Question

The combustion of benzene (l) gives $CO_2 (g) \: and \: H_{2}O (l)$. Given that heat of combustion of benzene at constant volume is -3263.9 kJ mol^-1 at 25°C ; heat of combustion (in kJ mol^-1) of benzene at constant pressure will be ( [tex] R = 8.314 J{K}^{-1} \: {mol}^{-1} ) ​

Answer 1

■ Key Idea

Calculate the heat of combustion with the help of following formula

$\boxed{ \boxed{\Delta H_p = \Delta U + \Delta n_g RT}}$

$\sf \red{Where ;}$

$\Delta H_p$ = Heat of combustion at constant pressure

$\Delta U$ = Heat at constant volume (It is also called $\Delta E$ )

$\Delta n_g$ = Change in number of moles (In gaseous state)

$R$ = Gas constant

$T$ = Temperature

$\\ \rm \red{ \: From \: the \: equation,}$

$C_6 H_6 (l) + \dfrac{15}{2} O_2 \longrightarrow \: 6CO_2 (g) + 3 H_2 O (l)$

$\rm \: Change \: in \: the \: number \: of \: gaseous \: moles \: i.e.$

$\Delta n_g = 6 - \dfrac{15}{2} = - \dfrac{3}{2} \: or \: - 1.5$

Now we have $\Delta n_g$ and other values given in the question are

$\rm\Delta U = -3263.9 \: kJ/mol$

$\rm \: T = 25 °C \: = 273 + 25 = 298 \: K$

$\rm \: R = 8.314 \: J \: K^{-1} \: mol^{-1}$

So,

$\rm \: \Delta H_p = (-3263.9) + (-1.5) × 8.314 × {10}^{-3} × 298$

$\boxed{ \boxed{ \rm \: \Delta H_p = -3267.6 \: kJ \: mol^{-1}}}$