Question

The combustion of one mole of benzene takes place at $298 \,K$ and $1$ atm. After combustion, $CO_{2\left(g\right)}$ and $H_{2}O_{\left(l\right)}$ are produced and $3267\, kJ$ of heat is liberated. Standard enthalpies of formation of $CO_{2\left(g\right)}$ and $H_{2}O_{\left(l\right)}$ are $-393.5 \,kJ\,mol^{-1}$ and $-285.83 \,kJ\, mol^{-1}$ respectively. The standard enthalpy of formation, $\Delta_{f} H^{\circ}$ of benzene is

• A. $45.5 \,kJ\, mol^{-1}$
• B. $20 \,kJ\,mol^{-1}$
• C. $48.51 \,kJ\,mol^{-1}$
• D. $15.52\,kJ\,mol^{-1}$

Answer 1

Answer: (C)

Required equation is
$6C_{\left(s\right)}+3H_{2\left(g\right)} \rightarrow C_{6}H_{6\left(l\right)} ; \Delta H_{f} =$ ?
Given:
(i) $C_{6}H_{6\left(l\right)}+\frac{15}{2}O_{2\left(g\right)}\rightarrow6CO_{2\left(g\right)}+3H_{2}O_{\left(l\right)}, \Delta H=-3267\,kJ\,mol^{-1}$
(ii) $C_{\left(s\right)}+O_{2\left(g\right)}\rightarrow, CO_{2\left(g\right)}, \Delta H =-393.5\, kJ\,mol^{-1}$
(iii) $H_{2\left(g\right)}+\frac{1}{2}O_{2\left(g\right)} \rightarrow H_{2}O_{\left(l\right)}, \Delta H=-285.83\,kJ\,mol^{-1}$
Operating $6 \times$ eqn. (ii) $+3\times$ eqn. (iii)-eqn. (i):
$6C_{\left(s\right)}+3H_{2\left(g\right)}\rightarrow C_{6}H_{6\left(l\right)}, \Delta H = 6\left(-395.5\right)+3\left(-285.83\right)-\left(-3267.0\right)$
$=-2361-857.49+3267.0$
$=48.51\,kJ\,mol^{-1}$