JEE Main 2022 · Chemistry · Chemical Thermodynamics · medium

56 g of benzene (liquid, M=78) is burnt completely in a bomb calorimeter with heat capacity 21.8 kJ K^-1 and the temperature rises by 3.0 K. Products…

Numerical answer

1.56\ \mathrm g

of benzene (liquid,

M=78

) is burnt completely in a bomb calorimeter with heat capacity

21.8\ \mathrm{kJ\,K^{-1}}

and the temperature rises by

3.0\ \mathrm K

. Products are

\mathrm{CO_2(g)}

and

\mathrm{H_2O(l)}

298\ \mathrm K

(

R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}

). The magnitude of

\Delta_cH^\circ

\mathrm{kJ\,mol^{-1}}

(nearest integer) is

Show answer and solution

Answer: 3274

Heat released

=21.8\times3.0=65.4

kJ for

1.56/78=0.020

mol, so

\Delta_cU=-3270\ \mathrm{kJ\,mol^{-1}}

. For

\mathrm{C_6H_6(l)+7.5O_2(g)\to6CO_2(g)+3H_2O(l)}

\Delta n_g=6-7.5=-1.5

\Delta H=\Delta U+\Delta n_gRT=-3270+(-1.5)(8.314\times10^{-3})(298)=-3270-3.72=-3273.7\ \mathrm{kJ\,mol^{-1}}

. Magnitude

\approx3274

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