Chem 201X In-Class Worksheet 8: Thermochemistry Answer Key Dr. Lara Baxley

1. If a 90.8 g piece of ice, originally at 0 ºC, is placed in a cup containing 506 g of water initially at 24.8 ºC, what will be the final temperature at thermal equilibrium? The ∆Hfus of water is 6.02 kJ/mol and specific heat capacity of liquid water is 4.184 J/gºC. [8.9 ºC]

qice = ni∆H + miC∆Ti ni = 90.8g/18.02 g/mol = 5.0388 mol qwater = mwC∆Tw –qwater = +qice – mwC∆Tw = ni∆H + miC∆Ti –(506g)(4.184J/gºC)(Tf – 24.8 ºC) = (5.0388mol)(6020J/mol) + (90.8g)(4.184J/gºC)(Tf – 0ºC) –2117.104(Tf –24.8) = 30333.576 + 379.9072(Tf – 0) –2117.104Tf + 52504.1792 = 30333.576 + 379.9072Tf 22170.6032 = 2497.0112Tf Tf = 8.878856 ºC = 8.9 ºC 2. Acetylene C2H2 is a gas that is used in welding torches because it produces a very hot flame.

The thermochemical equation for the combustion of acetylene is shown below.

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) ΔH = –2599.2 kJ

a. What is the change in heat when 12.0 moles of CO2 is produced? [–7.80 x 103 kJ] 12.0 mol CO2 | –2599.2 kJ = –7797.6 kJ = –7.80 x 103 kJ | 4 moles CO2

b. What is the change in heat when 56.4 g of acetylene burns? [–2.81 x 103 kJ] 56.4 g C2H2 | mol C2H2 | –2599.2 kJ = –2814.8 kJ = –2.81 x 103 kJ | 26.04 g C2H2 | 2 mole C2H2

c. If 56.4 g of acetylene burns in a calorimeter and all of the heat is transferred to 20.0 kg of water initially at 17.4 ºC, what will be the final temperature of the water? [51.0 ºC]

qw = –qrxn = –(–2814.8 kJ) = 2814.8 kJ = 2.8148 x 106 J qw = mC∆T 2.8148 x 106 J = (2.00 x 104 g)(4.184 J/gºC)(Tf – 17.4 ºC) Tf = 51.0 ºC

Chem 201X In-Class Worksheet 8: Thermochemistry Answer Key Dr. Lara Baxley

3. When calcium chloride dissolves in water, the ΔH of the process is shown below. In a coffee cup calorimeter, 15.2 g of calcium chloride is dissolved in 300.0 g of water at 22.0°C. What is the final temperature of the solution? Assume that the solution has a specific heat capacity of 4.184 J/g°C. [30.5 ºC]

½ CaCl2(s) → ½ Ca2+(aq) + Cl–(aq) ΔH = –40.8 kJ

15.2 g CaCl2 | mol CaCl2 = 0.13696 mol CaCl2 (I kept extra digits, but 3 are significant) | 110.98 g CaCl2

qrxn = 0.13696 mol CaCl2 | –40.8 kJ | rxn | 103 J = –1.1176 x 104 J | rxn | ½ mol CaCl2 | kJ

–qrxn = qwater qwater = +1.1176 x 104 J qwater = mC∆T +1.1176 x 104 J = (315.2 g)(4.18 J/gºC)(Tf – 22.0 ºC) Tf = 30.5 ºC

4. A 30.0 mL sample of 0.882 M H2O2 at 25.0 °C is placed in a coffee cup calorimeter and

allowed to decompose completely according the thermochemical equation shown below. The final temperature of the solution is 44.9 °C. Calculate the enthalpy of the reaction shown, in kJ. The mass of the solution is 30.0 g and the specific heat capacity of the solution is 4.18 J/g°C. [–189 kJ]

2 H2O2(aq) → O2(g) + 2 H2O(l) ∆H = ? kJ

qwater = mC∆T = (30.0 g)(4.18 J/gºC)(44.9 ºC–25.0 ºC) = +2495.46 J qwater = –qrxn qrxn = –2495.46 J mol H2O2 = 0.0300 L | 0.882 mol H2O2 = 0.02646 mol H2O2 | L ∆H = –2495.46 J | 2 mol H2O2 | kJ = –189 kJ 0.02646 mol H2O2 | rxn | 103 J

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Chem 201X In-Class Worksheet 8: Thermochemistry Answer Key Dr. Lara Baxley

5. Calculate the ∆H for the reaction ClF(g) + F2(g) → ClF3(g) given the following: [–108.7 kJ]

2ClF(g) + O2(g) → Cl2O(g) + F2O(g) ∆H = +167.4 kJ 2ClF3(g) + 2O2(g) → Cl2O(g) + 3F2O(g) ∆H = +341.4 kJ 2F2(g) + O2(g) → 2F2O(g) ∆H = –43.4 kJ

½ (#1): ClF(g) + ½ O2(g) → ½ Cl2O(g) + ½ F2O(g) ∆H = +83.7 kJ –½(#2): ½ Cl2O(g) + 3/2 F2O(g) → ClF3(g) + O2(g) ∆H = –170.7 kJ ½ (#3): F2(g) + ½ O2(g) → F2O(g) ∆H = –21.7 kJ ClF(g) + F2(g) → ClF3(g) ∆H = –108.7 kJ