# Chapter 5: thermochemstry. Internal Energy: E

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1 Chapter 5: thermochemstry tonight s goals Energy and Enthalpy Review Enthalpies of Reaction Calorimetry Hess Law Enthalpies of Formation Internal Energy: E E = The sum of all kinetic and potential energies of all components of the system. E after a reaction/process is: E = E final E initial If E is positive, energy was gained by the system If E is negative, energy was lost to the surroundings When energy is exchanged between the system and surroundings ( E), it is exchanged as either heat (q) or work (w): E = q + w 1

2 Work = force x distance In an open container in the lab, the only work done is by a gas pushing on the surroundings or by the surroundings pushing on a gas In a closed system with a movable piston, that work can be measured by the rising and falling of a piston Gas pushes piston up: work is done by the system; energy goes to surroundings Piston is pushed down: work is done on the system, energy is gained by system This is called PV work (Pressure and Volume are changing) When pressure is constant (our assumption); w = -P V Enthalpy Enthalpy (H) is defined as internal energy + pressure x volume (work): H = E + PV When the system changes at constant pressure, the change in enthalpy is H = E + P V Since E = q + w and w = -P V, we can substitute these into the enthalpy expression: H = E + P V = (q+w) w H = q (at constant pressure) Thus, enthalpy is a measure of the heat flow between a system and the surroundings 2

3 Enthalpy of Reaction, H The change in enthalpy, H, after a reaction is the enthalpy of the products minus the enthalpy of the reactants: H = H products H reactants H is called the enthalpy or heat of reaction H depends on the states of products and reactants. H is numerically the same, but opposite sign for reverse reaction Calorimetry: measurement of heat flow We can t know exact enthalpy of reactants and products, but we can measure H at constant pressure using calorimetry, the measurement of heat flow. If we perform a reaction in aqueous solution, the molecules/atoms involved in the reaction are the system and the water/container are the surroundings We can measure T of the surroundings easily T = T final T initial Using T and information on the mass and specific heat capacities of the water and container, we can calculate q surroundings and thus, H of the reaction 3

4 Heat Capacity and Specific Heat The amount of energy required to raise the temperature of a substance by 1 K (1 C) is its heat capacity. Specific heat capacity (or specific heat) is the energy required to raise the temperature of 1 g of a substance by 1 K. Other Substances Sand Soil Wood 1.7 (1.2 to 2.3) Specific Heat capacity Look at the units for specific heat: J/g-K Specific heat = heat transferred (J) Mass (m) temp change ( T) C s = q m T or q = C s m T 4

5 Back to the coffee cup By carrying out a reaction in aqueous solution in a calorimeter, we can estimate the heat change for the system by measuring ΔT for the contents & calorimeter (the surroundings). Δq sys = - Δq surr = ΔH rxn Δq surr = Δq contents + Δq cup Δq contents = C s (water) m contents ΔT Δq cup = C (cup) ΔT For today s lab; C s (water) = J/g- o C C (cup) = 50.0 J/g- o C Measuring ΔH Using a Coffee-Cup Calorimeter A student mixes 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee-cup calorimeter. The temperature of the solution increases from 21.0 C to 27.5 C. Calculate the ΔH for the reaction in kj/mol HCl, assuming: the calorimeter loses only a negligible quantity of heat, the total volume of the solution is 100 ml the solution density is 1.0 g/ml and its C s is 4.18 J/g-K. Solution Analyze: Write reaction equation: HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) We need to calculate the heat produced per mole of HCl, given the temperature increase of the solution, the number of moles of HCl and NaOH involved, and the density and specific heat of the solution. Plan: The total heat produced can be calculated using q rxn = -C s x m x Δ T. The # moles of HCl can be calculated from the volume and M of the solution and used to determine the heat produced per mol HCl. Solve: V of solution is 100 ml, mass is: The temperature change is: We can calculate q rxn 5

6 Measuring ΔH Using a Coffee-Cup Calorimeter (con t) To find ΔH on a molar basis, we need to calculate # mol HCl: Volume x M (mol/l) = # of mol HCl L x 1 mol/l = mol HCl ΔH rxn = -2.7 kj, so ΔH rxn /mol HCl = -2.7 kj/0.050 M HCl = -54 kj/mol HCl Reality check: The sign of ΔH rxn is negative, which it should be for an exothermic reaction Another cup of coffee? When 50.0 ml of M AgNO 3 and 50.0 ml of M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from C to C. The temperature increase is caused by the following reaction:. AgNO 3 (aq) + HCl(aq) AgCl(s) + HNO 3 (aq) Calculate ΔH for this reaction in kj/mol AgNO 3, assuming the solution has a mass of g and a specific heat of 4.18 J/g C. ΔH rxn = q rxn = -C s x m x ΔT = J/g C x g x 0.81 C = -339 J # mol AgNO 3 = L x M AgNO 3 = mol AgNO 3 ΔH rxn in kj/mol = 339 J/ mol AgNO 3 x 1kJ/1000 J = -68 kj/mol 6

7 hess law Hess s law states that If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps. Intermediate step is formation of carbon monoxide (incomplete combustion), followed by further combustion to carbon dioxide. More on hess law Hess' law allows ΔH rxn to be calculated even when it can t be measured directly. To do this, we perform arithmetic operations on chemical equations and known ΔH values. Chemical equations may be multiplied or divided by a whole number. When an equation is multiplied by a constant, so is ΔH If an equation is reversed, ΔH for reaction is reversed (-ΔH) Addition of chemical equations leads to a net equation If ΔH for each equation is added, the result will be the ΔH for the net equation. 7

8 let s try one Use the thermochemical equations shown below to determine the enthalpy for the reaction: 2NH 3 (g) N 2 (g) + 3H 2 (g) CH 2 O(g) + N 2 (g) + 3H 2 (g) N 2 H 4 (l) + CH 4 O(l) N 2 H 4 (l) + H 2 (g) 2NH 3 (g) CH 2 O(g) + H 2 (g) CH 4 O(l) N 2 H 4 (l) + CH 4 O (l) CH 2 O(g) + N 2 (g) + 3H 2 (g) 2NH 3 (g) N 2 H 4 (l) + H 2 (g) CH 2 O(g) + H 2 (g) CH 4 O(l) 2NH 3 (g) N 2 (g) + 3H 2 (g) ΔH=18.5 KJ Δ H=-9.0 KJ Δ H=32.5 KJ Δ H= KJ Δ H=+9.0 KJ Δ H= 32.5 KJ Δ H= 23.0 KJ For more fun like this: Scroll down the menu for the Hess Law DP (drill and practice) Given: Try some more Hess Math B 2 O 3 (s) + 3 H 2 O (g) 3 O 2 (g) + B 2 H 6 (g) H 2 O (l) H 2 O (g) H 2 (g) + (1/2) O 2 (g) H 2 O (l) ΔH = 2035 kj/mol ΔH = 44 kj/mol ΔH = -286 kj/mol 2 B (s) + 3 H 2 (g) B 2 H 6 (g) ΔH = 36 kj/mol Find the ΔH f of: 2 B (s) + (3/2) O 2 (g) B 2 O 3 (s) First we have to adjust the equations: (3/2) O 2 B 2 H 6 (g) + 3 O 2 (g) B 2 O 3 (s) + 3 H 2 O (g) ΔH = J/mol (reversed) 3 H 2 O (g) 3 H 2 O (l) 3x ΔH = -132 kj/mol (reversed) 3 H 2 O (l) 3 H 2 (g) + (3/2) O 2 (g) 3x ΔH = 858 kj/mol (reversed) 2 B (s) + 3 H 2 (g) B 2 H 6 (g) ΔH = 36 kj/mol Adding these equations and canceling out common terms, we get: 2 B (s) + (3/2) O 2 (g) B 2 O 3 (s) ΔH = kj/mol 8

9 Enthalpies of Formation Enthalpy of formation, H f, is defined as: the enthalpy change for a reaction in which a compound is made from its constituent elements in their element forms Standard enthalpies of formation, H f, are measured under standard conditions (25 C and 1.00 atm pressure) with the elements in their standard states The H f of the most stable form of an element is zero The standard enthalpy of formation is represented by a reaction where: each reactant is an element in its standard state the product is one mole of the compound. Example: 2Na(s) + ½ O 2 (g) Na 2 O (s) Which reaction represents the ΔH f reaction for NaNO 3? 1. Na + (aq) + NO 3- (aq) NaNO 3 (aq) 2. Na + (g) + NO 3- (g) NaNO 3 (s) 3. Na (s) + NO 3 (s) NaNO 3 (s) 4. 2 Na (s) + N 2 (g) + 3 O 2 (g) 2 NaNO 3 (s) 5. 2 Na (s) + 1/2 N 2 (g) + 3/2 O 2 (g) NaNO 3 (s) 9

10 Standard enthalpies of formation, ΔH f o Calculation of H using H f We can use Hess s law in this way: H = n H f products m H f reactants where n and m are the stoichiometric coefficients. To do this we look at an equation: Start with the reactants. Decompose them into elements, then rearrange the elements to form products. The overall H is the sum of the Hs for each step. 10

11 Calculation of H using H f C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (l) Decomposing into elements (note O 2 is elemental, so H f =0) C 3 H 8 (g) 3C(s) + 4H 2 (g) H 1 = H f [C 3 H 8 (g)] Form the products CO 2 and H 2 O from their elements: 3C(s) + 3O 2 (g) 3CO 2 (g) H 2 = 3 H f [CO 2 (g)] 4H 2 (g) + 2O 2 (g) 4H 2 O(l) H 3 = 4 H f [H 2 O(l)] We look up the values and add: H = [ΣnH f of Products] - [Σ mh f of Reactants] = [3( kj) + 4( kj)] [1( kj) + 5(0 kj)] = [( kj) + ( kj)] [( kj) + (0 kj)] = ( kj) ( kj) = kj Foods and Fuels Foods and fuels both involve combustion-type reactions to produce heat The term fuel value is used to express the quantity of heat released when a fuel is combusted Our body s fuels are carbohydrates, fats and proteins Fats yield more energy/gram than carbohydrates or proteins (38 kj/g vs. 17 kj/g) Carbohydrates and fats produce CO 2 and H 2 O when metabolized; proteins produce N-containing compounds Consume more than burn excess stored as fat Can use our knowledge of energy to calculate calories burned, etc 11

12 Energy in Fuels Most of the energy consumed in the US is from fossil fuels. The higher the % of H and C in a fuel, the more energy is produced per gram Energy consumption is increasing dramatically; research into more efficient use of earth s energy resources is critical Creative energy conversions Hybrid cars capture some of the waste energy during braking to charge electric battery, which can provide energy to move the car at slow speeds Wind turbines Solar panels 12

13 Upcoming events In lab: Standardization of NaOH Lab Report due Heats of reaction pre-lab due Chapter 4-5 review on Thursday Mastering Chemistry Chap 5 due Thur 5/20 Second Lecture Quiz (Chap 4-5) on 5/20 We ll start Chapter 6 next Tuesday 13

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