13 SPONTANEOUS & NONSPONTANEOUS PRECESSES Spontaneous Processes: Physical or chemical changes that occurs by themselves, without outside assistance Spontaneous Physical Change: (ball rolling from the top of a hill) high potential energy Spontaneous Chemical Change: ( reaction of solid potassium with liquid water) low potential energy spontaneous 2 K(s) + 2 H 2 O (l) 2 KOH(aq) + H 2 (g) In order for spontaneous changes to go in the opposite direction work would have to be expended, and the changes would be called non-spontaneous. Non-spontaneous Physical Change: (Rolling a ball uphill) Non-spontaneous Chemical Change Change: (Obtaining K(s) from KOH(aq)) non-spontaneous 2 KOH (aq) + H 2 (g) 2 K(s) + 2 H 2 O(l) The Second Law of Thermodynamics determines if a reaction is spontaneous or not. Exothermic Reactions (H 0) are not necessarily spontaneous and Endothermic Reactions (H 0) are not necessarily non-spontaneous 13

14 ENTROPY (S) Entropy is the thermodynamic quantity that is a measure of the randomness or disorder in a system. Entropy is a state function: - the quantity of entropy in a substance depends only on variables that determine the state of the substance (temperature and pressure) Entropy is measured in J/K (SI unit) Entropy increases when the disorder in a sample of the substance increases: Example: Consider the melting of ice at 0C and 1 atm. MELTING ICE H 2 O molecules occupy regular fixed positions (crystal lattice) Ordered crystalline structure LIQUID WATER H 2 O molecules move about freely, giving a disordered structure Less-ordered structure Lower Entropy Higher Entropy S i = Initial Entropy = Entropy of Ice (41 J/K) S f = Final Entropy = Entropy of liquid water (63 J/K) melting at 0C and 1 atm H 2 O(s) H 2 O(l) S = 22 J/K S i = 41 J/K S f = 63 J/K In any natural process the degree of disorder (Entropy) increases. Natural processes are changes that occur by themselves (spontaneous changes). Second Law of Thermodynamics in reference to the system and its surroundings The Total Entropy of a System and its Surroundings always increases for a Spontaneous Process. ENTROPY ENERGY is created during a spontaneous change can be neither created or destroyed during a spontaneous change 14

15 ENTROPY & 2 ND LAW OF THERMODYNAMICS The 2 nd Law of Thermodynamics states that for any spontaneous process, the entropy of the universe increases (S univ > 0). Therefore, the criterion for spontaneity is the entropy of the universe. Processes that increase the entropy of the universe those that result in greater dispersal or randomization of energy occur spontaneously. Consider the example below: the expansion of a gas into a vacuum (a spontaneous process with no change in enthalpy). When the valve between the flasks opens, the gas in the left flask spontaneously expands into a vacuum. Since the pressure against the expansion is zero, no work is done by the gas. However, even though the total energy of the gas does not change, the entropy does change. This can easily be seen by the various ways the gas can distribute itself in the flasks (3 of which are shown below). Probability analysis indicates that it would be more probable to find the atoms in state C compared to either states A or B. And it can also be seen that state C has greater entropy than either state A or B. Therefore, the change of entropy in transitioning from state A to state C is positive because state C has greater entropy than state A. S = S final S initial S > 0 15

16 ENTROPY CHANGE FOR A PHASE TRANSITION The entropy of a sample of matter increases as it changes state from a solid to a liquid or from a liquid to a gas. This is due to the increase in disorder of molecules in liquid compared to solid and gas compared to liquid. We can therefore predict the sign of S for processes involving changes of state (or phase change). In general, entropy increases ( S > 0) for each of the following: The phase transition from a solid to a liquid the phase transition from a solid to a gas the phase transition from a liquid to a gas an increase in the number of moles of a gas during a chemical reaction Examples: 1. Predict the sign of S for each of the following processes: a) the boiling of water S = b) I 2 (g) I 2 (s) S = c) CaCO 3 (s) CaO (s) + CO 2 (g) S = 16

17 HEAT TRANSFER & CHANGES IN ENTROPY OF SURROUNDINGS The criterion for spontaneity is an increase in the entropy of the universe. However, there are several spontaneous processes that include a decrease in entropy. For example, when water freezes below 0C, the entropy of the water decreases, yet the process is spontaneous. The 2 nd Law of Thermodynamics states that for a spontaneous process, the entropy of universe increases ( S univ > 0). In the example above, even though the entropy of water decreases, the entropy of the universe must increase in order for the process to be spontaneous. Similar to the distinction of a system and surroundings in thermodynamics, we can distinguish between the entropy of the system and the surroundings. In the example of the freezing water, S sys is the entropy change of water, and the S surr is the entropy change of the surroundings. The entropy change of the universe is then the sum of the entropy changes of the system and the surrounding. S univ = S sys + S surr If a spontaneous process includes a decrease in S sys, then it would follow that there must be a greater increase in S surr in order for the S univ to increase. Since freezing of water is an exothermic process, it would follow that the heat given off by the process increases the entropy of the surroundings by a greater amount than the decrease of entropy of the system, making it a spontaneous process. To summarize: An exothermic process increases the entropy of the surroundings. An endothermic process decreases the entropy of the surroundings. 17

18 TEMPERATURE DEPENDENCE OF S surr Because of the large increase in S surr, the freezing of water is spontaneous at low temperatures. However, freezing of water is NOT spontaneous at high temperatures. Why? Units of entropy are J/K. Therefore, entropy is a measure of energy dispersal (joules) per unit temperature (K). The higher the temperature, the lower the amount of entropy change for a given amount of energy dispersed. Water does not freeze spontaneously at high temperatures, because the increase in S surr is small compared to decrease in S sys, making S univ negative. S univ = S sys + S surr (for water freezing) 18

19 QUANTIFYING ENTROPY CHANGE OF SURROUNDINGS & SYSTEM We know that when a system exchanges heat with surroundings, it changes the entropy of the surroundings. At constant pressure, q sys can be used to quantify the entropy change of the surroundings. In general, A process that emits heat into the surroundings (q sys = negative) increases the entropy of the surroundings (positive S surr ) A process that absorbs heat from the surroundings (q sys = positive) decreases the entropy of the surroundings (negative S surr ) The magnitude of the change in entropy of the surroundings is proportional to the magnitude of the q sys. S surr α q sys It can also be shown that for a given amount of heat exchanged with the surroundings, the magnitude of S surr is inversely proportional to the temperature. S surr α 1/T Combining the above relationships: S = surr -q T sys At constant pressure, q sys = H sys. Therefore, S = surr - H T sys (constant P, T) 19

20 Examples: 1. Calculate S surr for the melting of ice (Heat absorbed = Heat of Fusion = H fus = 6.0 kj/mol) 3 - Hfus - 6.0x10 J S surr = = = -22 J/K T 273 K 2. Liquid ethanol, C 2 H 6 O (l), at 25 C has an entropy of 161 J/(mol x K). If the heat of vaporization (H vap ) at 25 C is 42.3 kj/mol, what is the entropy of the vapor in equilibrium with the liquid at 25 C? C 2 H 6 O (l) C 2 H 6 O (g) H vap = kj/mol S sys = (S gas S liq ) > 0 3 qsys H vap kj/mol 10 J S sys = = = x = 142 J/K mol T T 298 K 1 kj S gas = S liq + S sys = 161 J/K J/K = 303 J/K 3. Consider the combustion of propane gas: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) H sys = 2044 kj a) Calculate the S surr associated with this reaction occurring at 25C. b) Determine the sign of S sys c) Determine the sign of S univ. Will this reaction be spontaneous? 20

