JEE Advanced 2022 Paper 2 · Chemistry · Chemical Thermodynamics · medium

A bomb calorimeter measures the heat of complete combustion of liquid ethanol at 298 K as Δ U=-1364.5 kJ mol^-1, the products being CO2(g) and…

Single correct option

A bomb calorimeter measures the heat of complete combustion of liquid ethanol at

298\ \mathrm{K}

\Delta U=-1364.5\ \mathrm{kJ\,mol^{-1}}

, the products being

\mathrm{CO_2(g)}

and

\mathrm{H_2O(l)}

. Take

\Delta H_f^\circ[\mathrm{CO_2(g)}]=-393.5

and

\Delta H_f^\circ[\mathrm{H_2O(l)}]=-285.8\ \mathrm{kJ\,mol^{-1}}

, and

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

. The standard enthalpy of formation of

\mathrm{C_2H_5OH(l)}

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

is closest to

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Answer: (A) $-277.4$

The combustion is

\mathrm{C_2H_5OH(l)}+3\mathrm{O_2(g)}\to2\mathrm{CO_2(g)}+3\mathrm{H_2O(l)}

, so

\Delta n_g=2-3=-1

.

A bomb calorimeter measures

\Delta U

, so first convert to enthalpy:

\Delta H=\Delta U+\Delta n_gRT=-1364.5+(-1)(8.314\times10^{-3})(298)=-1364.5-2.48=-1367.0\ \mathrm{kJ\,mol^{-1}}

.

Now apply Hess's law:

\Delta H_{comb}=2\Delta H_f(\mathrm{CO_2})+3\Delta H_f(\mathrm{H_2O})-\Delta H_f(\mathrm{C_2H_5OH})

, so

\Delta H_f(\mathrm{C_2H_5OH})=2(-393.5)+3(-285.8)+1367.0=-787.0-857.4+1367.0=-277.4\ \mathrm{kJ\,mol^{-1}}

.

The other options come from typical slips:

-279.9

uses

\Delta U

directly as

\Delta H

;

-282.4

takes the correction with the wrong sign (

\Delta H=\Delta U+RT

); and

-1367.0

is just the enthalpy of combustion, not of formation.

Original question written in the JEE Advanced 2022 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake