Question

For the combustion of n-octane $C_{8}H_{18}+ \, O_{2} \rightarrow CO_{2}+H_{2}O$ at $\left(25\right)^{o}C \, \left(i g n o r e \, r e s o n a n c e \, i n \, C O_{2}\right)$

• A. $Δ \text{H} = Δ \text{E} - \text{5.5} \times \text{8.31} \times \text{0.298 in kJ/mol}$
• B. $Δ \text{H} = Δ \text{E} + \text{4.5} \times \text{8.31} \times \text{0.298 in kJ/mol}$
• C. $Δ \text{H} = Δ \text{E} - \text{4.5} \times \text{8.31} \times \text{298 in kJ/mol}$
• D. $\Delta \text{H} = \Delta \text{E} - \text{4.5} \times \text{8.31} \times \text{0.298 in kJ/mol}$

Answer 1

Answer: (D)

n-Octane existed as liquid from, $\left(\text{C}\right)_{8} \left(\text{H}\right)_{1 8 - \left(\textit{l}\right)} + \frac{2 5}{2} \left(\text{O}\right)_{2} \rightarrow 8 \left(\text{CO}\right)_{2 \left(\text{g}\right)} + 9 \left(\text{H}\right)_{2} \left(\text{O}\right)_{\left(\textit{l}\right)}$
$\Delta \text{n}_{\text{g}} = 8 - \text{12.5} = - \text{4.5}$
$\Delta \text{H}=\Delta \text{E}+\Delta \text{n}_{\text{g}}\text{RT}$
$\Delta \text{H} = \Delta \text{E} - \text{4.5} \times \text{8.31} \times \text{0.298 in kJ/mol}$