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Chapter 6: Problem 21

The enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when $\mathrm{H}_{2} \mathrm{O}(l)$ is formed is \(-891 \mathrm{kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed is $-803 \mathrm{kJ} / \mathrm{mol}$ . Use these data and Hess's law to determine the enthalpy of vaporization for water.

Short Answer

Expert verified
Using Hess's law and the given enthalpy of combustion for \(CH_{4}(g)\) when \(H_{2}O(l)\) and \(H_{2}O(g)\) are formed, the enthalpy of vaporization for water can be determined through the difference in enthalpy changes for these reactions: \(\Delta H_{vap} = \Delta H_{2} - \Delta H_{1}\). Calculating the difference yields \(\Delta H_{vap} = 88\,kJ/mol\) for 2 moles of water. To find the enthalpy of vaporization for 1 mole of water, divide by 2, resulting in an enthalpy of vaporization for water of \(44\,kJ/mol\).

Step by step solution

01

Combustion reactions of methane

Write down the combustion reactions for \(CH_{4}\) with the formation of \(H_{2}O(l)\) and \(H_{2}O(g)\): Reaction 1 (with \(H_{2}O(l)\)): \(CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)\) with enthalpy change, \(\Delta H_{1} = -891\,kJ/mol\) Reaction 2 (with \(H_{2}O(g)\)): \(CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)\) with enthalpy change, \(\Delta H_{2} = -803\,kJ/mol\)
02

Using Hess's Law

According to Hess's Law, the enthalpy change of a process is independent of the path taken. The overall enthalpy change of the target process, \(H_{2}O(l) \rightarrow H_{2}O(g)\), can be determined by the difference between Reaction 2 and Reaction 1. Specifically, we want to find the enthalpy change for the process, \(2H_{2}O(l) \rightarrow 2H_{2}O(g)\), abbreviated as \(\Delta H_{vap}\). The relationship between the given reactions and the target reaction is: \(\Delta H_{vap} = \Delta H_{2} - \Delta H_{1}\)
03

Calculating the enthalpy of vaporization of water

Plug in the known enthalpy change values for reactions 1 and 2. Then, solve for the enthalpy change of the vaporization process: \(\Delta H_{vap} = (-803\,kJ/mol) - (-891\,kJ/mol) = 88\,kJ/mol\) This is the enthalpy change for the process of vaporizing 2 moles of water. To find the enthalpy of vaporization for 1 mole of water, divide this value by 2: \(\Delta H_{vap, 1\,mol} = \frac{88\,kJ/mol}{2} = 44\,kJ/mol\) Therefore, the enthalpy of vaporization for water is \(44\,kJ/mol\).

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Most popular questions from this chapter

In a coffee-cup calorimeter, 150.0 \(\mathrm{mL}\) of 0.50 \(\mathrm{M}\) HCl is added to 50.0 \(\mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{NaOH}\) to make 200.0 \(\mathrm{g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\) . If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{kJ} / \mathrm{mol}\) , calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is 4.184 \(\mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assume no heat loss to the surroundings.

Given: $$ \begin{array}{ll}{2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CuO}(s)} & {\Delta H^{\circ}=-288 \mathrm{kJ}} \\\ {\mathrm{Cu}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuO}(s)+\mathrm{Cu}(s)} & {\Delta H^{\circ}=11 \mathrm{kJ}}\end{array} $$ Calculate the standard enthalpy of formation $\left(\Delta H_{f}^{\circ}\right)\( for \)\mathrm{CuO}(s) .$

Calculate \(\Delta H\) for the reaction: $$ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-1010 . \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-317 \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-623 \mathrm{kJ}\) $$ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-286 \mathrm{kJ}\)

The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by \(2.54^{\circ} \mathrm{C}\) . Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is 26.42 \(\mathrm{kJ} / \mathrm{g} .\) ) A \(0.2130-\mathrm{g}\) sample of vanillin \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) is then burned in the same calorimeter, and the temperature increases by $3.25^{\circ} \mathrm{C}$ . What is the energy of combustion per gram of vanillin? Per mole of vanillin?

A swimming pool, 10.0 \(\mathrm{m}\) by \(4.0 \mathrm{m},\) is filled with water to a depth of 3.0 \(\mathrm{m}\) at a temperature of \(20.2^{\circ} \mathrm{C}\) . How much energy is required to raise the temperature of the water to \(24.6^{\circ} \mathrm{C} ?\)
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