Solutions/Answers to Exercises
Exercise 1. Consider the following reaction:
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ΔH = -89.3 kJ
(2) Calculate ΔH, in kJ, for the production of 2.68 g of KCl.
\(\displaystyle 2.68\;g\times\frac{1\;mol}{74.55\;g}\times\frac{-89.3\;kJ}{2\;mol}\;=\;\mathbf{-1.61\;kJ}\)
Calculate ΔH, in kJ, for the decomposition of 0.682 g of KClO3.
\(\displaystyle 0.682\;g\times\frac{1\;mol}{122.55\;g}\times\frac{-89.3\;kJ}{2\;mol}\;=\;\mathbf{-0.248\;kJ}\)
Exercise 2. Consider the following reaction:
CaO (s) + 3 C (s) → CaC2 + CO (g) ΔHo = 464.6 kJ
2. Calculate ΔH if 423.26 g of carbon completely reacts.
\(\displaystyle 423.26\;g\;C\frac{1\;mol\;C}
{12.011\;g\;C}\frac{464.6\;kJ}{3\;mol\;C}\;=\;\mathbf{5455\;J}\)
Exercise 3. Use the following enthalpies of reaction
(2) C (s) + 2 F2 (g) → CF4 (g) ΔH = -679 kJ
(3) 2 C (s) + 2 H2 (g) → C2H4 (g) ΔH = +52.2 kJ
to calculate ΔH for the following reaction
C2H4 (g) + 6 F2 → 2 CF4 (g) + 4 HF (g) ΔH = ?
Multiply the coefficients of equations 1 and 2 by two and then multiply both ΔH values by two. Then reverse equation 3 and change the sign of ΔH. Finally, add the equations to obtain the overall equation and add the ΔH values.
(1) 2 H2 (g) + 2 F2 (g) → 4 HF (g) ΔH = 2 x (-536 kJ)
(2) 2 C (s) + 4 F2 (g) → 2 CF4 (g) ΔH = 2 x (-679 kJ)
(3) C2H4 (g) → 2 C (s) + 2 H2 (g) → ΔH = -(+52.2 kJ)
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C2H4 (g) + 6 F2 → 2 CF4 (g) + 4 HF (g) ΔH = -2482.2 kJ
Exercise 4. Use the data below to calculate ΔH for
N2O (g) + NO2 (g) → 3 NO (g) ΔH = ?
Data:
(2) 2 NO (g) + O2 (g) → 2 NO2 (g) ΔH = -113.2 kJ
(3) 2 N2O (g) → 2 N2 (g) + O2 (g) ΔH = -163.2 kJ
Leave equation (1) alone. Reverse and divide equation (2) by two. Divide equation (3) by two.
(1) N2 (g) + O2 (g) → 2 NO (g) ΔH = +180.6 kJ
(2) NO2 → NO (g) + 1/2 O2 (g) ΔH = -(-113.2) x 1/2 kJ
(3) N2O (g) → N2 (g) + 1/2O2 (g) ΔH = 1/2 x -163.2 kJ
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N2O (g) + NO2 (g) → 3 NO (g) ΔH = 155.7 kJ