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Chapter 5: Problem 74

Using values from Appendix \(\mathrm{C},\) calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(\mathrm{CaO}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{CaCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{CuO}(s)+\mathrm{NO}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{NO}_{2}(g)\) (d) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
The calculated values of ΔH⁰ for the given reactions are as follows: (a) ΔH⁰ = -217.5 kJ/mol (b) ΔH⁰ = -560.4 kJ/mol (c) ΔH⁰ = 86.5 kJ/mol (d) ΔH⁰ = -286.8 kJ/mol

Step by step solution

01

(a) CaO(s) + 2 HCl(g) -> CaCl2(s) + H2O(g)

First, we need to find the ΔH⁰ values of the reactants and products from Appendix C. ΔH⁰(CaO(s)) = -635.1 kJ/mol ΔH⁰(HCl(g)) = -92.3 kJ/mol ΔH⁰(CaCl2(s)) = -795.4 kJ/mol ΔH⁰(H2O(g)) = -241.8 kJ/mol Next, we will apply the formula: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) ΔH⁰(reaction) = [ΔH⁰(CaCl2(s)) + ΔH⁰(H2O(g))] - [ΔH⁰(CaO(s)) + 2 × ΔH⁰(HCl(g))] ΔH⁰(reaction) = [(-795.4) + (-241.8)] - [(-635.1) + 2 × (-92.3)] ΔH⁰(reaction) = -1037.2 + 635.1 + 184.6 ΔH⁰(reaction) = -217.5 kJ/mol
02

(b) 4 FeO(s) + O2(g) -> 2 Fe2O3(s)

First, we need to find the ΔH⁰ values of the reactants and products from Appendix C. ΔH⁰(FeO(s)) = -272.0 kJ/mol ΔH⁰(O2(g)) = 0 kJ/mol (as it is an element in its standard state) ΔH⁰(Fe2O3(s)) = -824.2 kJ/mol Next, we will apply the formula: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) ΔH⁰(reaction) = [2 × ΔH⁰(Fe2O3(s))] - [4 × ΔH⁰(FeO(s)) + ΔH⁰(O2(g))] ΔH⁰(reaction) = [2 × (-824.2)] - [4 × (-272.0) + 0] ΔH⁰(reaction) = -1648.4 + 1088.0 ΔH⁰(reaction) = -560.4 kJ/mol
03

(c) 2 CuO(s) + NO(g) -> Cu2O(s) + NO2(g)

First, we need to find the ΔH⁰ values of the reactants and products from Appendix C. ΔH⁰(CuO(s)) = -156.1 kJ/mol ΔH⁰(NO(g)) = 90.3 kJ/mol ΔH⁰(Cu2O(s)) = -168.6 kJ/mol ΔH⁰(NO2(g)) = 33.2 kJ/mol Next, we will apply the formula: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) ΔH⁰(reaction) = [ΔH⁰(Cu2O(s)) + ΔH⁰(NO2(g))] - [2 × ΔH⁰(CuO(s)) + ΔH⁰(NO(g))] ΔH⁰(reaction) = [(-168.6) + 33.2] - [2 × (-156.1) + 90.3] ΔH⁰(reaction) = -135.4 + 312.2 - 90.3 ΔH⁰(reaction) = 86.5 kJ/mol
04

(d) 4 NH3(g) + O2(g) -> 2 N2H4(g) + 2 H2O(l)

First, we need to find the ΔH⁰ values of the reactants and products from Appendix C. ΔH⁰(NH3(g)) = -45.9 kJ/mol ΔH⁰(O2(g)) = 0 kJ/mol (as it is an element in its standard state) ΔH⁰(N2H4(g)) = 50.6 kJ/mol ΔH⁰(H2O(l)) = -285.8 kJ/mol Next, we will apply the formula: ΔH⁰(reaction) = Σ ΔH⁰(products) - Σ ΔH⁰(reactants) ΔH⁰(reaction) = [2 × ΔH⁰(N2H4(g)) + 2 × ΔH⁰(H2O(l))] - [4 × ΔH⁰(NH3(g)) + ΔH⁰(O2(g))] ΔH⁰(reaction) = [2 × 50.6 + 2 × (-285.8)] - [4 × (-45.9) + 0] ΔH⁰(reaction) = 101.2 - 571.6 + 183.6 ΔH⁰(reaction) = -286.8 kJ/mol

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Most popular questions from this chapter

(a) Calculate the kinetic energy in joules of a \(1200-\mathrm{kg}\) automobile moving at \(18 \mathrm{~m} / \mathrm{s}\). (b) Convert this energy to calories. (c) What happens to this energy when the automobile brakes to a stop?

A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{~g} \mathrm{CO}_{2}(g), 4.47 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}(g),\) and \(311 \mathrm{~kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empirical-formula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

Given the data $$ \begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g) & & \Delta H=+180.7 \mathrm{~kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) & & \Delta H=-113.1 \mathrm{~kJ} \\ 2 \mathrm{~N}_{2} \mathrm{O}(g) & \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H &=-163.2 \mathrm{~kJ} \end{aligned} $$ use Hess's law to calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) $$

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is used as a fuel in race cars. (a) Write a balanced equation for the combustion of liquid methanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced by combustion per liter of methanol. Methanol has a density of \(0.791 \mathrm{~g} / \mathrm{mL}\). (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per kJ of heat emitted.

The heat of combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l),\) is \(-1367 \mathrm{~kJ} / \mathrm{mol}\). A batch of Sauvignon Blanc wine contains \(10.6 \%\) ethanol by mass. Assuming the density of the wine to be \(1.0 \mathrm{~g} / \mathrm{mL},\) what is the caloric content due to the alcohol (ethanol) in a 6 -oz glass of wine \((177 \mathrm{~mL}) ?\)
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