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College Chemistry : Standard Enthalpy and Entropy of Reactions

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Example Question #1 : Thermochemistry And Changes Of State

Use average bond enthalpies to estimate the enthalpy change, in kilojoules, of the combustion of one mole of hexane $(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3)$ to form liquid water and carbon dioxide gas.

Type of Bond	Average bond enthalpy $\left(\frac{kJ}{mol}\right)$
Carbon-hydrogen single bond	414
Carbon-oxygen double bond	736
Oxygen-oxygen double bond	498
Oxygen-hydrogen single bond	464
Carbon-carbon single bond	347

Possible Answers:

3066kJ

-3066kJ

4012kJ

-4012kJ

Correct answer:

-3066kJ

Explanation:

Start by writing the balanced equation for the combustion of hexane:

$\text{C}_6\text{H}_{14}(l)+\frac{19}{2}\text{O}_2(g)\rightarrow 6\text{CO}_2(g)+7\text{H}_2\text{O}(l)$

Recall the following equation:

$\Delta H_{rxn}=\Sigma(\Delta H \text{ bonds broken})+\Sigma(\Delta H \text{ bonds made})$

Remember that energy is required to break bonds, so $\Delta H\text{ bonds broken}$ should be positive. Energy is also released when bonds are made so $\Delta H\text{ bonds made}$ should be negative.

Next, draw out the Lewis structures of each molecule to figure out the number of bonds made or broken.

For the hexane, there are $\text{C-H}$ bonds that must be broken, and $\text{C-C}$ bonds that must be broken.

For each oxygen, there is only 1 oxygen double bond to break. However, we will need to multiply this number by its stoichiometric coefficient.

For each carbon dioxide, there are double bonds between carbon and oxygen to break. For the given equation, there are total of these double bonds because we have moles of carbon dioxide.

For each water, there are single bonds between hydrogen and oxygen. For the given equation, there are a total of of these bonds because we have moles of water.

Now, use the given information regarding to average bond enthalpies to find the change in enthalpy for the reaction.

$\Delta H_{rxn}=\Sigma(\Delta H \text{ bonds broken})+\Sigma(\Delta H \text{ bonds made})$

$\Delta H_{rxn}=14(414)+5(347)+\frac{19}{2}(498)+12(-736)+14(-464)=-3066\text{kJ}$

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College Chemistry : Standard Enthalpy and Entropy of Reactions

Study concepts, example questions & explanations for College Chemistry

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Example Question #1 : Thermochemistry And Changes Of State

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