CHEMICAL EQUILIBRIUM

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "CHEMICAL EQUILIBRIUM"

Transcription

1 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 and reverse reactions become equal, an equilibrium system is established. Stepwise view to an equilibrium system Step 1 A + B ¾¾ C + D Fast No Reaction ¾ C + D Step A + B ¾¾ C + D Forward reaction slows down. There are fewer A and B molecules available. A + B ¾ C + D Reverse reaction starts slowly at first. There are few C and D molecules available. Step 3 A + B ¾¾ C + D Forward reaction slows down further as the number of A and B molecules decreases. A + B ¾ C + D Reverse reaction speeds up as the number of C and D molecules increases. Step 4 A + B forward ¾¾¾ reverse ¾¾¾ C + D RATE OF FORWARD REACTION = RATE OF REVERSE REACTION 1

2 Chemistry 10 Chapter 14 CHEMICAL EQUILIBRIUM Characteristics of a Chemical Equilibrium System: 1. A mixture of Reactants and Products is present. The composition of the reaction mixture no longer changes: Concentration of reactants is constant Concentration of products is constant NOTE: Concentration of reactants ¹ Concentration of products 3. A Chemical Equilibrium is a Dynamic Equilibrium; both reactions (forward and reverse) are still going on 4. The Dynamic Equilibrium may be controlled (shifted to the right to favor products, or shifted to the left to favor reactants) by changing the conditions for the reaction. Definition of Chemical Equilibrium: A state reached by a reaction mixture when the rate of forward reaction and the rate of reverse reactions become equal.

3 Chemistry 10 Chapter 14 COMPOSITION OF AN EQUILIBRIUM MIXTURE In an equilibrium mixture, the amount of one substance is determined experimentally and the amounts of the other substances are calculated. Example 1: CO (g) + 3 H (g) ¾¾ ¾¾ CH 4 (g) + H O (g) When mol CO and mol H are placed in a L vessel at 97 C and allowed to come to equilibrium, the mixture is found to contain mol H O. How many moles of each substance are present at equilibrium? CO (g) + 3 H (g) ¾¾ ¾¾ CH 4 (g) + H O (g) Initial (0.387) Equilibrium ( ) ( ) Equilibrium mol mol mol mol Example : Consider the following dynamic equilibrium: CO (g) + H O (g) ¾¾ ¾¾ CO (g) + H (g) Suppose you start with a gaseous mixture containing 1.00 mol of CO and 1.00 mol of H O. When equilibrium is reached at 1000 C, the mixture contains 0.43 mol H. What is the molar composition of the equilibrium mixture? CO (g) + H O (g) ¾¾ ¾¾ CO (g) + H (g) Initial 1.00 mol 1.00 mol 0 0 x x + x + x Equilibrium (1.00 x) (1.00 x) Equilibrium 0.57 mol 0.57 mol 0.43 mol 0.43 mol 3

4 Chemistry 10 Chapter 14 Example 3: mol of N and.50 mol of H are placed into a reaction vessel at 450 C and 10.0 atm. N (g) + 3 H (g) ¾¾ ¾¾ NH 3 (g) What is the composition of the equilibrium mixture if you obtain mol of NH 3 from it? Initial Equilibrium Equilibrium N (g) + 3 H (g) ¾¾ ¾¾ NH 3 (g) 4

5 Chemistry 10 Chapter 14 THE EQUILIBRIUM CONSTANT Consider the following equilibrium system: N O 4 (g) forward ¾¾¾ reverse ¾¾¾ NO (g) NOTE: Both forward and reverse reactions are elementary reactions. At Equilibrium: Rate (forward reaction) = Rate(reverse reaction) k f [N O 4 ] = k r [NO ] k f [NO ] = = constant k [N O ] r 4 k f = K c = equilibrium constant k r implies molar concentration In General: aa + bb ¾¾ ¾¾ cc + dd c d [C] [D] c a b K = [A] [B] [C], [D] = molar concentrations (molarities) of the products at equilibrium [A], [B] = molar concentrations (molarities) of the reactants at equilibrium This relationship was postulated by two chemists in 1864 and called the law of mass action. 5

6 Chemistry 10 Chapter 14 EQUILIBRIUM CONSTANT The Equilibrium Constant, K c, is a constant for a particular reaction at a given temperature. NOTE: 1. If the overall reaction occurs by a multi step mechanism, one can show that K c for the overall reaction equals a product of ratios of the individual rate constants.. It is common practice to write K c without units Example 1: Write the equilibrium constant expression for the equation shown below: NO (g) + 7 H (g) ¾¾ ¾¾ NH 3 (g) + 4 H O(g) K c = Example : The Equilibrium Constant for a reaction is: [NH ] [O ] K = [NO] [H O] c 4 6 What is the Equilibrium Constant expression when the equation for this reaction is halved and then reversed? Original Equation: 4 NO + 6 H O ¾¾ ¾¾ 4 NH O Halved and Reversed: K c = 6

7 Chemistry 10 Chapter 14 CALCULATING K C FOR REACTIONS Molar concentrations (molarities) of products and reactants at equilibrium must be substituted in the expression of K c Example 1: CO (g) + 3 H (g) ¾¾ ¾¾ CH 4 (g) + H O (g) When 1.00 mol CO and 3.00 mol H are placed in a 10.0 L vessel at 97 C and allowed to come to equilibrium, the mixture is found to contain mol H O. Calculate the equilibrium constant for this reaction. CO (g) + 3 H (g) ¾¾ ¾¾ CH 4 (g) + H O (g) Initial 1.00 mol 3.00 mol Equilibrium mol 1.84 mol mol mol Equilibrium amounts: Equilibrium Concentrations(M) CO: mol mol/10.00 L = M H : 1.84 mol 1.84 mol/10.00 L = M CH 4 : mol mol/10.00 L = M H O: mol mol/10.00 L = M [H O] [CH ] (0.387 M)(0.387 M) K = = = c 3 3 [CO] [H ] ( M(0.184 M) Example : An 8.00 L reaction vessel at 491 C contained mol H, 0.75 mol I and 3.00 mol HI. Assuming that the substances are at equilibrium, find the value of K c at 491 C. The equilibrium is: H (g) + I (g) ¾¾ ¾¾ HI(g) Moles at Equilibrium: Molarities at Equilibrium: moles 0.75 moles 3.00 moles (moles/l) 8.00 L 8.00 L 8.00 L Molarities at Equilibrium: M M M [HI] (0.375 M) K c = = = 50.3 [H ] [I ] ( M)( M) 7

8 Chemistry 10 Chapter 14 Example 3: H S, a colorless gas dissociates on heating: H S ¾¾ ¾¾ H (g) + S (g) When mol of H S was put in a 10.0 L vessel and heated to 113 C, it gave an equilibrium mixture containing mol H. What is the value of K c at this temperature? Note: It is convenient to express equilibrium amounts of reactants and products in molar concentrations (molarities in mol/l) Molarity of H S = Molarity of H = Initial H S (g) ¾¾ ¾¾ H (g) + S (g) Equilibrium Equilibrium K c = Example 4: In the contact process, sulfuric acid is manufactured by first oxidizing SO to SO 3, which is then reacted with water. The reaction of SO with O is: SO (g) + O (g) ¾¾ ¾¾ SO 3 (g) A.00 L flask was filled with mol SO and mol O. At equilibrium, at 900 K, the flask contained mol SO 3. Obtain the value of K c. Molarity of SO = Molarity of O = Initial SO (g) + O (g) ¾¾ ¾¾ SO 3 (g) Equilibrium Equilibrium K c = 8

9 Chemistry 10 Chapter 14 GAS PHASE EQUILIBRIA (K P ) Gas Phase Equilibria refers to equilibrium systems where all reactants and products are gases. Concentrations of gases can be expressed in terms of partial pressures, since the concentration of a gas is proportional to its partial pressure. n mol P PV = nrt ( )= V L RT n 1 molar concentration of a gas = = P( V RT ) Constant at a given temperature K p is the equilibrium constant for a gaseous reaction expressed in terms of partial pressures. K p has a value different from K c K p = K c (RT) n sum of coefficients sum of coefficients n = of of gaseous products gaseous reactants Example 1: CO (g) + 3 H (g) ¾¾ ¾¾ CH 4 (g) + H O(g) P P CH4 H O p 3 P CO P H K = K c = 3.9 (at 100K) n = (1+1) (1+3) = K p = K c (RT) n = 3.9 [(0.081) (100)] = 4.04 x 10 4 Example : The equilibrium constant K c equals 10.5 for the following reaction at 7 C. CO(g) + What is the value of K p at this temperature? H (g) ¾¾ ¾¾ CH 3 OH (g) n = K p = K c (RT) n = 9

10 Chemistry 10 Chapter 14 OTHER EQUILIBRIUM CONSTANT If a given chemical equation can be obtained by taking the sum of other equations, the Equilibrium Constant for the overall equation equals the product of the equilibrium constants of the other equations. K overall = K 1 K Example 1: The following equilibria occur at 100K CO(g) + CH 4 (g) + CO(g) + 3 H (g) ¾¾ ¾¾ CH 4 (g) + H O(g) K 1 = 3.9 H S (g) ¾¾ ¾¾ CS (g) + 4 H (g) K = 3.3x10 4 H S (g) ¾¾ ¾¾ CS (g) + H O (g) + H (g) K 3 =??? [CH ] [H O] [CS ] [H ] K = K = [CO] [H ] [CH 4] [H S] [ CH ] [H O] 4 K1K = 3 [CO] [ H ] x 4 [CS ] [H ] [ CH ] [H S] 4 = K 3 The equilibrium constant for an overall equation is equal to the product of the equilibrium constants of the individual equations. K 3 = K 1 x K Because an equilibrium can be approached from either direction, the direction in which we write the chemical equation is arbitrary. For example, for the reaction shown below: N O 4 (g) ¾¾ ¾¾ NO (g) [NO ] K c = = 0.1 (at 100 C) [NO 4 ] When considering the equilibrium in the reverse direction: NO (g) ¾¾ ¾¾ N O 4 (g) [N O ] 1 K = = = 4.7 (at 100 C) [NO ] c K fwd = 1/K rev 10

11 Chemistry 10 Chapter 14 CLASSIFICATION OF CHEMICAL EQUILIBRIA Chemical Equilibria can be classified according to the physical state of the reactants and products present: I. Homogeneous Equilibrium An equilibrium that involves reactants and products in a single phase. Example: Catalytic methanation: CO(g) + 3 H (g) ¾¾ ¾¾ CH 4 (g) + H O (g) II. Heterogeneous Equilibrium An equilibrium involving reactants and products in more than one phase Example: 3 Fe (s) + 4 H O (g) ¾¾ ¾¾ Fe 3 O 4 (s) + 4 H (g) [H ] K = [H O] 4 c 4 NOTE: The concentrations of solids are omitted. Reason: The concentration of a pure solid or pure liquid is a constant at a given temperature. Equilibrium Constant expression can be written with pure solids included: [Fe O ] [H ] K c = [Fe] 3 [H 4 O] By rearrangement: 3 4 [Fe] [H ] K c = [Fe O ] [H O] constant factors variable factors [Fe] [H ] 3 4 ' c K c = 4 [Fe3O 4 ] [H O] K = The concentrations of pure solids and liquids are incorporated in the value of K c. The concentration of solvent is also omitted from the expression of K c for a homogeneous reaction (if constant). The equilibrium is not affected by pure solids, pure liquids, or solvents. 11

12 Chemistry 10 Chapter 14 Examples: Identify each of the following equilibriums as homogeneous or heterogeneous and write Kc expressions for each: 1. FeO (s) + H (g) ¾¾ ¾¾ Fe (s) + H O (g). CH 4 (g) + H S (g) ¾¾ ¾¾ CS (g) + 4 H O (g) 3. Ti (s) + Cl (g) ¾¾ ¾¾ TiCl 4 (l) 4. Given the equilibrium shown below, for which Kp = at 480 C: Cl (g) + H O (g) ¾¾ ¾¾ 4 HCl (g) + O (g) What is the value of Kp for the reaction shown below: HCl (g) + ½ O (g) ¾¾ ¾¾ Cl (g) + H O (g) 1

13 Chemistry 10 Chapter 14 MEANING OF THE EQUILIBRIUM CONSTANT The magnitude of the equilibrium constant indicates the extent to which the forward and reverse reactions take place. K c >>> 1 ì ü amount of ï í products at ï ý ïî equilibrium ï þ ì amount of ü ï ï íreactants at ý ïî equilibrium ï þ Products are favored at equilibrium N (g) + 3 H (g) ¾¾ ¾ NH 3 (g) Equilibrium Conc s: M M.0 M [NH 3 ] (.0 M) c 3 [N ][H ] (0.010 M)( M ) K = = = 4.1x10 8 K c <<< 1 ì ü amount of ï ï í products at ý ïî equilibrium ï þ ì amount of ü ï ï íreactants at ý ïî equilibrium ï þ Reactants are favored at equilibrium N (g) + O (g) ¾¾ ¾ NO (g) Equilibrium Conc s: 1.0 M 1.0 M 6.8x10 16 M K c 1 (0.5 10) 16 [NO] (6.8x10 M) K c = = = 4.6x 10 [N ][O ] (1.0 M)( 1.0 M) ì amount of ü ï í products at ï ý ì amount of ü ï ï íreactants at ý ïî equilibrium ï þ ïî equilibrium ï þ 31 Neither reactants, nor products are predominant CO (g) + 3 H (g) ¾¾ ¾ CH 4 (g) + H O (g) Equilibrium Conc s: M M M M [CH ][H O] ( M)( M) K = = = c 3 3 [CO][H ] ( M)(0.184 M) 13

14 Chemistry 10 Chapter 14 Example 1: The Equilibrium Constant for the reaction below equals 4.0 x at 5 0 C. NO (g) + O (g) ¾¾ ¾ NO (g) (a) Does the equilibrium mixture contain predominantly reactants or products? (b) If [NO] = [O ] =.0 x 10 6 M at equilibrium, what is the equilibrium concentration of [NO ]? [NO ] K = [NO ] = K [NO] [O ] c c [NO] [O ] [NO ] = K [NO] [O ] = (4.0x10 )(.0x10 ) (.0 x10 )= 1.8x c M NO (g) + O (g) ¾¾ ¾ NO (g) Equilibrium Conc s:.0 x10 6 M.0 x10 6 M 1.8 x10 M much smaller much larger Example : Sulfur dioxide reacts with oxygen at 5 C to reach equilibrium. K c for this reaction at this temperature is 8.0x SO (g) + O (g) ¾¾ ¾ SO 3 (g) a) From the magnitude of K c, which reaction is favored (forward or reverse) when equilibrium is established? b) If an equilibrium mixture is 1.0 M SO 3 and has equal concentrations of SO and O, what is the concentration of SO in the mixture? Does this result agree with what you expect from the magnitude of K c? 14

15 Chemistry 10 Chapter 14 PREDICTING THE DIRECTION OF REACTION Consider the following reaction and the following composition of gas mixture present. CO (g) + 3 H (g) ¾¾ ¾ CH 4 (g) + H O (g) K c = 3.93 (100 K) Composition of gas mixture: M M M M (NOT AT EQUIL) At 100K and in the presence of a catalyst, what would be more likely? 1. Would forward reaction be favored? (would products be favored?) OR. Would reverse reaction be favored? (would reactants be favored?) To answer this, a new quantity is introduced: Q c = the Reaction Quotient = has the same form as K c but the concentrations used to calculate its value are not equilibrium concentrations. 3 3 [CH 4][H O] (1.00x10 M)(1.00x10 M) c 3 [CO][H ] (.00x10 M)(.00x10 M) Q = = = 6.5 Q c = 6.5 K c = 3.9 should decrease to For the reaction mixture to reach equilibrium: Q c 3.9 Q c decreases if: [CH 4 ], [H O] decreases [CO], [H ] increases CO (g) + 3H (g) CH 4 (g) + H O(g) The reaction goes to the left Reverse reaction is favored Formation of reactants is favored Q c = 0 K c Q c = Forward Rxn Reverse Rxn Reactants Equilibrium Products Increasing Q c 15

16 Chemistry 10 Chapter 14 PREDICTING THE DIRECTION OF REACTION In General: For a reaction mixture that is NOT AT EQUILIBRIUM : aa + bb ¾¾ ¾ cc + dd c d [C] [D] Q = [A] [B] c a b Compare: Q c to K c If: Q c > K c Reaction will go the left ( ) If: Q c < K c Reaction will go to the right ( ) If: Q c = K c Reaction mixture is at equilibrium ( ) Examples: The following reaction has an equilibrium constant, K c, equal to 3.59 at 900 C, and the following composition of reaction mixture: CH 4 (g) + H S (g) ¾¾ ¾ CS (g) + 4 H (g) 1.6 M 1.3 M 1.43 M 1.1 M (a) Is the reaction mixture at equilibrium? [H ] [CS ] (1.1 M) (1.43 M) Q = = = c [CH 4][H S] (1.6 M)(1.3 M) Q c = 1.0 K c = 3.59 Q c K c Not at equilibrium (b) If not at equilibrium, in which direction will the reaction go to reach equilibrium? Q c < K c Reaction will go to the right Forward reaction is favored Formation of products is favored 16

17 Chemistry 10 Chapter 14 CALCULATING EQUILIBRIUM CONCENTRATIONS We can use equilibrium constant to calculate the equilibrium concentration of all the substances in the mixture. Some of the variations that these problems occur are shown in the examples below: Example 1: Nitric oxide, NO, is formed in automobile exhaust by the reaction of N and O (from air): N (g) + O (g) ¾¾ ¾ NO (g) K c for this reaction equals at 17 C. If an equilibrium mixture at 17 C contains 0.03 mol N and mol O per liter, what is the equilibrium concentration of NO? [NO] K = [NO] = K [N ][O ] c c [N ][O ] 3 [NO] = K c [N ][O ] = (0.005)(0.03)(0.031)= 1.3x10 M Example : The equilibrium shown below has a K p value of 1.45x10 5 at 500 C. In an equilibrium mixture of the three gases at this temperature, the partial pressure of H is 0.98 atm and that of N is 0.43 atm. What is the partial pressure of NH 3 in this mixture? N (g) + 3 H (g) ¾¾ ¾ NH 3 (g) K p = 17

18 Chemistry 10 Chapter 14 Example 3: For the reaction shown below, initially a mixture contains mol each of N and O in a 8.00 L vessel. Find the composition of the mixture when equilibrium is reached at 3900 C. K c = at 3900 C N (g) + O (g) ¾¾ ¾ NO (g) First: Obtain starting molar concentrations in mol/l mol [N ] = [O ] = = M 8.00 L N (g) + O (g) ¾¾ ¾ NO (g) Initial x x +x Equilibrium x x x [NO] (x) K c = = = [N ][O ] ( x)( x) Taking square roots of both sides: x = ( x) Rearranging: x = x = x Rearranging further: = x x = x 10 3 M Equilibrium concentrations are: [O ] = [N ] = x = = M [NO] = x = x x 10 3 M = M 18

19 Chemistry 10 Chapter 14 Example 4: Ammonium hydrogen sulfide decomposes at room temperature as shown below: NH 4 HS (s) ¾¾ ¾ NH 3 (g) + H S (g) K p = at 5 C A sample of ammonium hydrogen sulfide is placed in a flask at 5 C. After equilibrium has been reached, what is the total pressure of the flask? (Note: solids have no pressure) K p = = Example 5: At relatively high temperatures, the following reaction can be used to produce methyl alcohol: CO (g) + H (g) ¾¾ ¾ CH 3 OH (l) K c = 13.5 If the concentration of CO at equilibrium were found to be M, what would be the equilibrium concentration of H? 19

20 Chemistry 10 Chapter 14 CALCULATIONS INVOLVING QUADRATIC EQUATIONS Some of the problems involving equilibrium involve use of the quadratic equation in order to determine the unknown variable. The example below shown one such problem. Example 6: PCl 5 decomposes when heated: PCl 5 (g) ¾¾ ¾ PCl 3 (g) + Cl (g) If the initial concentration of PCl 5 is 1.00 M, what is the equilibrium composition of the gaseous mixture at 160 C? K c at 160 C is PCl 5 (g) ¾¾ ¾ PCl 3 (g) + Cl (g) Initial 1.00 M 0 0 x +x +x Equilibrium 1.00 x x x [PCl ][Cl ] (x)(x) = [PCl ] (1.00 x) 3 K c = = 5 (0.011)(1.00 x) = x x = x x x = 0 Quadratic Equation x x = 0 ax + bx + c = 0 x 1,x = b ± b 4ac a x = ± (0.011) 4( 0.011) = ±0.913 In theory: There are two mathematical solutions In practice: Only one solution makes physical sense 0

21 Chemistry 10 Chapter 14 x 1 = = x = = Equilibrium Concentrations are: correct reject (concentration cannot be negative) [PCl 5 ] = 1.00 M x = 1.00 M M = 0.86 M [PCl 3 ] = [Cl ] = x = M Example 7: Calculate the composition of the gaseous mixture obtained when 1.5 mol of carbon dioxide is exposed to hot carbon at 800 C in a 1.5 L vessel. The equilibrium constant K c at 800 C is 14.0 for the reaction: Initial Equilibrium CO (g) + C (s) ¾¾ ¾ Cl (g) [CO ] = K c = 14.0 = 1

