JEE MAIN - Chemistry (2018 (Offline) - No. 23)
The combustion of benzene(l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25oC; heat of combustion (in kJ mol–1) of benzene at constant pressure will be :
(R = 8.314 JK–1 mol–1)
(R = 8.314 JK–1 mol–1)
–3267.6
4152.6
–452.46
3260
Explanation
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 = 25o 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
$$\Delta H = \Delta u\,\, + \,\,\Delta ng\,\,RT$$
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 = 25o 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
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