21 GIBBS FREE ENERGY The relationships between enthalpy change in a system and the entropy change in the surroundings is shown below: S = surr - H T sys Also, for any process, the entropy change of the universe is the sum of the entropy change of the system and the entropy change of the surroundings. S univ = S sys + S surr Combining the two relationships above leads to the following relationship: S = S - univ sys H T sys Multiplying equation above with T, we arrive at the expression: T S univ = H sys T S sys Willard Gibbs (American physicist) combined together the concepts of enthalpy (H) and entropy (S) into a single thermodynamic function called Gibbs Free Energy (G), defined as shown below: G = H TS The change in free energy ( G) is expressed as: G = H T S Since S univ is a criterion for spontaneity, G is also a criterion for spontaneity (but opposite in sign), and is commonly called chemical potential because it is analogous to mechanical potential energy. Just as mechanical systems tend toward lower potential energy, chemical systems tend toward lower Gibbs free energy. 21

22 GIBBS FREE ENERGY Summarizing Gibbs Free Energy (at constant T & P): G is proportional to the negative of S univ. A decrease in Gibbs free energy ( G < 0) corresponds to a spontaneous process. An increase in Gibbs free energy ( G > 0) corresponds to a nonspontaneous process. 0 + G < 0 G > 0 Reaction is spontaneous Product-favored reaction Forward reaction is favored Reaction is non-spontaneous Reactant-favored reaction Reverse reaction is favored Conclusion: G gives a composite of the two factors that contribute to spontaneity (H and S) 22

23 SPONTANEITY & FREE ENERGY Consider the following situations: Case 1: H is negative and S is positive (exothermic) (increase in entropy) Reaction is spontaneous at all temperatures Case 2: 2 N 2 O (g) 2 N 2 (g) + O 2 (g) H = kj (2 mol gas) (3 mol gas) H is positive and S is negative (endothermic) (decrease in entropy) Reaction is nonspontaneous at all temperatures 3 O 2 (g) 2 O 3 (g) H = kj (3 mol gas) (2 mol gas) Case 3: H is negative and S is negative (exothermic) (decrease in entropy) Reaction is spontaneous only at low temperatures Case 4: H 2 O (l) H 2 O (s) H = 6.01 kj H is positive and S is positive (endothermic) (increase in entropy) Reaction is spontaneous only at high temperatures H 2 O (l) H 2 O (g) H = kj (at 100C) 23

24 SPONTANEITY & FREE ENERGY Examples: 1. Predict the conditions (high temperature, low temperature, all temperatures or no temperatures) under which each reaction below is spontaneous: a) H 2 O (g) H 2 O (l) b) CO 2 (s) CO 2 (g) c) 2 NO 2 (g) 2 NO (g) + O 2 (g) (endothermic) 2. For the reaction shown below, calculate G at 25C and determine whether the reaction is spontaneous. If the reaction is not spontaneous at 25C, determine at what temperature (if any) the reaction becomes spontaneous. CCl 4 (g) C (s, graphite) + 2 Cl 2 (g) H = kj S = J/K 24

25 ENTROPY CHANGES IN CHEMICAL REACTIONS Similar to the H for a reaction, we can define the standard entropy change for a reaction (S) as the entropy for a process in which all reactants and products are in their standard states. The standard states are defined as: Gas: pure gas at pressure exactly 1 atm. Liquid or Solid: pure substance at most stable form and pressure of 1 atm and temperature of 25C. Solution: concentration of 1 M. Since entropy is a state function, the standard change in entropy of a reaction is: S = S products S reactants The standard molar entropy (S) of a substance is needed to calculate the S for a reaction. The standard molar entropy of a substance is measured against the absolute value of zero as defined by the 3 rd law of thermodynamics. Third Law of Thermodynamics: The third law of thermodynamics states that: The entropy of a perfect crystal at absolute zero (0 K) is zero A perfect crystal at 0 K has only one possible way to arrange its components. Standard molar entropy values for substances are tabulated in textbooks and available in Appendix II. 25

26 STANDARD MOLAR ENTROPY The standard entropy of any substance is affected by the state of the substance, molar mass, molecular complexity and the extent of dissolution. As the molar mass of a substance increases, the value of standard molar entropy also increases. For example, the standard molar entropy of noble gases increases progressively from He to Xe. Similarly, the standard molar entropy of a substance depends on its state, as shown in the example below: In general, S gas > S liq > S solid For a given state of matter, standard molar entropy also increases with molecular complexity. For example, consider the following: Examples: 1. Rank each set of substances below in order of increasing standard molar entropy (S): a) I 2 (g), F 2 (g), Br 2 (g), Cl 2 (g) b) NH 3 (g), Ne (g), SO 2 (g), CH 3 CH 2 OH (g), He (g) 26

27 CALCULATING S FOR A REACTION Recall that the entropy usually increases in the following situations: (a) A reaction in which a molecule is broken into two or more smaller molecules. Example: fermentation of glucose (grape sugar) to alcohol: C 6 H 12 O 6 (s) 2 C 2 H 5 OH (l) + 2 CO 2 (g) 1 large molecule 4 smaller molecules (b) A reaction in which there is an increase in moles of gas. This may result from a molecules breaking up, in which case Rule (a) and Rule (b) are related. Example: the burning of acetylene in oxygen: 2 C 2 H 2 (g) + 3 O 2 (g) 4 CO 2 (g) + 2 H 2 O (g) 5 moles of gas 6 moles of gas n gas = 6 5 = +1 The standard entropy change (S) for a reaction can also be calculated from the standard entropies of reactants and products as shown below: S = ns (products) ns (reactants) Examples: 1. For each pair of substances, choose the one with the higher expected value of standard molar entropy. Explain the reasons for your choices. a) CO (g) CO 2 (g) b) CH 3 OH (l) CH 3 OH (g) c) Ar (g) CO 2 (g) d) NO 2 (g) CH 3 CH 2 CH 3 (g) 27

28 2. Predict the sign of S 0 for each reaction shown below. If you cannot predict the sign for any reaction, state why. a) C 2 H 2 (g) + 2 H 2 (g) C 2 H 6 (g) S = b) N 2 (g) + O 2 (g) 2 NO (g) S = c) 2 C (s) + O 2 (g) 2 CO (g) S = 3. Using tabulated S values in your text, calculate S for the reactions shown below: 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) 28

29 CALCULATING STANDARD FREE ENERGY CHANGES The standard free energy change (G) for a reaction is the criterion for its spontaneity, and can be calculated by 3 different methods. Method 1: Earlier in this chapter we learned how to use tabulated values of H f and S to calculate H and S for a reaction. We can now use the H and S values to calculate the G value for a reaction, using the equation: G = H TS Since H f and S values are tabulated at 25C, the equation above should only be valid at 25C. However, since H f and S values are affected very little by temperature, we can estimate the G values for temperatures other than 25C. Examples: 1. Given H f and S values shown below, calculate G for the reaction shown below, and determine if the reaction is spontaneous. SO 2 (g) + ½ O 2 (g) SO 3 (g) 2. Use the data in Appendix IIA of your text to calculate H, S and G at 25C for the reaction shown below. Is the reaction spontaneous? If not, would a change in temperature make it spontaneous? If so, should the temperature be raised or lowered from 25C? NH 4 Cl (s) HCl (g) + NH 3 (g) 29