22 Chemistry 10 Chapter 14 LE CHATELIER S PRINCIPLE The effect of changes on the equilibrium can be predicted using the Le Chatelier s principle. Le Chatelier s principle states that: If a stress is applied to a system at equilibrium, the system will respond in such a way as to relieve the stress and restore a new equilibrium under a new set of conditions. Note sequence of events: 1. Stress applied. Equilibrium system response (equilibrium shift) 3. New equilibrium Stress is a change in any of the following: A) Concentration of Reactants or Products B) Pressure C) Temperature A) Effect of Concentration Change on Equilibrium (Adding or Removing Reactants or Products) Consider: Stress: A + B ¾¾ ¾ C + D increased to B* Response: Forward reaction speeds up Equilibrium shifts to the right Products are favored New Equil.: A + B ¾¾ ¾ C + D decreased increased increased increased A < A B<B <B* C >C D >D

23 Chemistry 10 Chapter 14 Example 1: In General: EFFECT OF CONCENTRATION CHANGE ON EQUILIBRIUM Adding Reagent Equilibrium always shifts in the direction that tends to reduce the concentration of the added reacting species. When concentration of reactant is increased equilibrium shifts forward ( ¾¾ ) When concentration of product is increased equilibrium shifts reverse ( ¾ ) Removing Reagent Equilibrium always shifts in the direction that tends to increase the concentration of the removed reacting species. When concentration of reactant is decreased equilibrium shifts reverse ( ¾ ) When concentration of product is decreased equilibrium shifts forward ( ¾¾ ) H (g) + I (g) ¾¾ ¾ HI (g) (700K) Starting conc s 1.00 M 1.00 M 0 Change: 0.79 M 0.79 M M Equil Conc s: 0.1 M 0.1 M 1.58 M Stress: M Response: Equilibrium shifts forward to use up added reactant H (g) + I (g) ¾¾ ¾ HI (g) (700K) New Equil. Conc s: 0.15 M 0.35 M M 0.1M<0.35 M<0.41 M 3

24 Chemistry 10 Chapter 14 Example : Cl (g) + H O (l) ¾¾ ¾ HOCl (aq) + H 3 O + (aq) + Cl (aq) Reagent Species Change in concentration Equilibrium Shift Cl Cl H O H O HOCl HOCl Cl Cl increase decrease increase decrease increase decrease increase decrease Example 3: CO (g) + 3 H (g) ¾¾ ¾ CH 4 (g) + H O (g) Molar Equil. Composition: mol mol mol mol What changes in the amount of reagents would produce more CH 4? increase amount of CO or H remove CH 4 or H O Most practical and least expensive solution is to remove H O (cool reaction mixture to condense water vapor) CO (g) + 3 H (g) ¾¾ ¾ CH 4 (g) + H O (g) Molar Equil. Composition: mol mol mol mol Stress: mol Response: Equilibrium shifts New Equilibrium Composition: mol mol mol 0.1 mol decreased decreased increased decreased Equilibrium shift can also be predicted from an evaluation of Q c (Reaction Quotient) 4

25 Chemistry 10 Chapter 14 EFFECT OF PRESSURE CHANGE ON EQUILIBRIUM A change in pressure has an effect on equilibrium only when the following two conditions exist: 1. At least one of the reacting species (reactant or product) is a gas.. Total number of moles of gaseous reactants Total number of moles of gaseous products The Pressure of a gas may be: increased by decreasing the volume, at constant temperature (achieved by decreasing the size of the reaction vessel) decreased by increasing the volume, at constant temperature (achieved by increasing the size of the reaction vessel) To predict the equilibrium shift caused by a change in pressure, consider the following: The effect of increasing the pressure (decreasing the volume) of the equilibrium system = Increasing the concentration of gaseous reacting species (reactants and products) Example 1: (One reacting species is a gas) CaCO 3 (s) ¾¾ ¾ CaO (s) + CO (g) Stress Decrease in volume Increase in volume Pressure Change increased decreased Direction of Equilibrium shift Reverse ( ¾ ) (removes gas) Forward ( ¾¾ ) (adds gas) Amounts of Reacting Species CaCO 3 CaO CO increased decreased decreased decreased increased increased 5

26 Chemistry 10 Chapter 14 Example : (All reacting species are gases) Number of moles of reactants Number of moles of products N (g) + 3 H (g) ¾¾ ¾ NH 3 (g) 4 moles moles Stress: Pressure is increased (Volume is decreased) Response: Equilibrium shifts to the side that has the fewer number of moles ( ¾¾ ) Reason: Fewer moles of gas will exert less pressure (provided temperature is constant) New Equil: N (g) + 3 H (g) ¾¾ ¾ NH 3 (g) decreased decreased increased Stress Decrease in volume Increase in volume Pressure Change increased decreased Direction of Equilibrium shift Forward ( ¾¾ ) (fewer mol of gas) Reverse ( ¾ ) (more mol of gas) Amounts of Reacting Species N H NH 3 decreased decreased increased increased increased decreased Example 3: (Some (not all) reacting species are gases) C (s) + 1 CO (g) ¾¾ ¾ CO (g) Stress Decrease in volume Increase in volume Pressure Change Direction of Equilibrium shift Amounts of Reacting Species C CO CO 6

27 Chemistry 10 Chapter 14 Example 4: (Evaluating Q c to predict Equilibrium Shift) CO (g) + 3 H (g) ¾¾ ¾ CH 4 (g) + H O(g) 4 moles moles Stress: Equilibrium mixture is compressed to half its volume, at constant temperature Pressure Change: Result: Pressure is doubled Partial pressure of all gaseous reactants (concentrations) are doubled K = [CH ][H O] 4 c 3 [CO][H ] Q = ( [CH ])( [H O] ( [CO])( [H 4 c 3 ]) Q = c 4 [CH ][H O] 16 Q K 4 c 3 c [CO][H ] 4 = Q c < K c Reaction proceeds in the forward direction Equilibrium shifts ¾¾ Products are favored CONCLUSIONS: At constant temperature: 1. If the pressure is increased (Volume is decreased), the reaction shifts in the direction of fewer moles of gas. If the pressure is decreased (Volume is increased), the reaction shifts in the direction of more moles of gas. 7

28 Chemistry 10 Chapter 14 EFFECT OF TEMPERATURE CHANGE ON EQUILIBRIUM 1. Effect on Rates of Reactions An increase in temperature: increases the rate of both forward and reverse reactions causes the system to reach equilibrium faster. Effect on Equilibrium Depends on the type of reaction (exothermic or endothermic) Meaning: N (g) + 3 H ¾¾ ¾ NH 3 (g) H = 91.8 kj ¾¾¾¾ ¾¾¾¾ N (g) + 3 H exothermic endothermic NH 3 (g) kj An increase in temperature favors the endothermic reaction ( ¾ ) Reason: the endothermic reaction absorbs heat and thus lowers the temperature decreases K c of the equilibrium system A decrease in temperature favors the exothermic reaction ( ¾¾ ) Reason: the exothermic reaction gives off heat and thus increases the temperature increases K c of the equilibrium system N (g) + 3 H (g) ¾¾ ¾ NH 3 (g) H = 91.8 kj Stress Increase in temp. (Heat added) Decrease in temp. (Heat removed) Equilibrium Shift Amounts of Reacting Species N H NH 3 ¾ increased increased decreased decreased ¾¾ decreased decreased increased increased K c CONCLUSIONS: 1. Equilibrium Shift with Temperature Change In order to increase the amount of products obtained: (a) the temperature must be increased if the reaction is endothermic ( H 0 > 0) (b) the temperature must be decreased if the reaction is exothermic ( H 0 < 0). Effect of Temperature on Kc If the amount of products is to be increased, K c should also increase: (a) For an endothermic reaction, K c is larger at higher temperatures (b) For an exothermic reaction, K c is larger at lower temperatures. 8

29 Chemistry 10 Chapter 14 EFFECT OF CATALYS ON EQUILIBRIUM A catalyst is a substance that increases the rate of a reaction but is not consumed by it. SO (g) + O (g) ¾¾ ¾ SO 3 (g) K c = 1.7 x 10 6 K c is very large, therefore the reaction should go almost to completion. However, both forward and reverse reactions are very slow (if un catalyzed) and it takes a very long time to reach equilibrium with very little SO 3 obtained in a reasonable amount of time. The use of a Pt catalyst speeds up both forward and reverse reactions and as a result equilibrium is reached fast. The equilibrium mixture contains mostly SO 3 due to the large K c. A CATALYST: 1. Has no effect on the equilibrium composition. Speeds up the attainment of equilibrium 3. Is very useful for reactions: that are slow, if un catalyzed that have a large K c, 4. Is not useful for reactions with a small K c 5. May have an effect on the amount of product obtained if it affects the rate of one reaction out of several possible reactions: GOAL: production of NO (used to obtain HNO 3 ) by the oxidation of NH 3 (a) 4 NH 3 (g) + 5 O (g) ¾¾ ¾ 4 NO + 6 H O (g) Large K c (slow) (b) 4NO ¾¾ ¾ N (g) + O (g) Large K c (slow) Overall: 4 NH 3 (g) + 3O (g) ¾¾ ¾ N (g) + 6 H O (g) NOTE: No NO is produced SOLUTION: Use of a Pt catalyst speeds up reaction (a) but not reaction (b). Therefore use of Pt catalyst at moderate temperatures produces NO selectively, since decomposition of NO is too slow at these temperatures to be significant. RESULT: Equilibrium mixture will contain mostly NO 9

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

Chapter 14. CHEMICAL EQUILIBRIUM

Chapter 14. CHEMICAL EQUILIBRIUM Chapter 14. CHEMICAL EQUILIBRIUM 14.1 THE CONCEPT OF EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT Many chemical reactions do not go to completion but instead attain a state of chemical equilibrium. Chemical

More information

CHAPTER 14 (MOORE) CHEMICAL EQUILIBRIUM

CHAPTER 14 (MOORE) CHEMICAL EQUILIBRIUM CHAPTER 14 (MOORE) CHEMICAL EQUILIBRIUM This chapter deals with chemical equilibrium, or how far chemical reactions proceed. Some reactions convert reactants to products with near 100% efficiency but others

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

Chemical Equilibrium. Rate Forward Reaction = Rate Reverse Reaction. Chapter 14. Hill, Petrucci, McCreary & Perry 4 th. Ed.

Chemical Equilibrium. Rate Forward Reaction = Rate Reverse Reaction. Chapter 14. Hill, Petrucci, McCreary & Perry 4 th. Ed. Chapter 14 Chemical Equilibrium Hill, Petrucci, McCreary & Perry 4 th Ed. Chemical Equilibrium Many Reactions seem to STOP before all the reactants are used up. The Concentrations of Reactants and Products

More information

Chemistry 4th Edition McMurry/Fay

Chemistry 4th Edition McMurry/Fay 13 Ch a pt e r Chemical Equilibrium Chemistry 4th Edition McMurry/Fay Dr. Paul Charlesworth Michigan Technological University The Equilibrium State 01 Chemical Equilibrium: A state achieved when the rates

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

A k 1. At equilibrium there is no net change in [A] or [B], namely d[a] dt

A k 1. At equilibrium there is no net change in [A] or [B], namely d[a] dt Chapter 15: Chemical Equilibrium Key topics: Equilibrium Constant Calculating Equilibrium Concentrations The Concept of Equilibrium Consider the reaction A k 1 k 1 B At equilibrium there is no net change

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

Chapter 15: Chemical Equilibrium: How Much Product Does a Reaction Really Make?

Chapter 15: Chemical Equilibrium: How Much Product Does a Reaction Really Make? Chapter 15: Chemical Equilibrium: How Much Product Does a Reaction Really Make? End-of-Chapter Problems: 15.1-15.10, 15.13-15.14, 15.17-15.99, 15.102-15.104 Example: Ice melting is a dynamic process: H

More information

Chapter 14 Chemical Equilibrium

Chapter 14 Chemical Equilibrium Chapter 14 Chemical Equilibrium Forward reaction H 2 (g) + I 2 (g) 2HI(g) Reverse reaction 2HI(g) H 2 (g) + I 2 (g) At equilibrium H 2 (g) + I 2 (g) 2HI(g) Chemical equilibrium is reached when reactants

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

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

AP* Chemistry CHEMICAL EQUILIBRIA: GENERAL CONCEPTS

AP* Chemistry CHEMICAL EQUILIBRIA: GENERAL CONCEPTS AP* Chemistry CHEMICAL EQUILIBRIA: GENERAL CONCEPTS THE NATURE OF THE EQUILIBRIUM STATE: Equilibrium is the state where the rate of the forward reaction is equal to the rate of the reverse reaction. At

More information

Name AP Chemistry / / Chapter 13 Collected AP Exam Free Response Questions 1980 2010 Answers

Name AP Chemistry / / Chapter 13 Collected AP Exam Free Response Questions 1980 2010 Answers Name AP Chemistry / / Chapter 13 Collected AP Exam Free Response Questions 1980 2010 Answers 1980 - #6 NH 4 Cl(s) NH 3 (g) + HCl(g) ΔH = +42.1 kilocalories Suppose the substances in the reaction above

More information

Chemical Equilibrium

Chemical Equilibrium Chemical Equilibrium Chapter 14 1 Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions

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

We will be looking at: A. Judging the extent of a reaction B. Predicting the direction of a reaction C. Calculating equilibrium concentrations

We will be looking at: A. Judging the extent of a reaction B. Predicting the direction of a reaction C. Calculating equilibrium concentrations 13.5 Using the Equilibrium Constant We will be looking at: A. Judging the extent of a reaction B. Predicting the direction of a reaction C. Calculating equilibrium concentrations A. Judging the extent

More information

Chapter 13 Chemical Equilibrium. Equilibrium is Dynamic. The Equilibrium Constant. Equilibrium and Catalysts. Characteristics of Chemical Equilibrium

Chapter 13 Chemical Equilibrium. Equilibrium is Dynamic. The Equilibrium Constant. Equilibrium and Catalysts. Characteristics of Chemical Equilibrium Characteristics of Chemical Equilibrium John W. Moore Conrad L. Stanitski Peter C. Jurs http://academic.cengage.com/chemistry/moore Chapter 13 Chemical Equilibrium Many reactions fail to go to completion.

More information

Chemical Equilibrium-A Dynamic Equilibrium

Chemical Equilibrium-A Dynamic Equilibrium Chemical Equilibrium-A Dynamic Equilibrium Page 1 When compounds react, they eventually form a mixture of products and (unreacted) reactants, in a dynamic equilibrium Much like water in a U-shape tube,

More information

Gas Phase Equilibrium

Gas Phase Equilibrium Gas Phase Equilibrium Chemical Equilibrium Equilibrium Constant K eq Equilibrium constant expression Relationship between K p and K c Heterogeneous Equilibria Meaning of K eq Calculations of K c Solving

More information

Principles of Reactivity: Chemical Equilibria

Principles of Reactivity: Chemical Equilibria Principles of Reactivity: Chemical Equilibria This chapter addresses the principle of equilibrium equilibrium What can you do to reestablish equilibrium? non-equilibrium Whose principle supports this?

