Thermodynamics and Thermochemistry - Result Question 20
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 $(R=8.314 $ $JK^{-1} $ $mol^{-1})$
(2018 Main)
(a) $4152.6$
(b) $-452.46$
(c) $3260$
(d) $-3267.6$
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Answer:
Correct Answer: 20. (d)
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)+\dfrac{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-\dfrac{15}{2}=-\dfrac{3}{2} \text { or }-1.5 $
Now we have $\Delta n _g$ and other values given in the question are
$\Delta U =-3263.9$ $ kJ / mol$
$T =25^{\circ} C=273+25=298 K$
$R =8.314 $ $JK^{-1}$ $ mol^{-1}$
So, $\Delta H _p=(-3263.9)+(-1.5) \times 8.314 \times 10^{-3} \times 298$
$ =-3267.6$ $ kJ$ $ mol^{-1} $
BETA
