JEE Main 2018 · Chemistry · Chemical Thermodynamics · medium

39 g of liquid benzene (M=78 g mol^-1) is burnt completely in a bomb calorimeter whose total heat capacity is 10.0 kJ K^-1 and the temperature rises…

Single correct option

0.39\ \mathrm{g}

of liquid benzene (

M=78\ \mathrm{g\,mol^{-1}}

) is burnt completely in a bomb calorimeter whose total heat capacity is

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

and the temperature rises by

1.62\ \mathrm{K}

300\ \mathrm{K}

. For

\mathrm{C_6H_6(l)+\tfrac{15}{2}O_2(g)\to 6CO_2(g)+3H_2O(l)}

\Delta H

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

is (

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

)

Show answer and solution

Answer: (C) $-3243.7$

Heat released

=10.0\times1.62=16.2

kJ for

0.39/78=0.005

mol, so

\Delta U=-16.2/0.005=-3240

kJ mol

^{-1}

(constant volume). Gas moles change:

\Delta n_g=6-7.5=-1.5

\Delta H=\Delta U+\Delta n_gRT=-3240+(-1.5)(8.314\times10^{-3})(300)=-3240-3.74=-3243.7\ \mathrm{kJ\,mol^{-1}}

. Using

+1.5

by mistake gives

-3236.3

; forgetting the correction gives

-3240

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake