Question

Calculate ΔHrxn for the following reaction:7C(s)+8H2(g)→C7H16(l)

Calculate ΔHrxn for the following reaction:

7C(s)+8H2(g)→C7H16(l)

Use the following reactions and given ΔH values:

C7H16(l)+11O2(g)→7CO2(g)+8H2O(g),ΔHC(s)+O2(g)→CO2(g),ΔH2H2(g)+O2(g)→2H2O(g),

ΔH===−4464.3kJ−393.5kJ−483.5kJ

Express your answer to four significant figures in kilojoules.

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Answer #1

\(\mathrm{C}_{7} \mathrm{H}_{16}(\mathrm{I})+11 \mathrm{O}_{2}(\mathrm{~g}) \Rightarrow 7 \mathrm{CO}_{2}(\mathrm{~g})+8 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})(1) ; \Delta H_{1}=-4464.3 \mathrm{KJ} / \mathrm{mol} .\)

\(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})(2) ; \Delta H_{2}=-393.5 \mathrm{KJ} / \mathrm{mol} .\)

\(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})(3) ; \Delta \mathrm{H}_{3}=-483.5 \mathrm{KJ} / \mathrm{mol}\)

Now

\((\) Eq. \(2 \times 7+\) Eq.3x4) - Eq.1

\(7 \mathrm{C}+11 \mathrm{O}_{2}+8 \mathrm{H}_{2}+7 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{C}_{7} \mathrm{H}_{16}+11 \mathrm{O}_{2}+7 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\)

or, \(7 \mathrm{C}(\mathrm{s})+8 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{7} \mathrm{H}_{16}(\mathrm{I})\)

so, \(\Delta H_{\text {rxn }}=7 \times \Delta H_{2}+4 \times \Delta H_{3}-\Delta H_{1}\)

\(=7 \times(-393.5)+4 \times(483.5)-(-4464.3)\)

\(=-2754.5-1934+4464.3\)

\(=-224.2 \mathrm{KJ} / \mathrm{mol}\)

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