30 Method 2: CALCULATING STANDARD FREE ENERGY CHANGES Because free energy is a state function, change in free energy or reactions (G) can also be calculated from the free energy of the reactants and products. In order to do this, we define the free energy of formation (G f ) for a substance as follows: The free energy of formation (G f ) is the change in free energy when 1 mole of a compound in its standard state forms from its constituent elements in their standard states. The free energy of formation of pure elements in their standard states is zero. The free energy change of a reaction can then be calculated using the equation below: G = ng f (products) mg f (reactants) Examples: 1. Calculate the standard free energy change (G) for the following reaction at 25 C, using standard free energies of formation given below: Na 2 CO 3 (s) + H + (aq) 2 Na + (aq) + HCO 3 (aq) G f (kj/mol) : G = [(2) x ( 261.9) + ( 587.1)] [( ) + 0] kj = 62.8 kj 2. Calculate the standard free-energy change (G) for the reaction shown below, and determine whether it is spontaneous or not. (Use G f values in your textbook). 2 NOCl (g) 2 NO (g) + Cl 2 (g) 3. Calculate the standard free-energy change (G) for the reaction shown below, and compare with the value obtained by method 1 on the previous page. (Use G f values in your textbook). NH 4 Cl (s) HCl (g) + NH 3 (g) 30

31 Method 3: CALCULATING STANDARD FREE ENERGY CHANGES Since free energy is a state function, the following relationships can be used to calculate G for a reaction: If a chemical equation is multiplied by some factor, then G for the reaction is also multiplied by the same factor. If a chemical equation is reversed, then G changes sign. If a chemical equation can be expressed as the sum of a series of steps, then G for the overall equation is the sum of the free energies of reactions for each step. Examples: 1. Find G for the reaction shown below: 3 C (s) + 4 H 2 (g) C 3 H 8 (g) Use the following reactions with known G values: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) G = 2074 kj C (s) + O 2 (g) CO 2 (g) G = kj 2 H 2 (g) + O 2 (g) 2 H 2 O (g) G = kj 2. Determine G for the reaction shown below: Fe 2 O 3 (s) + 3 CO (g) 2 Fe (s) + 3 CO 2 (g) Use the following reactions with known G values: 2 Fe (s) + 3/2 O 2 (g) Fe 2 O 3 (s) G = kj CO (g) + ½ O 2 (g) CO 2 (g) G = kj 31

32 FREE ENERGY and USEFUL WORK In a spontaneous reaction: the free energy is lowered as reactants change to products, the change in free energy is released as free-energy change( G) this G can be harnessed to perform useful work A spontaneous reaction can be used to obtain useful work The change in free energy of a chemical reaction represents the maximum amount of energy available (or free) to do work (if G is negative). Maximum useful work = w max = G completely In an ideal situation: G Useful work converted into converted In real situations: G Some Useful Work + Some Entropy into Conclusions: 1. In theory, all of the free-energy decrease liberated during a spontaneous chemical change can be used to do work (this would be w max ) 2. In practice, less work is obtained and the difference appears as an increase in entropy. C (s, graphite) + 2 H 2 (g) CH 4 (g) H = 74.6 kj S = 80.8 J/K G = 50.5 kj Why is only 50.5 kj available as free energy, even though 74.6 kj is produced in the reaction? How is this reaction spontaneous ( G< 0) when entropy decreases? 32

33 FREE ENERGY CHANGES FOR NONSTANDARD STATES The standard free energy changes for a reaction ( G) are only useful for a very narrow set of conditions where reactants and products are in their standard states. For example, consider the standard free energy change for evaporation of liquid water: H 2 O (l) H 2 O (g) G = kj/mol Since G for this process is positive, it is nonspontaneous. Yet when water is spilled under ordinary conditions, it evaporates spontaneously. Why? This is because ordinary conditions are not standard states. For example, for water vapor, standard conditions are those were P H2O would be 1 atm. Based on the G value given above, under these conditions, the water vapor would condense spontaneously. Free energy changes under nonstandard conditions ( G) can be calculated from G according to the equation below: G = G + RT lnq Q = Thermodynamic Reaction Quotient for reactions involving gases Q is obtained from partial pressures. for reactions in solutions, Q is obtained from molar concentrations R = Gas Law Constant in units of energy = J/mol K G = Standard Free-Energy Change G = Free-Energy Change when Reactants in nonstandard states are changed to products in non-standard states 33

34 FREE ENERGY CHANGES FOR NONSTANDARD STATES The relationship of free energy change for evaporation of water and pressure of water can be seen from diagram below: Under standard condition, Q = P H2O = 1 atm and G = G, as expected. G = G + RT ln (1) = kj/mol At equilibrium, where water vapor has a pressure of atm, G = 0 as shown below: G = G + RT ln (0.0313) = (8.314 J/molK)(298) ln (0.0313) = kj/mol 8.59 kj/mol = 0 Under nonstandard condition, P H2O = atm and G can be calculated as shown below: G = G + RT ln (0.005) = (8.314 J/molK)(298) ln (0.005) = kj/mol 13.1 kj/mol = 4.5 kj/mol Under these conditions, since G< 0, evaporation would be spontaneous. 34

35 FREE ENERGY CHANGES FOR NONSTANDARD STATES Examples: 1. For the reaction shown below: 2 NO (g) + O 2 (g) 2 NO 2 (g) G = 71.2 kj Calculate G under the following conditions: P NO = atm P O2 = atm P NO2 = 2.00 atm 2. Calculate G for the reaction shown below if the reaction mixture consists of 1.0 atm N 2, 3.0 atm H 2 and 1.0 atm NH 3. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G = kj 35

36 FREE ENERGY AND EQUILIBRIUM G determines the spontaneity of a reaction when reactants and products are in their standard states. Previously we learned that equilibrium constant (K) determines how far a reaction proceeds, also a measure of spontaneity. Therefore, it would follow that free energy changes and equilibrium constant are related. This relationship is shown below: G = G + RT ln Q At equilibrium 0 = G + RT ln K G = RT ln K It follows that: 1. For reactions involving only gases: K = K p 2. For reactions involving only solutes in liquid solutions: K = K c 3. For net ionic equations: K = K sp 36

37 RELATIONSHIP OF G, K, AND SPONTANEITY G = RT ln K When K > 1 When K 1 When K <1 Reactants Products Reactants Products Reactants Products log K > 0 log K 0 log K < 0 G < 0 G 0 G > 0 Reaction is spontaneous Reaction gives an equilibrium mixture Forward reaction is non-spontaneous Reverse reaction is spontaneous Examples: 1. Use the standard free energies of formation given to determine the equilibrium constant (K) for the following reaction at 25 C: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G f (kj/mol) Use the standard free energies in your text to calculate G and K for the following reaction at 25 C: 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 37

38 CHANGE OF FREE ENERGY WITH TEMPERATURE How can G be found for temperatures other than standard temperatures (25 C)? An approximate method used to calculate G T is based on the assumption that both H and S are constant with respect to temperature (only approximately true) Then: G T = H - TS (a convenient approximation for G T ) NOTE: G T is strongly temperature dependent G T = Change in free Energy for a substance: at 1 atm of pressure (standard pressure) and at the specified temperature, T (nonstandard temperature) depends on the value of determined by SPONTANEITY G T TEMPERATURE Meaning: It follows that SPONTANEITY IS TEMPERATURE DEPENDENT! Some chemical changes may be non-spontaneous at one temperature but spontaneous at another temperature. Examples: 1. Sodium Carbonate, Na 2 CO 3, can be prepared by heating sodium hydrogen carbonate, NaHCO 3. Given the H f and S values below, estimate the temperature at which NaHCO 3 decomposes to products at 1 atm. 2 NaHCO 3 (s) Na 2 CO 3 + H 2 O (g) + CO 2 (g) H f (kj/mol) : S (J/K) :

