Unit 5 Thermodynamics
5.4 Exercises
OpenStax
Section 5.4 Exercises
- The following sequence of reactions occurs in the commercial production of aqueous nitric acid:
[latex]\begin{array}{l l} 4 \text{NH}_3(g) + 5\text{O}_2(g) \longrightarrow 4 \text{NO}(g) + 6\text{H}_2 \text{O}(l) & \Delta H = -907 \;\text{kJ} \\[1em] 2 \text{NO}(g) + \text{O}_2(g) \longrightarrow 2\text{NO}_2(g) & \Delta H = -113 \;\text{kJ} \\[1em] 3\text{NO}_2(g) + \text{H}_2 \text{O}(l) \longrightarrow 2\text{HNO}_3(aq) + \text{NO}(g) & \Delta H = -139 \;\text{kJ} \end{array}[/latex]Determine the total energy change for the production of one mole of aqueous nitric acid by this process, using ammonia and oxygen as the only reactants.
- From the molar heats of formation (Source: OpenStax Chemistry 2e), determine how much heat is required to evaporate one mole of water: [latex]\text{H}_2 \text{O}(l) \longrightarrow \text{H}_2 \text{O}(g)[/latex]
- Calculate ΔH°298 for the process [latex]\text{Sb}(s) + \frac{5}{2} \text{Cl}_2(g) \longrightarrow \text{SbCl}_5(g)[/latex]
from the following information:
[latex]\begin{array}{l l} \text{Sb}(s) + \frac{3}{2}\text{Cl}_2(g) \longrightarrow \text{SbCl}_3(g) & \Delta H^{\circ}_{298} = -314 \;\text{kJ} \\[1em] \text{SbCl}_3(s) + \text{Cl}_2(g) \longrightarrow \text{SbCl}_5(g) & \Delta H^{\circ}_{298} = -80 \;\text{kJ} \end{array}[/latex]
- Calculate ΔH for the process [latex]\text{Hg}_2 \text{Cl}_2(s) \longrightarrow 2\text{Hg}(l) + \text{Cl}_2(g)[/latex]
from the following information:
[latex]\begin{array}{l l} \text{Hg}(l) + \text{Cl}_2(g) \longrightarrow \text{HgCl}(s) & \Delta H = -224 \;\text{kJ} \\[1em] \text{Hg}(l) + \text{HgCl}_2(s) \longrightarrow \text{Hg}_2 \text{Cl}_2(s) & \Delta H = -41.2 \;\text{kJ} \end{array}[/latex]
- Calculate the standard molar enthalpy of formation of NO(g) from the following data:
[latex]\begin{array}{l l} \text{N}_2(g) + 2\text{O}_2(g) \longrightarrow 2\text{NO}_2(g) & \Delta H^{\circ}_{298} = 66.4 \;\text{kJ} \\[1em] 2\text{NO}(g) + \text{O}_2(g) \longrightarrow 2\text{NO}_2(g) & \Delta H^{\circ}_{298} = -114.1 \;\text{kJ} \end{array}[/latex] - Using the data (Source: OpenStax Chemistry 2e), calculate the standard molar enthalpy change for each of the following reactions:
(a) [latex]\text{Si}(s) + 2\text{F}_2(g) \longrightarrow \text{SiF}_4(g)[/latex]
(b) [latex]2\text{C}(s) + 2\text{H}_2(g) \longrightarrow \text{CH}_3 \text{CO}_2 \text{H}(l)[/latex]
(c) [latex]\text{CH}_4(g) + \text{N}_2(g) \longrightarrow \text{HCN}(g) + \text{NH}_3(g)[/latex]
(d) [latex]\text{CS}_2(g) + 3\text{Cl}_2(g) \longrightarrow \text{CCl}_4(g) + \text{S}_2 \text{Cl}_2(g)[/latex]
- The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Using the data (Source: OpenStax Chemistry 2e), determine how much heat is produced by the decomposition of exactly 1 mole of H2O2 under standard conditions. [latex]2\text{H}_2 \text{O}_2(l) \longrightarrow 2\text{H}_2 \text{O}(g) + \text{O}_2(g)[/latex]
- Calculate the enthalpy of combustion of butane, C4H10(g) for the formation of H2O(g) and CO2(g). The enthalpy of formation of butane is −126 kJ/mol.
- The white pigment TiO2 is prepared by the reaction of titanium tetrachloride, TiCl4, with water vapour in the gas phase: [latex]\text{TiCl}_4(g) + 2\text{H}_2 \text{O}(g) \longrightarrow \text{TiO}_2(s) + 4\text{HCl}(g)[/latex].
How much heat is evolved in the production of exactly 1 mole of TiO2(s) under standard state conditions?
- In the early days of automobiles, illumination at night was provided by burning acetylene, C2H2. Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC2:[latex]\text{CaC}_2(s) + \text{H}_2 \text{O}(l) \longrightarrow \text{Ca(OH)}_2(s) + \text{C}_2 \text{H}_2(g)[/latex] (unbalanced equation). Calculate the standard enthalpy of the reaction. The ΔH°f of CaC2 is −63.35 kJ/mol.
Solutions
- −494 kJ. There are multiple ways to solve this. Keep the first reaction unchanged, add 2× the second reaction and (4/3)× the third reaction.
- 44.01 kJ
- −394 kJ
- 265 kJ
- 90.3 kJ mol−1 of NO
- (a) −1615.0 kJ mol−1; (b) −484.3 kJ mol−1; (c) 164.2 kJ mol−1; (d) −232.1 kJ mol−1
- −54.04 kJ
- −2.66 × 103 kJ mol−1
- 66.3 kJ evolved
- −122.8 kJ