QJEEM18Ch83b1: JEE Main 2018 Chemistry - Class 11 JEEM18Ch83b1 Solution: The combustion of benzene(l) gives CO 2 (g) and H 2 O(l). Given …

Formula of Heat of combination is

$\Delta H = \Delta u\,\, + \,\,\Delta ng\,\,RT$

Where, $\Delta H$ $=$ Heat of combination at constant pressure

$\Delta u\, =$ Heat at constant volume

$\Delta {n_g}$ change in number of moles for gaseous molecule.

R = gas constant

T = Temperature.

The Required reaction

C₆H₆(l) + ${{15} \over 2}$ O₂(g) $\to$ 6CO₂(g) + 3H₂O(l)

Here O₂ and CO₂ are gaseous molecules so to calculate $\Delta {n_g}$ we only consider those.

$\therefore\,\,\,$ $\Delta {n_g}$ $=$ 6 $-$ ${{15} \over 2}$ = $-$ ${3 \over 2}$

Given $\Delta$ u = $-$ 3263.9 kJ mol ^{$-$ 1} R = 8.314 JK $-$ mol ^{$-$ 1}

R = 8.314 JK ^{$-$ 1} mol ^{$-$ 1}

= 8.314 $\times$ 10 ^{$-$ 3} kJ K ^{$-$ 1} mol ^{$-$ 1}

T = 25^o C

= 25 + 273 K

= 298 K

So, $\Delta H$ = $-$ 3263.9 + $\left( { - {3 \over 2}} \right)$ $\times$ 8.314 $\times$ 10 ^{$-$ 3} $\times$ 298

= $-$ 3267.6 kJ mol ^{$-$ 1}