39 CHANGE OF FREE ENERGY WITH TEMPERATURE Examples: 2. Given H and S values at 25C, estimate G at 400 K for the reaction shown below: 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) H = kj S = J/K 3. Methanol can be prepared from reaction of CO and H 2, as shown below: CO (g) + 2 H 2 (g) CH 3 OH (g) H = 90.7 kj S = J/K a) Is this reaction spontaneous under standard conditions at 25C? b) Calculate G at 500 K. Is the reaction spontaneous under standard conditions at this temperature? 39

40 Answers to In-Chapter Problems: Page 5 Example No. Answer 1 positive 2 negative 3 positive 7 1 E = +71 J 2 E = 727 kj kj 9 3a 3b kj 0 J J 10 5 E = 998 J 6 d a 2b 1a 1b 1c 3a H = kj H = kj positive negative positive S surr = +6.86x10 3 J/K 20 3b S sys = c 1a 1b 1c 2 S univ = + ; reaction is spontaneous spontaneous at low temperature; nonspontaneous at high temperature nonspontaneous at low temperature; spontaneous at high temperature nonspontaneous at low temperature; spontaneous at high temperature G = kj; reaction is not spontaneous The reaction becomes spontaneous at T = 673 K 40

41 Page Example No. 1a 1b 1a 1b 1c 1d 2a 2b 2c Answer CO 2 (g) ; greater molar mass and complexity CH 3 OH (g) ; gas phase CO 2 (g); greater molar mass and complexity CH 3 CH 2 CH 3 ; greater complexity S = negative S = inconclusive S = positive 3 S = J/K G = 70.9 kj ; spontaneous 2 H = kj S = J/K G = 91.2 kj nonspontaneous; becomes spontaneous at high temperature 2 G = kj ; nonspontaneous 3 G = 91.2 kj ; value is comparable to that calculated by method 1 1 G = 23 kj 2 G = 29.4 kj 1 G = 50.7 kj 2 G = 41.5 kj 1 K= 6.9x K= 1.39x T= K 2 G T = kj 39 3a 3b G = 24.7 kj ; reaction is spontaneous at 25C G T = kj ; reaction is nonspontaneous at 500 K 41

THERMOCHEMISTRY & DEFINITIONS

THERMOCHEMISTRY & DEFINITIONS THERMOCHEMISTRY & DEFINITIONS Thermochemistry is the study of the study of relationships between chemistry and energy. All chemical changes and many physical changes involve exchange of energy with the

More information

The first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.

The first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work. The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed

More information

Chapter 20. Thermodynamics p. 811 842. Spontaneity. What have we learned about spontaneity during this course?

Chapter 20. Thermodynamics p. 811 842. Spontaneity. What have we learned about spontaneity during this course? Chapter 20 p. 811 842 Spontaneous process: Ex. Nonspontaneous process: Ex. Spontaneity What have we learned about spontaneity during this course? 1) Q vs. K? 2) So.. Spontaneous process occurs when a system

More information

Spontaneity of a Chemical Reaction

Spontaneity of a Chemical Reaction Spontaneity of a Chemical Reaction We have learned that entropy is used to quantify the extent of disorder resulting from the dispersal of matter in a system. Also; entropy, like enthalpy and internal

More information

Standard Free Energies of Formation at 298 K. Average Bond Dissociation Energies at 298 K

Standard Free Energies of Formation at 298 K. Average Bond Dissociation Energies at 298 K 1 Thermodynamics There always seems to be at least one free response question that involves thermodynamics. These types of question also show up in the multiple choice questions. G, S, and H. Know what

More information

Chemical Thermodynamics

Chemical Thermodynamics Chemical Thermodynamics David A. Katz Department of Chemistry Pima Community College Tucson, AZ 85709, USA First Law of Thermodynamics The First Law of Thermodynamics was expressed in the study of thermochemistry.

More information

Thermochemistry. Thermochemistry 1/25/2010. Reading: Chapter 5 (omit 5.8) As you read ask yourself

Thermochemistry. Thermochemistry 1/25/2010. Reading: Chapter 5 (omit 5.8) As you read ask yourself Thermochemistry Reading: Chapter 5 (omit 5.8) As you read ask yourself What is meant by the terms system and surroundings? How are they related to each other? How does energy get transferred between them?

More information

3. Of energy, work, enthalpy, and heat, how many are state functions? a) 0 b) 1 c) 2 d) 3 e) 4 ANS: c) 2 PAGE: 6.1, 6.2

3. Of energy, work, enthalpy, and heat, how many are state functions? a) 0 b) 1 c) 2 d) 3 e) 4 ANS: c) 2 PAGE: 6.1, 6.2 1. A gas absorbs 0.0 J of heat and then performs 15.2 J of work. The change in internal energy of the gas is a) 24.8 J b) 14.8 J c) 55.2 J d) 15.2 J ANS: d) 15.2 J PAGE: 6.1 2. Calculate the work for the

More information

Name AP CHEM / / Collected AP Exam Essay Answers for Chapter 16

Name AP CHEM / / Collected AP Exam Essay Answers for Chapter 16 Name AP CHEM / / Collected AP Exam Essay Answers for Chapter 16 1980 - #7 (a) State the physical significance of entropy. Entropy (S) is a measure of randomness or disorder in a system. (b) From each of

More information

Name Date Class THERMOCHEMISTRY. SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510)

Name Date Class THERMOCHEMISTRY. SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510) 17 THERMOCHEMISTRY SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510) This section explains the relationship between energy and heat, and distinguishes between heat capacity and specific heat.

More information

Chapter 5. Thermochemistry

Chapter 5. Thermochemistry Chapter 5. Thermochemistry THERMODYNAMICS - study of energy and its transformations Thermochemistry - study of energy changes associated with chemical reactions Energy - capacity to do work or to transfer

More information

Thermodynamics. Thermodynamics 1

Thermodynamics. Thermodynamics 1 Thermodynamics 1 Thermodynamics Some Important Topics First Law of Thermodynamics Internal Energy U ( or E) Enthalpy H Second Law of Thermodynamics Entropy S Third law of Thermodynamics Absolute Entropy

More information

Chapter 18 Homework Answers

Chapter 18 Homework Answers Chapter 18 Homework Answers 18.22. 18.24. 18.26. a. Since G RT lnk, as long as the temperature remains constant, the value of G also remains constant. b. In this case, G G + RT lnq. Since the reaction

More information

Bomb Calorimetry. Example 4. Energy and Enthalpy

Bomb Calorimetry. Example 4. Energy and Enthalpy Bomb Calorimetry constant volume often used for combustion reactions heat released by reaction is absorbed by calorimeter contents need heat capacity of calorimeter q cal = q rxn = q bomb + q water Example

More information

Sample Exercise 15.1 Writing Equilibrium-Constant Expressions

Sample Exercise 15.1 Writing Equilibrium-Constant Expressions Sample Exercise 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for K c for the following reactions: Solution Analyze: We are given three equations and are asked to write

More information

Name Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question.

Name Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. Assessment Chapter Test A Chapter: States of Matter In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. 1. The kinetic-molecular

More information

Test Review # 9. Chemistry R: Form TR9.13A

Test Review # 9. Chemistry R: Form TR9.13A Chemistry R: Form TR9.13A TEST 9 REVIEW Name Date Period Test Review # 9 Collision theory. In order for a reaction to occur, particles of the reactant must collide. Not all collisions cause reactions.

More information

Reading. Spontaneity. Monday, January 30 CHEM 102H T. Hughbanks

Reading. Spontaneity. Monday, January 30 CHEM 102H T. Hughbanks Thermo Notes #3 Entropy and 2nd Law of Thermodynamics Monday, January 30 CHEM 102H T. Hughbanks Reading You should reading Chapter 7. Some of this material is quite challenging, be sure to read this material

More information

Thermodynamics: First Law, Calorimetry, Enthalpy. Calorimetry. Calorimetry: constant volume. Monday, January 23 CHEM 102H T.

Thermodynamics: First Law, Calorimetry, Enthalpy. Calorimetry. Calorimetry: constant volume. Monday, January 23 CHEM 102H T. Thermodynamics: First Law, Calorimetry, Enthalpy Monday, January 23 CHEM 102H T. Hughbanks Calorimetry Reactions are usually done at either constant V (in a closed container) or constant P (open to the

More information

Thermochemistry. r2 d:\files\courses\1110-20\99heat&thermorans.doc. Ron Robertson

Thermochemistry. r2 d:\files\courses\1110-20\99heat&thermorans.doc. Ron Robertson Thermochemistry r2 d:\files\courses\1110-20\99heat&thermorans.doc Ron Robertson I. What is Energy? A. Energy is a property of matter that allows work to be done B. Potential and Kinetic Potential energy

More information

Thermodynamics Worksheet I also highly recommend Worksheets 13 and 14 in the Lab Manual

Thermodynamics Worksheet I also highly recommend Worksheets 13 and 14 in the Lab Manual Thermodynamics Worksheet I also highly recommend Worksheets 13 and 14 in the Lab Manual 1. Predict the sign of entropy change in the following processes a) The process of carbonating water to make a soda

More information

States of Matter CHAPTER 10 REVIEW SECTION 1. Name Date Class. Answer the following questions in the space provided.

States of Matter CHAPTER 10 REVIEW SECTION 1. Name Date Class. Answer the following questions in the space provided. CHAPTER 10 REVIEW States of Matter SECTION 1 SHORT ANSWER Answer the following questions in the space provided. 1. Identify whether the descriptions below describe an ideal gas or a real gas. ideal gas

More information

Thermodynamics- Chapter 19 Schedule and Notes

Thermodynamics- Chapter 19 Schedule and Notes Thermodynamics- Chapter 19 Schedule and Notes Date Topics Video cast DUE Assignment during class time One Review of thermodynamics 1_thermo_review AND Review of thermo Wksheet 2.1ch19_intro Optional: 1sc_thermo

More information

11 Thermodynamics and Thermochemistry

11 Thermodynamics and Thermochemistry Copyright ç 1996 Richard Hochstim. All rights reserved. Terms of use.» 37 11 Thermodynamics and Thermochemistry Thermodynamics is the study of heat, and how heat can be interconverted into other energy

More information

Thermodynamics and Equilibrium

Thermodynamics and Equilibrium Chapter 19 Thermodynamics and Equilibrium Concept Check 19.1 You have a sample of 1.0 mg of solid iodine at room temperature. Later, you notice that the iodine has sublimed (passed into the vapor state).

More information

Thermodynamics explores the connection between energy and the EXTENT of a reaction but does not give information about reaction rates (Kinetics).

Thermodynamics explores the connection between energy and the EXTENT of a reaction but does not give information about reaction rates (Kinetics). Thermodynamics explores the connection between energy and the EXTENT of a reaction but does not give information about reaction rates (Kinetics). Rates of chemical reactions are controlled by activation

More information

ENERGY. Thermochemistry. Heat. Temperature & Heat. Thermometers & Temperature. Temperature & Heat. Energy is the capacity to do work.

ENERGY. Thermochemistry. Heat. Temperature & Heat. Thermometers & Temperature. Temperature & Heat. Energy is the capacity to do work. ENERGY Thermochemistry Energy is the capacity to do work. Chapter 6 Kinetic Energy thermal, mechanical, electrical, sound Potential Energy chemical, gravitational, electrostatic Heat Heat, or thermal energy,

More information

AP* Chemistry THERMOCHEMISTRY

AP* Chemistry THERMOCHEMISTRY AP* Chemistry THERMOCHEMISTRY Terms for you to learn that will make this unit understandable: Energy (E) the ability to do work or produce heat ; the sum of all potential and kinetic energy in a system

More information

Chemistry B11 Chapter 4 Chemical reactions

Chemistry B11 Chapter 4 Chemical reactions Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl

More information

Unit 19 Practice. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.

Unit 19 Practice. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question. Name: Class: Date: Unit 19 Practice Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The first law of thermodynamics can be given as. A) E = q + w B) =

More information

Thermodynamics. S (reactants) S S (products) AP Chemistry. Period Date / / R e v i e w. 1. Consider the first ionization of sulfurous acid:

Thermodynamics. S (reactants) S S (products) AP Chemistry. Period Date / / R e v i e w. 1. Consider the first ionization of sulfurous acid: AP Chemistry Thermodynamics 1. Consider the first ionization of sulfurous acid: H 2SO 3(aq) H + (aq) + HSO 3 - (aq) Certain related thermodynamic data are provided below: H 2SO 3(aq) H + (aq) HSO 3 - (aq)

More information

UNIT 1 THERMOCHEMISTRY

UNIT 1 THERMOCHEMISTRY UNIT 1 THERMOCHEMISTRY THERMOCHEMISTRY LEARNING OUTCOMES Students will be expected to: THERMOCHEMISTRY STSE analyse why scientific and technological activities take place in a variety individual and group

More information

Reading: Moore chapter 18, sections 18.6-18.11 Questions for Review and Thought: 62, 69, 71, 73, 78, 83, 99, 102.

Reading: Moore chapter 18, sections 18.6-18.11 Questions for Review and Thought: 62, 69, 71, 73, 78, 83, 99, 102. Thermodynamics 2: Gibbs Free Energy and Equilibrium Reading: Moore chapter 18, sections 18.6-18.11 Questions for Review and Thought: 62, 69, 71, 73, 78, 83, 99, 102. Key Concepts and skills: definitions

More information

Review - After School Matter Name: Review - After School Matter Tuesday, April 29, 2008

Review - After School Matter Name: Review - After School Matter Tuesday, April 29, 2008 Name: Review - After School Matter Tuesday, April 29, 2008 1. Figure 1 The graph represents the relationship between temperature and time as heat was added uniformly to a substance starting at a solid

More information

Mr. Bracken. Multiple Choice Review: Thermochemistry

Mr. Bracken. Multiple Choice Review: Thermochemistry Mr. Bracken AP Chemistry Name Period Multiple Choice Review: Thermochemistry 1. If this has a negative value for a process, then the process occurs spontaneously. 2. This is a measure of how the disorder

More information

AP Practice Questions

AP Practice Questions 1) AP Practice Questions The tables above contain information for determining thermodynamic properties of the reaction below. C 2 H 5 Cl(g) + Cl 2 (g) C 2 H 4 Cl 2 (g) + HCl(g) (a) Calculate ΔH for

More information

4. Using the data from Handout 5, what is the standard enthalpy of formation of BaO (s)? What does this mean?

4. Using the data from Handout 5, what is the standard enthalpy of formation of BaO (s)? What does this mean? HOMEWORK 3A 1. In each of the following pairs, tell which has the higher entropy. (a) One mole of liquid water or one mole of water vapor (b) One mole of dry ice or one mole of carbon dioxide at 1 atm

More information

R = J/mol K R = L atm/mol K

R = J/mol K R = L atm/mol K version: master Exam 1 - VDB/LaB/Spk This MC portion of the exam should have 19 questions. The point values are given with each question. Bubble in your answer choices on the bubblehseet provided. Your

More information

Chemistry 13: States of Matter

Chemistry 13: States of Matter Chemistry 13: States of Matter Name: Period: Date: Chemistry Content Standard: Gases and Their Properties The kinetic molecular theory describes the motion of atoms and molecules and explains the properties

More information

Chemical Reactions Practice Test

Chemical Reactions Practice Test Chemical Reactions Practice Test Chapter 2 Name Date Hour _ Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The only sure evidence for a chemical reaction

More information

7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.

7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790. CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions,

More information

CHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g)

CHEM 105 HOUR EXAM III 28-OCT-99. = -163 kj/mole determine H f 0 for Ni(CO) 4 (g) = -260 kj/mole determine H f 0 for Cr(CO) 6 (g) CHEM 15 HOUR EXAM III 28-OCT-99 NAME (please print) 1. a. given: Ni (s) + 4 CO (g) = Ni(CO) 4 (g) H Rxn = -163 k/mole determine H f for Ni(CO) 4 (g) b. given: Cr (s) + 6 CO (g) = Cr(CO) 6 (g) H Rxn = -26

More information

SUPPLEMENTARY TOPIC 3 ENERGY AND CHEMICAL REACTIONS

SUPPLEMENTARY TOPIC 3 ENERGY AND CHEMICAL REACTIONS SUPPLEMENTARY TOPIC 3 ENERGY AND CHEMICAL REACTIONS Rearranging atoms. In a chemical reaction, bonds between atoms in one or more molecules (reactants) break and new bonds are formed with other atoms to

More information

Energy and Chemical Reactions. Characterizing Energy:

Energy and Chemical Reactions. Characterizing Energy: Energy and Chemical Reactions Energy: Critical for virtually all aspects of chemistry Defined as: We focus on energy transfer. We observe energy changes in: Heat Transfer: How much energy can a material

More information

AP Chemistry 2007 Scoring Guidelines Form B

AP Chemistry 2007 Scoring Guidelines Form B AP Chemistry 2007 Scoring Guidelines Form B The College Board: Connecting Students to College Success The College Board is a not-for-profit membership association whose mission is to connect students to

More information

CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM Chemistry 10 Chapter 14 CHEMICAL EQUILIBRIUM Reactions that can go in both directions are called reversible reactions. These reactions seem to stop before they go to completion. When the rate of the forward

More information

Answers: Given: No. [COCl 2 ] = K c [CO][Cl 2 ], but there are many possible values for [CO]=[Cl 2 ]

Answers: Given: No. [COCl 2 ] = K c [CO][Cl 2 ], but there are many possible values for [CO]=[Cl 2 ] Chemical Equilibrium What are the concentrations of reactants and products at equilibrium? How do changes in pressure, volume, temperature, concentration and the use of catalysts affect the equilibrium

More information

5. Which temperature is equal to +20 K? 1) 253ºC 2) 293ºC 3) 253 C 4) 293 C

5. Which temperature is equal to +20 K? 1) 253ºC 2) 293ºC 3) 253 C 4) 293 C 1. The average kinetic energy of water molecules increases when 1) H 2 O(s) changes to H 2 O( ) at 0ºC 3) H 2 O( ) at 10ºC changes to H 2 O( ) at 20ºC 2) H 2 O( ) changes to H 2 O(s) at 0ºC 4) H 2 O( )

More information

Chemical system. Chemical reaction A rearrangement of bonds one or more molecules becomes one or more different molecules A + B C.

Chemical system. Chemical reaction A rearrangement of bonds one or more molecules becomes one or more different molecules A + B C. Chemical system a group of molecules that can react with one another. Chemical reaction A rearrangement of bonds one or more molecules becomes one or more different molecules A + B C Reactant(s) Product(s)

More information

Chem 1A Exam 2 Review Problems

Chem 1A Exam 2 Review Problems Chem 1A Exam 2 Review Problems 1. At 0.967 atm, the height of mercury in a barometer is 0.735 m. If the mercury were replaced with water, what height of water (in meters) would be supported at this pressure?

More information

Chemistry 110 Lecture Unit 5 Chapter 11-GASES

Chemistry 110 Lecture Unit 5 Chapter 11-GASES Chemistry 110 Lecture Unit 5 Chapter 11-GASES I. PROPERITIES OF GASES A. Gases have an indefinite shape. B. Gases have a low density C. Gases are very compressible D. Gases exert pressure equally in all

More information

Entropy Changes & Processes

Entropy Changes & Processes Entropy Changes & Processes Chapter 4 of Atkins: he Second Law: he Concepts Section 4.3, 7th edition; 3.3, 8th edition Entropy of Phase ransition at the ransition emperature Expansion of the Perfect Gas

More information

K c = [C]c [D] d [A] a [B] b. k f [NO 2 ] = k r [N 2 O 4 ] = K eq = The Concept of Equilibrium. Chapter 15 Chemical Equilibrium

K c = [C]c [D] d [A] a [B] b. k f [NO 2 ] = k r [N 2 O 4 ] = K eq = The Concept of Equilibrium. Chapter 15 Chemical Equilibrium Chapter 15 Chemical Equilibrium Learning goals and key skills: Understand what is meant by chemical equilibrium and how it relates to reaction rates Write the equilibrium-constant expression for any reaction

More information

Chemical Equilibrium

Chemical Equilibrium Chapter 13 Chemical Equilibrium Equilibrium Physical Equilibrium refers to the equilibrium between two or more states of matter (solid, liquid and gas) A great example of physical equilibrium is shown

More information

ESSAY. Write your answer in the space provided or on a separate sheet of paper.

ESSAY. Write your answer in the space provided or on a separate sheet of paper. Test 1 General Chemistry CH116 Summer, 2012 University of Massachusetts, Boston Name ESSAY. Write your answer in the space provided or on a separate sheet of paper. 1) Sodium hydride reacts with excess

More information

DETERMINING THE ENTHALPY OF FORMATION OF CaCO 3

DETERMINING THE ENTHALPY OF FORMATION OF CaCO 3 DETERMINING THE ENTHALPY OF FORMATION OF CaCO 3 Standard Enthalpy Change Standard Enthalpy Change for a reaction, symbolized as H 0 298, is defined as The enthalpy change when the molar quantities of reactants

More information

Advanced Higher Chemistry. THERMODYNAMICS (Reaction Feasibility) Learning Outcomes Questions & Answers

Advanced Higher Chemistry. THERMODYNAMICS (Reaction Feasibility) Learning Outcomes Questions & Answers Advanced Higher Chemistry Unit 2 - Chemical Reactions THERMODYNAMICS (Reaction Feasibility) Learning Outcomes Questions & Answers KHS Chemistry Nov 2006 page 1 4. THERMODYNAMICS (Reaction Feasibility)

More information

Chapter 7: Chemical Equations. Name: Date: Period:

Chapter 7: Chemical Equations. Name: Date: Period: Chapter 7: Chemical Equations Name: Date: Period: 7-1 What is a chemical reaction? Read pages 232-237 a) Explain what a chemical reaction is. b) Distinguish between evidence that suggests a chemical reaction

More information

= T T V V T = V. By using the relation given in the problem, we can write this as: ( P + T ( P/ T)V ) = T

= T T V V T = V. By using the relation given in the problem, we can write this as: ( P + T ( P/ T)V ) = T hermodynamics: Examples for chapter 3. 1. Show that C / = 0 for a an ideal gas, b a van der Waals gas and c a gas following P = nr. Assume that the following result nb holds: U = P P Hint: In b and c,

More information

The value of a state function is independent of the history of the system.

