Use the following data at 298 K… | JEE Advanced 2023 P2

Use the following data at 298 K (assume Δ H^° and Δ S^° are independent of temperature): ΔfH^°(CaCO3,s)=-1207, ΔfH^°(CaO,s)=-635, ΔfH^°(CO2,g)=-394…

Single correct option

Use the following data at

298\ \mathrm{K}

(assume

\Delta H^\circ

and

\Delta S^\circ

are independent of temperature):

\Delta_fH^\circ(\mathrm{CaCO_3,s})=-1207

\Delta_fH^\circ(\mathrm{CaO,s})=-635

\Delta_fH^\circ(\mathrm{CO_2,g})=-394\ \mathrm{kJ\,mol^{-1}}

;

S^\circ(\mathrm{CaCO_3,s})=93

S^\circ(\mathrm{CaO,s})=40

S^\circ(\mathrm{CO_2,g})=214\ \mathrm{J\,K^{-1}mol^{-1}}

;

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

. Excess

\mathrm{CaCO_3(s)}

is heated in an evacuated rigid vessel and the equilibrium

\mathrm{CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)}

is established. At what temperature is the equilibrium pressure of

\mathrm{CO_2}

equal to

0.10\ \mathrm{bar}

Show answer and solution

Answer: (C) $988\ \mathrm{K}$

Reaction enthalpy and entropy from Hess's law:

\Delta H^\circ=(-635-394)-(-1207)=+178\ \mathrm{kJ}

and

\Delta S^\circ=(40+214)-93=+161\ \mathrm{J\,K^{-1}}

. Pure solids have unit activity, so

K_p=p_{\mathrm{CO_2}}/p^\circ=0.10

. Setting

\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=-RT\ln K_p

gives

T=\dfrac{\Delta H^\circ}{\Delta S^\circ-R\ln K_p}=\dfrac{178000}{161-8.314\ln 0.10}=\dfrac{178000}{161+19.14}\approx 988\ \mathrm{K}

. The value

1106\ \mathrm{K}

\Delta H^\circ/\Delta S^\circ

, which corresponds to

K_p=1

(1 bar). Using

\log_{10}

in place of

\ln

gives

1051\ \mathrm{K}

, and a sign slip in

R\ln K_p

gives

1255\ \mathrm{K}

. Hence

988\ \mathrm{K}

Original question written in the JEE Advanced 2023 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