Chem 201X In-Class Worksheet 7: Thermochemistry Answer Key
Dr. Lara Baxley
1. A student measures 59.4 g of water into a coffee cup calorimeter and measures the
temperature as 21.4 ºC. The student then adds to the water a 2.164 g piece of metal that has
been stored in a 160.0 ºC oven. The temperature of the water rises to a final temperature of
22.4 ºC.
a. What is the change in heat for the water? [248 J]
qw = mC∆T = (59.4 g)(4.184 J/gºC)(22.4 ºC – 21.4 ºC) = 248 J
b. What is the change in heat for the metal? [-248 J]
qm = –qw = –248 J
c. What is the specific heat capacity of the metal? [0.83 J/gºC]
qm = mC∆T
–248 J = (2.164 g)C(22.4 ºC – 160.0 ºC)
C = 0.83 J/gºC
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 7: 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
cont’ on the next page
Chem 201X In-Class Worksheet 7: 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
∆H = –43.4 kJ
2F2(g) + O2(g) → 2F2O(g)
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