The value of a state function is independent of the history of the system. 1 THERMODYNAMICS - The study of energy in matter - Thermodynamics allows us to predict whether a chemical reaction occurs or not. - Thermodynamics tells us nothing about how fast a reaction occurs. - i.

More information

1. solid, vapor, critical point correct. 2. solid, liquid, critical point. 3. liquid, vapor, critical point. 4. solid, liquid, triple point

1. solid, vapor, critical point correct. 2. solid, liquid, critical point. 3. liquid, vapor, critical point. 4. solid, liquid, triple point mcdonald (pam78654) HW 7B: Equilibria laude (89560) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0

More information

CHEM 1332 CHAPTER 14

CHEM 1332 CHAPTER 14 CHEM 1332 CHAPTER 14 1. Which is a proper description of chemical equilibrium? The frequencies of reactant and of product collisions are identical. The concentrations of products and reactants are identical.

More information

APS Science Curriculum Unit Planner

APS Science Curriculum Unit Planner Grade Level/Subject APS Science Curriculum Unit Planner Enduring Understanding Chemistry Stage 1: Desired Results Topic 3: Kinetics: The Kinetic Theory can explain the phases of matter, the energetics

More information

Chapter 6 Thermodynamics: The First Law

Chapter 6 Thermodynamics: The First Law Key Concepts 6.1 Systems Chapter 6 Thermodynamics: The First Law Systems, States, and Energy (Sections 6.1 6.8) thermodynamics, statistical thermodynamics, system, surroundings, open system, closed system,

More information

Enthalpy of Reaction and Calorimetry worksheet

Enthalpy of Reaction and Calorimetry worksheet Enthalpy of Reaction and Calorimetry worksheet 1. Calcium carbonate decomposes at high temperature to form carbon dioxide and calcium oxide, calculate the enthalpy of reaction. CaCO 3 CO 2 + CaO 2. Carbon

More information

CHEM 36 General Chemistry EXAM #1 February 13, 2002

CHEM 36 General Chemistry EXAM #1 February 13, 2002 CHEM 36 General Chemistry EXAM #1 February 13, 2002 Name: Serkey, Anne INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show

More information

Exam 4 Practice Problems false false

Exam 4 Practice Problems false false Exam 4 Practice Problems 1 1. Which of the following statements is false? a. Condensed states have much higher densities than gases. b. Molecules are very far apart in gases and closer together in liquids

More information

Equilibrium. Equilibrium 1. Examples of Different Equilibria. K p H 2 + N 2 NH 3 K a HC 2 H 3 O 2 H C 2 H 3 O 2 K sp SrCrO 4 Sr CrO 4

Equilibrium. Equilibrium 1. Examples of Different Equilibria. K p H 2 + N 2 NH 3 K a HC 2 H 3 O 2 H C 2 H 3 O 2 K sp SrCrO 4 Sr CrO 4 Equilibrium 1 Equilibrium Examples of Different Equilibria K p H 2 + N 2 NH 3 K a HC 2 H 3 O 2 H + - + C 2 H 3 O 2 K sp SrCrO 4 Sr 2+ 2- + CrO 4 Equilibrium deals with: What is the balance between products

More information

Energy Flow in Marine Ecosystem

Energy Flow in Marine Ecosystem Energy Flow in Marine Ecosystem Introduction Marin ecosystem is a functional system and consists of living groups and the surrounding environment It is composed of some groups and subgroups 1. The physical

More information

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10

The Gas Laws. Our Atmosphere. Pressure = Units of Pressure. Barometer. Chapter 10 Our Atmosphere The Gas Laws 99% N 2 and O 2 78% N 2 80 70 Nitrogen Chapter 10 21% O 2 1% CO 2 and the Noble Gases 60 50 40 Oxygen 30 20 10 0 Gas Carbon dioxide and Noble Gases Pressure Pressure = Force

More information

FORMA is EXAM I, VERSION 1 (v1) Name

FORMA is EXAM I, VERSION 1 (v1) Name FORMA is EXAM I, VERSION 1 (v1) Name 1. DO NOT TURN THIS PAGE UNTIL DIRECTED TO DO SO. 2. These tests are machine graded; therefore, be sure to use a No. 1 or 2 pencil for marking the answer sheets. 3.

More information

Unit 5 Practice Test. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.

Unit 5 Practice Test. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question. Name: Class: Date: Unit 5 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The internal energy of a system is always increased by. A) adding

More information

Chapter 15 Chemical Equilibrium

Chapter 15 Chemical Equilibrium Chapter 15 Chemical Equilibrium Chemical reactions can reach a state of dynamic equilibrium. Similar to the equilibrium states reached in evaporation of a liquid in a closed container or the dissolution

More information

Gas Laws. The kinetic theory of matter states that particles which make up all types of matter are in constant motion.

Gas Laws. The kinetic theory of matter states that particles which make up all types of matter are in constant motion. Name Period Gas Laws Kinetic energy is the energy of motion of molecules. Gas state of matter made up of tiny particles (atoms or molecules). Each atom or molecule is very far from other atoms or molecules.

More information

Intermolecular Forces

Intermolecular Forces Intermolecular Forces: Introduction Intermolecular Forces Forces between separate molecules and dissolved ions (not bonds) Van der Waals Forces 15% as strong as covalent or ionic bonds Chapter 11 Intermolecular

More information

k is change in kinetic energy and E

k is change in kinetic energy and E Energy Balances on Closed Systems A system is closed if mass does not cross the system boundary during the period of time covered by energy balance. Energy balance for a closed system written between two

More information

Test 5 Review questions. 1. As ice cools from 273 K to 263 K, the average kinetic energy of its molecules will

Test 5 Review questions. 1. As ice cools from 273 K to 263 K, the average kinetic energy of its molecules will Name: Thursday, December 13, 2007 Test 5 Review questions 1. As ice cools from 273 K to 263 K, the average kinetic energy of its molecules will 1. decrease 2. increase 3. remain the same 2. The graph below

More information

Problem # 2 Determine the kinds of intermolecular forces present in each element or compound:

Problem # 2 Determine the kinds of intermolecular forces present in each element or compound: Chapter 11 Homework solutions Problem # 2 Determine the kinds of intermolecular forces present in each element or compound: A. Kr B. NCl 3 C. SiH 4 D. HF SOLUTION: Kr is a single atom, hence it can have

More information

ENTHALPY CHANGES FOR A CHEMICAL REACTION scaling a rxn up or down (proportionality) quantity 1 from rxn heat 1 from Δ r H. = 32.

