Solutions/Answers to Exercises
Exercise 1. Write a formation reaction for K2CrO4 (s).
2 K (s) + Cr (s) + 2 O2 (g) → K2CrO4 (s)
Exercise 2. What is the value of ΔHof for LiOH (s)? Write a formation equation for LiOH (s).
ΔHof = -487.5 kJ/mol
Li (s) + 1/2 O2 (g) + 1/2 H2 (g) → LiOH (s)
Exercise 3. Calculate ΔHorxn, in kJ for the following reaction. (Table of Thermodynamic Quantities)
Look up the ΔHof values of reactant and products in the table
ΔHof [Fe2O3] = -824.2 kJ/mol
ΔHof [CO] = -110.5 kJ/mol
ΔHof [Fe(s)] = 0 kJ/mol
ΔHof [CO2] = -393.5 kJ/mol \(\displaystyle \Bigl [2\;mol\;Fe\times\;0\frac{kJ}{mol}\;+\;3\;mol\;CO_2\times\;-393.5\frac{kJ}{mol}\Bigr]\;-\;\) \(\displaystyle\Bigl [1\;mol\;Fe_2O_3\times\;-824.2\frac{kJ}{mol}\;+\;3\;mol\;CO\times\;-110.5\frac{kJ}{mol}\Bigr]\;=\;\mathbf{-24.8\;kJ}\)
Exercise 4. Methylcyclopentane, C6H12, undergoes combustion according to the following equation:
Use the standard heat of reaction, ΔHorxn, and the ΔHof values from the Table of Thermodynamic Quantities to calculate the standard enthalpy of formation, ΔHof for methylcyclopentane.
Look up the ΔHof values for O2, CO2, and H2O (g) in the table of Thermodynamic Quantities.
ΔHof [O2 (g)] = 0 kJ/mol
ΔHof [CO2 (g)] = -393.5 kJ/mol
ΔHof [H2O (g)] = -241.8 kJ/mol
ΔHorxn = -3672 kJ/mol. The equation is in terms of 1 mol of C6H12.
ΔHof = [ΔHof[CO2] + ΔHof [H2O]] – [ΔHof [O2 + ΔHof [C6H12]]
Solve for ΔHof [C6H12].
– ΔHof [C6H12] = -3672 kJ + 6 mol CO2 x 393.5 kJ/mol + 6 mol H2O x 241.8 kJ/mol = -139.8 kJ per mole of C6H12.