Thermodynamics and Thermochemistry - Result Question 95
####20. 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 $25^{\circ} C$; heat of combustion (in $kJ$ $mol^{-1}$ ) of benzene at constant pressure will be $\left(R=8.314 JK^{-1} mol^{-1}\right)$
(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,
$$ C _6 H _6(l)+\frac{15}{2} O _2(g) \longrightarrow 6 CO _2(g)+3 H _2 O(l) $$
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
$$ \begin{aligned} \Delta U & =-3263.9 kJ / mol \ T & =25^{\circ} C=273+25=298 K \ R & =8.314 JK^{-1} mol^{-1} \end{aligned} $$
So, $\Delta H _p=(-3263.9)+(-1.5) \times 8.314 \times 10^{-3} \times 298$
$$ =-3267.6 kJ mol^{-1} $$
