Thermodynamics and Thermochemistry 1 Question 17
20. The combustion of benzene ( $l$ ) gives $\mathrm{CO}{2}(g)$ and $\mathrm{H}{2} \mathrm{O}(l)$. Given that heat of combustion of benzene at constant volume is $-3263.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $25^{\circ} \mathrm{C}$; heat of combustion (in $\mathrm{kJ}$ $\mathrm{mol}^{-1}$ ) of benzene at constant pressure will be $\left(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
(2018 Main)
(a) 4152.6
(b) -452.46
(c) 3260
(d) -3267.6
(2017 Main)
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Solution:
- Key idea Calculate the heat of combustion with the help of following formula
$$ \Delta H_{p}=\Delta U+\Delta n_{g} R T $$
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.
From the equation,
$$ \mathrm{C}{6} \mathrm{H}{6}(l)+\frac{15}{2} \mathrm{O}{2}(g) \longrightarrow 6 \mathrm{CO}{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$
Change in the number of gaseous moles i.e.
$$ \Delta n_{g}=6-\frac{15}{2}=-\frac{3}{2} \text { or }-1.5 $$
Now we have $\Delta n_{g}$ and other values given in the question are
$$ \begin{aligned} \Delta U & =-3263.9 \mathrm{~kJ} / \mathrm{mol} \ T & =25^{\circ} \mathrm{C}=273+25=298 \mathrm{~K} \ R & =8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned} $$
So, $\Delta H_{p}=(-3263.9)+(-1.5) \times 8.314 \times 10^{-3} \times 298$
$$ =-3267.6 \mathrm{~kJ} \mathrm{~mol}^{-1} $$