More information

Chemical Equilibrium - Chapter 14

Chemical Equilibrium - Chapter 14 Chemical Equilibrium - Chapter 14 1. Dynamic Equilibrium a A + b B c C + d D At Equilibrium: Reaction is proceeding in both directions at the same rate. There is no net change in concentrations of reactants

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

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

Chapter 13. Chemical Equilibrium

Chapter 13. Chemical Equilibrium Chapter 13 Chemical Equilibrium Section 13.1 The Equilibrium Condition Section 13.1 The Equilibrium Condition Section 13.1 The Equilibrium Condition Section 13.1 The Equilibrium Condition Section 13.1

More information

EQUILIBRIUM. Consider the reversible system initially consisting of reactants only.

EQUILIBRIUM. Consider the reversible system initially consisting of reactants only. EQUILIBRIUM When non reversible chemical reactions proceed to completion, the concentration of the reactants gradually decrease, until there is NO limiting reactant remaining. Most chemical reactions,

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

EQUILIBRIUM. Academic Success Center

EQUILIBRIUM. Academic Success Center EQUILIBRIUM Academic Success Center Definition Equilibrium is a state where the concentrations of the reactants and products no longer change with time. This doesn t mean there is no movement between the

More information

Equilibrium Practice Problems

Equilibrium Practice Problems Equilibrium Practice Problems 1. Write the equilibrium expression for each of the following reactions: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 I 2 (s) + Cl 2 (g) 2 ICl (g) K = [ICl]

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

Chemical Equilibrium. Chapter 13

Chemical Equilibrium. Chapter 13 Chemical Equilibrium Chapter 13 Chemical Equilibrium When neither the products nor the reactant concentrations change any more with time. Chemical Equilibrium When the forward rate of reaction is equal

More information

Ch14 Chemical Equilibrium. Modified by Dr. Cheng-Yu Lai

Ch14 Chemical Equilibrium. Modified by Dr. Cheng-Yu Lai Ch14 Chemical Equilibrium Modified by Dr. Cheng-Yu Lai CHEMICAL EQUILIBRIUM Chemical Equilibrium: Chemical Equilibrium When the rate of the forward reaction equals the rate of the reverse reaction and

More information

Chapter 13 - Chemical Equilibrium

Chapter 13 - Chemical Equilibrium Chapter 1 - Chemical Equilibrium Intro A. Chemical Equilibrium 1. The state where the concentrations of all reactants and products remain constant with time. All reactions carried out in a closed vessel

More information

CHAPTER 14 CHEMICAL EQUILIBRIUM

CHAPTER 14 CHEMICAL EQUILIBRIUM CHATER 14 CHEMICAL EQUILIBRIUM roblem Categories Biological: 14.98. Conceptual: 14.1, 14., 14.9, 14.5, 14.54, 14.55, 14.56, 14.57, 14.58, 14.59, 14.60, 14.61, 14.6, 14.66, 14.67, 14.68, 14.69, 14.81, 14.91,

More information

Equilibrium Lecture #1. Schweitzer

Equilibrium Lecture #1. Schweitzer Equilibrium Lecture #1 Schweitzer What is equilibrium? Remember Equilibrium process between to competing reactions. At equilibrium the forward process is equal to the reverse process. *** It appears that

More information

Chemical Equilibrium. Chemical Equilibrium

Chemical Equilibrium. Chemical Equilibrium Chemical Equilibrium When some types of chemical reactions occur in the gas or solution phases, these reaction attain chemical equilibrium, i.e., the reaction does not go to completion, but the reaction

More information

Consider: N 2 (g) + 3H 2 (g) 2NH 3 (g) G = -32.90 kj/mol. conc. time

Consider: N 2 (g) + 3H 2 (g) 2NH 3 (g) G = -32.90 kj/mol. conc. time 5.111 Lecture Summary #19 CHEMICAL EQUILIBRIUM (Chapter 9 Section 9.0-9.9) Topics Nature of Chemical Equilibrium Meaning of K Relationship between Equilibrium Expressions External Effects on K 19.1 Chemical

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

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. A.P. Chemistry Practice Test - Ch. 13: Equilibrium Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) At equilibrium,. A) the rates of the forward

More information

5. Which of the following reactions is a nonredox decomposition reaction? A) 2CuO 2Cu + O2 B) 2KClO3 2KCl + 3 O2 C) CaCO3 CaO + CO2 D) SO2 + H2O H2SO3

5. Which of the following reactions is a nonredox decomposition reaction? A) 2CuO 2Cu + O2 B) 2KClO3 2KCl + 3 O2 C) CaCO3 CaO + CO2 D) SO2 + H2O H2SO3 CHEM 120 Online Course Chapter 9. Date: 1. Which of the following general reaction types is characterized by there being a single reactant? A) combination B) decomposition C) single-replacement D) double-replacement

More information

Worked solutions to student book questions Chapter 16 Controlling the yield of reactions

Worked solutions to student book questions Chapter 16 Controlling the yield of reactions E1. Write an equation to show the equilibrium that exists between NaI(s) and Na + (aq) and I (aq). AE1. NaI(s) Na + (aq) + I (aq) E. a Sketch a graph of the change in the radioactivity of the solution

More information

Guide to Chapter 13. Chemical Equilibrium

Guide to Chapter 13. Chemical Equilibrium Guide to Chapter 13. Chemical Equilibrium We will spend five lecture days on this chapter. During the first class meeting we will focus on how kinetics makes a segue into equilibrium. We will learn how

More information

CH 223 Chapter Thirteen Concept Guide

CH 223 Chapter Thirteen Concept Guide CH 223 Chapter Thirteen Concept Guide 1. Writing Equilibrium Constant Expressions Write the equilibrium constant (K c ) expressions for each of the following reactions: (a) Cu(OH) 2 (s) (b) Cu(NH 3 ) 4

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

Chemical Equilibria & the Application of Le Châtelier s Principle to General Equilibria. Example of Equilibrium. !A(g) + "B(g)!

Chemical Equilibria & the Application of Le Châtelier s Principle to General Equilibria. Example of Equilibrium. !A(g) + B(g)! Chemical Equilibria & the Application of Le Châtelier s Principle to General Equilibria CHEM 102H T. Hughbanks Example of Equilibrium N 2 + 3H 2! 2 NH 3 Reactions can occur, in principle, in either direction.

