JEE Main 2020 · Chemistry · Chemical Thermodynamics · medium

A 0.640 g sample of naphthalene (C10H8, M=128) is burnt in a bomb calorimeter with excess oxygen, producing CO2(g) and H2O(l). The calorimeter (heat…

Numerical answer

0.640\ \mathrm{g}

sample of naphthalene (

\mathrm{C_{10}H_8}

M=128

) is burnt in a bomb calorimeter with excess oxygen, producing

\mathrm{CO_2(g)}

and

\mathrm{H_2O(l)}

. The calorimeter (heat capacity

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

) shows a temperature rise of

2.50\ \mathrm{K}

. At

298\ \mathrm{K}

(

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

), the magnitude of the standard enthalpy of combustion of naphthalene in

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

, to the nearest integer, is

Show answer and solution

Answer: 5155

Heat released

=10.30\times2.50=25.75

kJ for

0.640/128=0.00500

mol, so

\Delta U=-5150\ \mathrm{kJ\,mol^{-1}}

. Reaction:

\mathrm{C_{10}H_8(s)+12O_2(g)\to10CO_2(g)+4H_2O(l)}

\Delta n_g=10-12=-2

\Delta H=\Delta U+\Delta n_gRT=-5150-2\times8.314\times298\times10^{-3}=-5150-4.955=-5154.96\approx-5155

. Magnitude

5155

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