### Chapter 8: Thermochemistry

```Chapter 8: Thermochemistry
6. Refer to Problem 1.16, Section 8.1, and Example 8.1.
a. In the solution process, heat is absorbed. Thus, it is not exothermic, it is endothermic.
b. Heat is absorbed from the water. Thus, heat exits the water and the process for the water
is exothermic. Thus, q H 2O = −27.6 kJ .
c.
250.0 mL ×
− 27.6 kJ ×
1.00 g
= 250.0 g
1 mL
1000 J
= −27600 J
1 kJ
q = mc∆t
-27600 J = (250.0 g)( 4.18 J/g°C) ∆t
∆t = -26.4°C
∆t = tf - ti
-26.4°C = tf – 30.0°C
tf = 3.6°C
d. t°F = 1.8t°C + 32
ti = 30.0°C
t°F = 1.8(30.0°C) + 32
ti = 86.0°F
tf = 3.6°C
t°F = 1.8(3.6°C) + 32
tf = 38.5°F
16. Refer to Sections 8.3 and 8.4, Figure 8.4, and Example 8.4.
a. CaO(s) + 3C(s) → CO(g) + CaC2(s)
∆H = +464.8 kJ
b. Heat is absorbed, therefore the reaction is endothermic.
c.
46
4.
8
kJ
CO + CaC 2
∆H
The enthalpy of the products is
higher than the enthalpy of the
reactants.
CaO + 3C
reaction path
d. The formation of one mole of CaC2 consumes 464.8 kJ, therefore:
1.00 g CaC 2 x
1 mol. CaC 2
464.8 kJ
x
= 7.25 kJ
64.098 g CaC 2 1 mol. CaC 2
e. From the balanced equation, we know there are 3 mol. C per mol. CaC2, thus:
20.00 kJ x
1 mol. CaC 2
12.01 g C
3 mol. C
x
x
= 1.550 g C
464.8 kJ
1 mol. CaC 2 1 mol. C
26. Refer to Section 8.4 and Tables 8.1 and 8.2.
Calculate the heat evolved in condensing benzene gas to liquid, then the amount evolved in
cooling the liquid to 25.00°C.
C6H6 (g, 80.00°C) → C6H6 (l, 80.00°C)
∆ H vap = 100.00 g x
∆H = -30.8 kJ/mol.
1 mol. C 6 H 6 - 30.8 kJ
x
= - 39.4 kJ
78.1 g C 6 H 6
1 mol.
C6H6 (l, 80.00°C) → C6H6 (l, 25.00°C)
q = mc∆t = (100.00 g)(1.72 J/g°C)(25.00°C – 80.00°C) = -9460 J = -9.46 kJ
∆H = ∆H (condensation) + q (cooling)
∆H = (-39.4 kJ) + (-9.46 kJ) = -48.9 kJ
34. Refer to Section 8.5 and Example 8.7.
a. ∆Hreaction = Σ ∆Hfº (products) - Σ ∆Hfº (reactants)
Recall that ∆Hfº for elements in their standard states is zero.
314.6 kJ = [2mol.∆Hfº Cu(s) + 1mol.∆Hfº O2(g)] – [2mol.∆Hfº CuO(s)]
314.6 kJ = 2mol.∆Hfº CuO(s)
-157.3 kJ = ∆Hfº CuO(s)
b. Use the answer in part a. as a conversion factor.
∆H ° = 13.58 g CuO x
1 mol. CuO
- 157.3 kJ
x
= - 26.85 kJ
79.55 g CuO 1 mol. CuO
```