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

IN CHAPTER 13, we examined how fast a chemical reaction occurs. In this chapter we. Chemical Equilibrium

IN CHAPTER 13, we examined how fast a chemical reaction occurs. In this chapter we. Chemical Equilibrium 14 Chemical Equilibrium Every system in chemical equilibrium, under the influence of a change of any one of the factors of equilibrium, undergoes a transformation... [that produces a change]... in the

More information

AP Chemistry Unit 7- Homework Problems Equilibrium and K sp

AP Chemistry Unit 7- Homework Problems Equilibrium and K sp AP Chemistry Unit 7- Homework Problems Equilibrium and K sp Nature of the Equilibrium State 1. Draw on this graph where equilibrium has been reached. [X] equilibrium time 2. What are three qualities of

More information

Chemical Equilibrium. Chapter 17: Keeping the chemical themes straight. Chapter 17 Homework Problems. N2O4 (g) 2NO2 (g)

Chemical Equilibrium. Chapter 17: Keeping the chemical themes straight. Chapter 17 Homework Problems. N2O4 (g) 2NO2 (g) Chapter 17: Chemical Equilibrium Equilibrium: The Extent of Chemical Reactions 17.1 The Dynamic Nature of the Equilibrium State 17. The Reaction Quotient and the Equilibrium Constant 17.3 Expressing Equilibria

More information

Chapter 16 Review Packet

Chapter 16 Review Packet Chapter 16 Review Packet AP Chemistry Chapter 16 Practice Multiple Choice Portion 1. For which process is ΔS negative? Note: ΔS = S final S initial therefore, if ΔS is positive, S final > S initial if

More information

Rate of Reaction and the Collision Theory. Factors that Affect the Rate of a Chemical Reaction

Rate of Reaction and the Collision Theory. Factors that Affect the Rate of a Chemical Reaction Chemical Kinetics and Thermodynamics Chemical Kinetics- concerned with: 1. Rates of Chemical Reactions- # of moles of reactant used up or product formed Unit time Or 2. Reaction Mechanisms- Rate of Reaction

More information

Equilibrium. Ron Robertson

Equilibrium. Ron Robertson Equilibrium Ron Robertson Basic Ideas A. Extent of Reaction Many reactions do not go to completion. Those that do not are reversible with a forward reaction and reverse reaction. To be really correct we

More information

Equilibria Involving Acids & Bases

Equilibria Involving Acids & Bases Week 9 Equilibria Involving Acids & Bases Acidic and basic solutions Self-ionisation of water Through reaction with itself: The concentration of water in aqueous solutions is virtually constant at about

More information

Chemical Reactions: Energy, Rates and Equilibrium

Chemical Reactions: Energy, Rates and Equilibrium Chemical Reactions: Energy, Rates and Equilibrium Chapter 7 Heat Changes During Chemical Reactions Bond Dissociation Energy- The amount of energy that must be supplied to break a bond and separate the

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

SECTION 14 CHEMICAL EQUILIBRIUM

SECTION 14 CHEMICAL EQUILIBRIUM 1-1 SECTION 1 CHEMICAL EQUILIBRIUM Many chemical reactions do not go to completion. That is to say when the reactants are mixed and the chemical reaction proceeds it only goes to a certain extent, and

More information

Chapter 14 Chemical Kinetics

Chapter 14 Chemical Kinetics Chapter 14 Chemical Kinetics Chemical kinetics: the area of chemistry dealing with the speeds or rates at which reactions occur. Rates are affected by several factors: The concentrations of the reactants:

More information

IB Chemistry. DP Chemistry Review

IB Chemistry. DP Chemistry Review DP Chemistry Review Topic 1: Quantitative chemistry 1.1 The mole concept and Avogadro s constant Assessment statement Apply the mole concept to substances. Determine the number of particles and the amount

More information

CHEM 102 CLASS NOTES Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 71272 CHAPTER 14, Chemical Equilibrium Chapter

CHEM 102 CLASS NOTES Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 71272 CHAPTER 14, Chemical Equilibrium Chapter CHEM 10 CLASS NOTES Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 717 CHAPTER 14, Chemical Equilibrium Chapter 14. Chemical Equilibrium 14.1 Characteristics of Chemical

More information

System. System, Boundary and surroundings: Nature of heat and work: Sign convention of heat: Unit-7 Thermodynamics

System. System, Boundary and surroundings: Nature of heat and work: Sign convention of heat: Unit-7 Thermodynamics Unit-7 Thermodynamics Introduction: The term Thermo means heat and dynamics means flow or movement.. So thermodynamics is concerned with the flow of heat. The different forms of the energy are interconvertible

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

CHEMICAL THERMODYNAMICS

CHEMICAL THERMODYNAMICS CHEMICAL THERMODYNAMICS Spontaneous reactions Nonspontaneous reactions Enthalpy change (ΔH) Exothermic and endothermic reactions Entropy change (ΔS) Gibbs free energy change, (ΔG) Free energy of formation

More information

STOICHIOMETRY. - the study of the quantitative aspects of chemical

STOICHIOMETRY. - the study of the quantitative aspects of chemical STOICHIOMETRY - the study of the quantitative aspects of chemical GENERAL PLAN FOR STOICHIOMETRY Mass reactant Mass product Moles reactant Stoichiometric factor Moles product STOICHIOMETRY It rests on

More information

CHAPTER 12 GASES AND THEIR BEHAVIOR

CHAPTER 12 GASES AND THEIR BEHAVIOR Chapter 12 Gases and Their Behavior Page 1 CHAPTER 12 GASES AND THEIR BEHAVIOR 12-1. Which of the following represents the largest gas pressure? (a) 1.0 atm (b) 1.0 mm Hg (c) 1.0 Pa (d) 1.0 KPa 12-2. Nitrogen

More information

3A Energy. What is chemical energy?

3A Energy. What is chemical energy? 3A Energy What is chemical energy? Chemical energy is a form of potential energy which is stored in chemical bonds. Chemical bonds are the attractive forces that bind atoms together. As a reaction takes

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

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

Problem Solving. Stoichiometry of Gases

Problem Solving. Stoichiometry of Gases Skills Worksheet Problem Solving Stoichiometry of Gases Now that you have worked with relationships among moles, mass, and volumes of gases, you can easily put these to work in stoichiometry calculations.

More information

Calculations with Chemical Reactions

Calculations with Chemical Reactions Calculations with Chemical Reactions Calculations with chemical reactions require some background knowledge in basic chemistry concepts. Please, see the definitions from chemistry listed below: Atomic

More information

b. Calculate the value of the equilibrium constant at 127ºC for the reaction 2NH 3 (g) N 2 (g) + 3H 2 (g)

b. Calculate the value of the equilibrium constant at 127ºC for the reaction 2NH 3 (g) N 2 (g) + 3H 2 (g) 1. Write the equilibrium expression for the following reaction: 4NH 3 (g) + 7O 2 (g) 4NO 2 (g) + 6H 2 O(g) 2. The following equilibrium concentrations were observed for this reaction at 127ºC: N 2 (g)

More information

Standard States. Standard Enthalpy of formation

Standard States. Standard Enthalpy of formation Standard States In any thermochemical equation, the states of all reactants and products must be specified; otherwise it becomes difficult for scientists to understand the experimental results of other

More information

CHEMICAL EQUILIBRIUM Chapter 13

CHEMICAL EQUILIBRIUM Chapter 13 Page 1 1 hemical Equilibrium EMIAL EQUILIBRIUM hapter 1 The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium

More information

CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS

CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS CHATER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS 17.1 If the rate of the forward reaction exceeds the rate of reverse reaction, products are formed faster than they are consumed. The change in reaction

More information

Chemistry Guide

Chemistry Guide 551534 - Chemistry Guide 1- Contents Question Item Objective Type Skill 1 0102 M03.02.04 Multiple-choice answer Mastery of Problem Solving 2 0099 M03.03.02 Multiple-choice answer Mastery of Concepts 3

More information

CHEMICAL EQUILIBRIUM (ICE METHOD)

CHEMICAL EQUILIBRIUM (ICE METHOD) CHEMICAL EQUILIBRIUM (ICE METHOD) Introduction Chemical equilibrium occurs when opposing reactions are proceeding at equal rates. The rate at which the products are formed from the reactants equals the

More information

Chapter 14. Review Skills

Chapter 14. Review Skills Chapter 14 The Process of Chemical Reactions ave you ever considered becoming a chemical engineer? The men and women in this profession develop industrial processes for the large-scale production of the

More information

Chemical Equations & Stoichiometry

Chemical Equations & Stoichiometry Chemical Equations & Stoichiometry Chapter Goals Balance equations for simple chemical reactions. Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of the term

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

Equilibrium, Acids and Bases Unit Summary:

Equilibrium, Acids and Bases Unit Summary: Equilibrium, Acids and Bases Unit Summary: Prerequisite Skills and Knowledge Understand concepts of concentration, solubility, saturation point, pressure, density, viscosity, flow rate, and temperature

More information

V. POLYPROTIC ACID IONIZATION. NOTICE: K a1 > K a2 > K a3 EQUILIBRIUM PART 2. A. Polyprotic acids are acids with two or more acidic hydrogens.

V. POLYPROTIC ACID IONIZATION. NOTICE: K a1 > K a2 > K a3 EQUILIBRIUM PART 2. A. Polyprotic acids are acids with two or more acidic hydrogens. EQUILIBRIUM PART 2 V. POLYPROTIC ACID IONIZATION A. Polyprotic acids are acids with two or more acidic hydrogens. monoprotic: HC 2 H 3 O 2, HCN, HNO 2, HNO 3 diprotic: H 2 SO 4, H 2 SO 3, H 2 S triprotic:

More information

CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS

CHAPTER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS CHATER 17 EQUILIBRIUM: THE EXTENT OF CHEMICAL REACTIONS END OF CHATER ROBLEMS 17.1 If the rate of the forward reaction exceeds the rate of reverse reaction, products are formed faster than they are consumed.

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

Figure 10.3 A mercury manometer. This device is sometimes employed in the laboratory to measure gas pressures near atmospheric pressure.

Figure 10.3 A mercury manometer. This device is sometimes employed in the laboratory to measure gas pressures near atmospheric pressure. Characteristics of Gases Practice Problems A. Section 10.2 Pressure Pressure Conversions: 1 ATM = 101.3 kpa = 760 mm Hg (torr) SAMPLE EXERCISE 10.1 Converting Units of Pressure (a) Convert 0.357 atm to

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

Form A. CORRECT: As gases mix, the disorder or number of microstates with the same energy increases. As a result, entropy increases as well.

Form A. CORRECT: As gases mix, the disorder or number of microstates with the same energy increases. As a result, entropy increases as well. Chem 130 Name Exam 3, Ch 7, 19, and a little 14 November 11, 2011 100 Points Please follow the instructions for each section of the exam. Show your work on all mathematical problems. Provide answers with

More information

Reaction Mechanism (continued) CHEMICAL KINETICS. Pt 2. Reaction Mechanisms. Often Used Terms. Reaction Mechanisms. The reaction

Reaction Mechanism (continued) CHEMICAL KINETICS. Pt 2. Reaction Mechanisms. Often Used Terms. Reaction Mechanisms. The reaction Reaction Mechanism (continued) CHEMICAL KINETICS Pt The reaction C H O + 5O 6CO + 4H O 3 4 3 has many steps in the reaction mechanism. Objectives! Be able to describe the collision and transition-state

More information

Chapter 5 Chemical Quantities and Reactions

Chapter 5 Chemical Quantities and Reactions Chapter 5 Chemical Quantities and Reactions 1 Avogadro's Number Small particles such as atoms, molecules, and ions are counted using the mole. 1 mole = 6.02 x 10 23 items Avogadro s number 602 000 000

More information

INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION. 37 74 20 40 60 80 m/e

INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION. 37 74 20 40 60 80 m/e CHM111(M)/Page 1 of 5 INTI COLLEGE MALAYSIA A? LEVEL PROGRAMME CHM 111: CHEMISTRY MOCK EXAMINATION: DECEMBER 2000 SESSION SECTION A Answer ALL EIGHT questions. (52 marks) 1. The following is the mass spectrum

More information

Chemistry: Chemical Equations

Chemistry: Chemical Equations Chemistry: Chemical Equations Write a balanced chemical equation for each word equation. Include the phase of each substance in the equation. Classify the reaction as synthesis, decomposition, single replacement,

More information

AP Chemistry 2009 Scoring Guidelines

AP Chemistry 2009 Scoring Guidelines AP Chemistry 2009 Scoring Guidelines The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900,

More information

Chapter 6 Quantities in Chemical Reactions

Chapter 6 Quantities in Chemical Reactions Chapter 6 Quantities in Chemical Reactions The Meaning of a Balanced Chemical Equation Mole-Mole Conversions Mass-Mass Conversions Limiting Reactants Percent Yield Energy Changes Copyright The McGraw-Hill

More information

Energy Changes in Chemical Reactions. System loses heat (negative); gains heat (positive) Describe the difference between the two.

Energy Changes in Chemical Reactions. System loses heat (negative); gains heat (positive) Describe the difference between the two. Energy Changes in Chemical Reactions Most reactions give off or absorb energy Energy is the capacity to do work or supply heat. Heat: transfer of thermal (kinetic) energy between two systems at different

More information

Thermodynamics. Energy can be used * to provide heat * for mechanical work * to produce electric work * to sustain life

Thermodynamics. Energy can be used * to provide heat * for mechanical work * to produce electric work * to sustain life Thermodynamics Energy can be used * to provide heat * for mechanical work * to produce electric work * to sustain life Thermodynamics is the study of the transformation of energy into heat and for doing

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

Chemistry Final Exam Review

Chemistry Final Exam Review Name: Date: Block: Chemistry Final Exam Review 2012-2013 Unit 1: Measurement, Numbers, Scientific Notation, Conversions, Dimensional Analysis 1. Write 0.000008732 in scientific notation 8.732x10-6 2. Write

More information

Ch 3. Rate Laws and Stoichiometry

Ch 3. Rate Laws and Stoichiometry Ch 3. Rate Laws and Stoichiometry How do we obtain r A = f(x)? We do this in two steps 1. Rate Law Find the rate as a function of concentration, r A = k fn (C A, C B ). Stoichiometry Find the concentration

More information

REVIEW QUESTIONS Chapter 5. 1. Determine the pressure of the gas (in mmhg) in the diagram below, given atmospheric pressure= 0.975 atm.

REVIEW QUESTIONS Chapter 5. 1. Determine the pressure of the gas (in mmhg) in the diagram below, given atmospheric pressure= 0.975 atm. Chemistry 101 ANSWER KEY REVIEW QUESTIONS Chapter 5 1. Determine the pressure of the gas (in mmhg) in the diagram below, given atmospheric pressure= 0.975 atm. atm = 0.975 atm h = 5 cmhg gas atm 760 mmhg

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 5 Thermochemistry

Chapter 5 Thermochemistry Chapter 5 Thermochemistry I. Nature of Energy Energy units SI unit is joule, J From E = 1/2 mv 2, 1J = 1kg. m 2 /s 2 Traditionally, we use the calorie as a unit of energy. 1 cal = 4.184J (exactly) The

More information