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Chapter 7: Problem 17

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
The enthalpy of vaporization for water (ΔHvap) can be determined using Hess's Law and the given enthalpy of combustion values for CH4(g). By subtracting the first reaction (when H2O(l) is formed) from the second reaction (when H2O(g) is formed), we obtain the reaction 2H2O(l) -> 2H2O(g) with an enthalpy change of ΔHvap = ΔH2 - ΔH1. Substituting the given values, we find that ΔHvap = (-803 kJ/mol) - (-891 kJ/mol), which gives us an enthalpy of vaporization for water of 88 kJ/mol.

Step by step solution

01

Write down the given reactions

First, let's write down the given reactions and their enthalpies: 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 And we need to determine the enthalpy of vaporization for water (ΔHvap).
02

Understand Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, no matter whether it happens in one step or multiple steps. So, if we can create a reaction that represents the enthalpy of vaporization for water using the given reactions, then we can find the ΔHvap for water.
03

Create a new reaction representing the enthalpy of vaporization

We can achieve this by subtracting the first reaction from the second reaction. This will give us a new reaction representing the vaporization of H2O: Reaction 3: 2H2O(l) -> 2H2O(g) ΔH3 = ΔHvap = ΔH2 - ΔH1 Now we just need to calculate ΔHvap by substituting the values of ΔH1 and ΔH2.
04

Calculate ΔHvap for water

ΔHvap = ΔH2 - ΔH1 ΔHvap = (-803 kJ/mol) - (-891 kJ/mol)
05

Solve for ΔHvap

ΔHvap = 88 kJ/mol So, the enthalpy of vaporization for water is 88 kJ/mol.

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Most popular questions from this chapter

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of \(20 .\) kg sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g) \Delta H=5640 \mathrm{kJ}$$ calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis.

One of the components of polluted air is NO. It is formed in the high- temperature environment of internal combustion engines by the following reaction: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{kJ}$$ Why are high temperatures needed to convert \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to NO?

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing \(150.0 \mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\). After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C} .\) If the specific heat capacity of nickel is \(0.444 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) \(30.0 \mathrm{L})\) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V:\) Calculate the work (in units of J) associated with the two pathways. Is work a state function? Explain.

Consider the reaction $$\mathrm{B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-2035 \mathrm{kJ}$$ Calculate the amount of heat released when \(54.0 \mathrm{g}\) of diborane is combusted.
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