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(A) The combustion reaction for $n$-octane $(C_8H_{18})$ is: $C_8H_{18}(l) + \frac{25}{2} O_2(g) \rightarrow 8CO_2(g) + 9H_2O(l)$ The change in the nu

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$-11.15$

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(A) The combustion reaction for $n$-octane $(C_8H_{18})$ is: $C_8H_{18}(l) + \frac{25}{2} O_2(g) ightarrow 8CO_2(g) + 9H_2O(l)$ The change in the number of moles of gaseous species is $\Delta n_g = n_{products(g)} - n_{reactants(g)} = 8 - 12.5 = -4.5$. The relationship between $\Delta H$ and $\Delta U$ is given by: $\Delta H = \Delta U + \Delta n_g RT$ $\Delta H - \Delta U = \Delta n_g RT$ Given $T = 298 \ K$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$: $\Delta H - \Delta U = (-4.5) 	imes 8.314 	imes 298 \ J \ mol^{-1}$ $= -11149.97 \ J \ mol^{-1} \approx -11.15 \ kJ \ mol^{-1}$.

Calculate the difference between $\Delta H$ and $\Delta U$ for the combustion of $n$-octane at $25^{\circ}C$. (in $kJ$)

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