ENTHALPY CHANGES FOR A CHEMICAL REACTION scaling a rxn up or down (proportionality) quantity 1 from rxn heat 1 from Δ r H. = 32. CHEMISTRY 103 Help Sheet #10 Chapter 4 (Part II); Sections 4.6-4.10 Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob http://www.chem.wisc.edu/areas/clc (Resource page) Nuggets: Enthalpy

More information

1. Thermite reaction 2. Enthalpy of reaction, H 3. Heating/cooling curves and changes in state 4. More thermite thermodynamics

1. Thermite reaction 2. Enthalpy of reaction, H 3. Heating/cooling curves and changes in state 4. More thermite thermodynamics Chem 105 Fri 10-23-09 1. Thermite reaction 2. Enthalpy of reaction, H 3. Heating/cooling curves and changes in state 4. More thermite thermodynamics 10/23/2009 1 Please PICK UP your graded EXAM in front.

More information

Chemistry 212 EXAM 1 January 27, 2004

Chemistry 212 EXAM 1 January 27, 2004 1 Chemistry 212 EXAM 1 January 27, 2004 _100 (of 100) KEY Name Part 1: Multiple Choice. (1 point each, circle only one answer, 1. Consider the following rate law: Rate = k[a] n [B] m How are the exponents

More information

Chemistry 151 Final Exam

Chemistry 151 Final Exam Chemistry 151 Final Exam Name: SSN: Exam Rules & Guidelines Show your work. No credit will be given for an answer unless your work is shown. Indicate your answer with a box or a circle. All paperwork must

More information

Chapter 13. Chemical Equilibrium

Chapter 13. Chemical Equilibrium Chapter 13 Chemical Equilibrium Chapter 13 Preview Chemical Equilibrium The Equilibrium condition and constant Chemical equilibrium, reactions, constant expression Equilibrium involving Pressure Chemical

More information

Equilibrium Notes Ch 14:

Equilibrium Notes Ch 14: Equilibrium Notes Ch 14: Homework: E q u i l i b r i u m P a g e 1 Read Chapter 14 Work out sample/practice exercises in the sections, Bonus Chapter 14: 23, 27, 29, 31, 39, 41, 45, 51, 57, 63, 77, 83,

More information

In the box below, draw the Lewis electron-dot structure for the compound formed from magnesium and oxygen. [Include any charges or partial charges.

In the box below, draw the Lewis electron-dot structure for the compound formed from magnesium and oxygen. [Include any charges or partial charges. Name: 1) Which molecule is nonpolar and has a symmetrical shape? A) NH3 B) H2O C) HCl D) CH4 7222-1 - Page 1 2) When ammonium chloride crystals are dissolved in water, the temperature of the water decreases.

More information

INTRODUCTORY CHEMISTRY Concepts and Critical Thinking

INTRODUCTORY CHEMISTRY Concepts and Critical Thinking INTRODUCTORY CHEMISTRY Concepts and Critical Thinking Sixth Edition by Charles H. Corwin Chapter 13 Liquids and Solids by Christopher Hamaker 1 Chapter 13 Properties of Liquids Unlike gases, liquids do

More information

= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm

= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm Chapter 13 Gases 1. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases have volumes that depend on their conditions, and can be compressed or expanded by

More information

Problem Set 3 Solutions

Problem Set 3 Solutions Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions 1 (a) One mole of an ideal gas at 98 K is expanded reversibly and isothermally from 10 L to 10 L Determine the amount of work in Joules We start

More information

Chemistry 212 EXAM 2 February 17, 2004 KEY

Chemistry 212 EXAM 2 February 17, 2004 KEY 1 Chemistry 212 EXAM 2 February 17, 2004 100_ (of 100) KEY Name Part 1: Multiple Choice. (1 point 4. The equilibrium constant for the each, circle only one answer, reaction 1. For the equilibrium that

More information

Electrochemistry. Chapter 17 Electrochemistry GCC CHM152. Ox # examples. Redox: LEO the lion goes GER. Oxidation Numbers (Chapter 4).

Electrochemistry. Chapter 17 Electrochemistry GCC CHM152. Ox # examples. Redox: LEO the lion goes GER. Oxidation Numbers (Chapter 4). Chapter 17 Electrochemistry GCC CHM152 Electrochemistry Electrochemistry is the study of batteries and the conversion between chemical and electrical energy. Based on redox (oxidation-reduction) reactions

More information

CHEMISTRY STANDARDS BASED RUBRIC ATOMIC STRUCTURE AND BONDING

CHEMISTRY STANDARDS BASED RUBRIC ATOMIC STRUCTURE AND BONDING CHEMISTRY STANDARDS BASED RUBRIC ATOMIC STRUCTURE AND BONDING Essential Standard: STUDENTS WILL UNDERSTAND THAT THE PROPERTIES OF MATTER AND THEIR INTERACTIONS ARE A CONSEQUENCE OF THE STRUCTURE OF MATTER,

More information

Chapter 3: Stoichiometry

Chapter 3: Stoichiometry Chapter 3: Stoichiometry Key Skills: Balance chemical equations Predict the products of simple combination, decomposition, and combustion reactions. Calculate formula weights Convert grams to moles and

More information

CHEM1612 2014-N-2 November 2014

CHEM1612 2014-N-2 November 2014 CHEM1612 2014-N-2 November 2014 Explain the following terms or concepts. Le Châtelier s principle 1 Used to predict the effect of a change in the conditions on a reaction at equilibrium, this principle

More information

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily.

IB Chemistry 1 Mole. One atom of C-12 has a mass of 12 amu. One mole of C-12 has a mass of 12 g. Grams we can use more easily. The Mole Atomic mass units and atoms are not convenient units to work with. The concept of the mole was invented. This was the number of atoms of carbon-12 that were needed to make 12 g of carbon. 1 mole

More information

www.chemsheets.co.uk 17-Jul-12 Chemsheets A2 033 1

www.chemsheets.co.uk 17-Jul-12 Chemsheets A2 033 1 www.chemsheets.co.uk 17-Jul-12 Chemsheets A2 033 1 AS THERMODYNAMICS REVISION What is enthalpy? It is a measure of the heat content of a substance Enthalpy change ( H) = Change in heat content at constant

More information

CHEMISTRY The Molecular Nature of Matter and Change

CHEMISTRY The Molecular Nature of Matter and Change CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 13 The Properties of Mixtures: Solutions and Colloids Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction

More information

87 16 70 20 58 24 44 32 35 40 29 48 (a) graph Y versus X (b) graph Y versus 1/X

87 16 70 20 58 24 44 32 35 40 29 48 (a) graph Y versus X (b) graph Y versus 1/X HOMEWORK 5A Barometer; Boyle s Law 1. The pressure of the first two gases below is determined with a manometer that is filled with mercury (density = 13.6 g/ml). The pressure of the last two gases below

More information

Thermochemical equations allow stoichiometric calculations.

Thermochemical equations allow stoichiometric calculations. CHEM 1105 THERMOCHEMISTRY 1. Change in Enthalpy ( H) Heat is evolved or absorbed in all chemical reactions. Exothermic reaction: heat evolved - heat flows from reaction mixture to surroundings; products

More information

CHAPTER 4: MATTER & ENERGY

CHAPTER 4: MATTER & ENERGY CHAPTER 4: MATTER & ENERGY Problems: 1,3,5,7,13,17,19,21,23,25,27,29,31,33,37,41,43,45,47,49,51,53,55,57,59,63,65,67,69,77,79,81,83 4.1 Physical States of Matter Matter: Anything that has mass and occupies

More information

Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT

Chapter 3. Chemical Reactions and Reaction Stoichiometry. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT Lecture Presentation Chapter 3 Chemical Reactions and Reaction James F. Kirby Quinnipiac University Hamden, CT The study of the mass relationships in chemistry Based on the Law of Conservation of Mass

More information