JEE Advanced 2022 Paper 1 · Chemistry · Chemical Thermodynamics · hard

The calorimeter constant of a bomb calorimeter is determined by burning 0.610 g of benzoic acid (M=122 g mol^-1, Δ Uc=-3227 kJ mol^-1), which raises…

Numerical answer

The calorimeter constant of a bomb calorimeter is determined by burning

0.610\ \mathrm{g}

of benzoic acid (

M=122\ \mathrm{g\,mol^{-1}}

\Delta U_c=-3227\ \mathrm{kJ\,mol^{-1}}

), which raises the temperature by

1.00\ \mathrm{K}

. In the same calorimeter,

0.920\ \mathrm{g}

of liquid ethanol (

M=46\ \mathrm{g\,mol^{-1}}

) burns completely to

\mathrm{CO_2(g)}

and

\mathrm{H_2O(l)}

with a temperature rise of

1.69\ \mathrm{K}

298\ \mathrm{K}

. Given

\Delta_fH^\circ(\mathrm{CO_2,g})=-393.5

and

\Delta_fH^\circ(\mathrm{H_2O,l})=-285.8\ \mathrm{kJ\,mol^{-1}}

and

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

, the magnitude of

\Delta_fH^\circ

of ethanol(l), in

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

, is (give the answer correct to two decimal places)

Show answer and solution

Answer: 278.51

Calibration:

0.610/122=0.00500

mol benzoic acid releases

0.00500\times3227=16.135

kJ, so

C_{cal}=16.135/1.00=16.135\ \mathrm{kJ\,K^{-1}}

. Ethanol:

0.920/46=0.0200

mol; heat released

=16.135\times1.69=27.268

kJ, hence

\Delta U_c=-27.268/0.0200=-1363.41\ \mathrm{kJ\,mol^{-1}}

. For

\mathrm{C_2H_5OH(l)+3O_2(g)\to2CO_2(g)+3H_2O(l)}

\Delta n_g=2-3=-1

, so

\Delta H_c=\Delta U_c+\Delta n_gRT=-1363.41-8.314\times10^{-3}\times298=-1365.89\ \mathrm{kJ\,mol^{-1}}

. Then

\Delta H_c=2(-393.5)+3(-285.8)-\Delta_fH^\circ(\mathrm{C_2H_5OH})

, giving

\Delta_fH^\circ=-787.0-857.4+1365.89=-278.51\ \mathrm{kJ\,mol^{-1}}

. Magnitude: 278.51.

Original question written in the JEE Advanced 2022 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake