(3)/(2) O_2(g) O_3(g) ; K_p for this reaction is 2.47 × 10^ -29 . At 298 ~K, Δ_r G^ for conversion of oxygen to ozone will be

Correct answer: C. 163 ~kJ ~mol^ -1 aligned Δ_r G^ = & 2.303 log~Kp(R=8.314 JK^ -1 ~mol^ -1 ) Δ_r G^ = & -2.303 × 8.314 JK^ -1 ~mol^ -1 × 298 ~K & × 2.47 × 10^ -29 = & 163000 Jmol^ -1 =163 ~kJ mol^ -1 aligned

JIPMER 2013

Exam: JIPMER
Year: 2013
Subject: Chemistry
Chapter: Thermodynamics (C)

JIPMER 2013 · Chemistry · Thermodynamics (C)

(3)/(2) O_2(g) O_3(g) ; K_p for this reaction is 2.47 × 10^ -29 ....

\frac{3}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{O}_3(g) ; K_p

for this reaction is

2.47 \times 10^{-29}

. At

298 \mathrm{~K}, \Delta_r G^{\circ}

for conversion of oxygen to ozone will be

Options

A $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $163 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $2303 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Correct answer

C. $163 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Step-by-step solution

\begin{aligned} \Delta_r G^{\circ}= & 2.303 \operatorname{log~Kp}\left(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) \\ \Delta_r G^{\circ}= & -2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K} \\ & \times \log 2.47 \times 10^{-29} \\ = & 163000 \mathrm{Jmol}^{-1}=163 \mathrm{~kJ} \mathrm{mol}^{-1}\end{aligned}

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