Question

Suppose the enthalpy of formation of $CO_{2}\left(\right.g\left.\right)$ and $H_{2}O\left(\right.l\left.\right)$ are $-394kJ/mol$ and -286 kJ/mol respectively. The standard enthalpy of formation of octane $\left(\right.C_{8}H_{18}\left.\right)$ is $-250kJ/mol$ . Then calculate the standard enthalpy of combustion of $C_{8}H_{18}$ :

• A. $-5200kJ/mol$
• B. $-5726kJ/mol$
• C. $-5476kJ/mol$
• D. $-5310kJ/mol$

Answer 1

Answer: (C)

The Balanced reaction from the combustion of $C_{8}H_{18}$ :
$C_{8}H_{18}+\frac{25}{2}O_{2} \rightarrow 8CO_{2}+9H_{2}O$
^{$\left(\Delta \right)_{r}H^{0} \, = \, 8\times \left(\left(ΔΗ\right)^{0}\right)_{\left(CO\right)_{2}} \, + \, 9\times \left(\left(ΔΗ\right)^{0}\right)_{H_{2} O} \, - \, \left(\right. \left(\left(ΔΗ\right)^{0}\right)_{C_{8} H_{18}} + \, \frac{25}{2} \left(\left(ΔΗ\right)^{0}\right)_{\left(\right. O_{2} \left.\right)} \left.\right) \\ \\ \therefore \, \left(\left(ΔΗ\right)^{0}\right)_{O_{2}}= \, 0$} (standard enthalpy of formation of $O_{2}$ )
(standard enthalpy of reaction = standard enthalpy of combustion)
$\left(\Delta \right)_{r}H^{o}=8\times \left(- 394\right)+9\times \left(- 286\right)-\left(- 250\right)$
$=-5476 \, k J / m o l$