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Chapter 6: Problem 17

The enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{~kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) 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
The enthalpy of vaporization for water can be determined using Hess's law and the given enthalpy of combustion values for methane when water is formed as a liquid and a gas. By subtracting the enthalpy of combustion for the liquid reaction from the enthalpy of combustion for the gas reaction, we find that the enthalpy of vaporization for water is \(ΔH₃ = 88 \: \text{kJ/mol}\).

Step by step solution

01

Write down the given information

The given information is the enthalpy of combustion of methane when water is formed as a liquid and as a gas: 1. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH₁ = -891 kJ/mol 2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), ΔH₂ = -803 kJ/mol
02

Write the reaction for the enthalpy of vaporization of water

The enthalpy of vaporization describes the conversion of one mole of water from the liquid phase to the gas phase: 3. H₂O(l) → H₂O(g), ΔH₃ = ΔH(vap)
03

Apply Hess's law

Hess's law states that the total enthalpy change in a reaction is equal to the sum of individual enthalpy changes in the reactions forming the desired reaction. We need to use reaction 1 and 2 to obtain reaction 3. To obtain reaction 3, we can visualize subtracting reaction 1 from reaction 2 (eliminating the CH₄ and O₂ molecules): ΔH₃ = ΔH₂ - ΔH₁ = (-803 kJ/mol) - (-891 kJ/mol)
04

Calculate ΔH₃

Now we can calculate the enthalpy of vaporization for water (ΔH₃): ΔH₃ = 891 kJ/mol - 803 kJ/mol = 88 kJ/mol Hence, the enthalpy of vaporization for water is 88 kJ/mol.

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Most popular questions from this chapter

The heat capacity of a bomb calorimeter was determined by burning \(6.79 \mathrm{~g}\) methane (energy of combustion \(=-802 \mathrm{~kJ} / \mathrm{mol} \mathrm{CH}_{4}\) ) in the bomb. The temperature changed by \(10.8^{\circ} \mathrm{C}\). a. What is the heat capacity of the bomb? b. A \(12.6-\mathrm{g}\) sample of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), produced a temperature increase of \(16.9^{\circ} \mathrm{C}\) in the same calorimeter. What is the energy of combustion of acetylene (in \(\mathrm{kJ} / \mathrm{mol}\) )?

The enthalpy of combustion of solid carbon to form carbon dioxide is \(-393.7 \mathrm{~kJ} / \mathrm{mol}\) carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is \(-283.3 \mathrm{~kJ} / \mathrm{mol}\) CO. Use these data to calculate \(\Delta H\) for the reaction $$ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) $$

A piston performs work of \(210 . \mathrm{L} \cdot \mathrm{atm}\) on the surroundings, while the cylinder in which it is placed expands from \(10 . \mathrm{L}\) to 25 L. At the same time, \(45 \mathrm{~J}\) of heat is transferred from the surroundings to the system. Against what pressure was the piston working?

Some automobiles and buses have been equipped to burn propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Compare the amounts of energy that can be obtained per gram of \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\) and per gram of gasoline, assuming that gasoline is pure octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .\) (See Example \(6.11 .\) ) Look up the boiling point of propane. What disadvantages are there to using propane instead of gasoline as a fuel?

Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{aligned} 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) & \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1010 . \mathrm{kJ} \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-317 \mathrm{~kJ} \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-143 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \Delta H &=-286 \mathrm{~kJ} \end{aligned} $$
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