Chemical And Ionic Equilibrium
- Chemical And Ionic Equilibrium Result Question 1
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- Chemical and Ionic Equilibrium - Result Question 1
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- Chemical and Ionic Equilibrium - Result Question 8
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- Chemical And Ionic Equilibrium Result Question 43 — JEE PYQ Chapterwise
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- Chemical and Ionic Equilibrium - Result Question 9
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- Chemical And Ionic Equilibrium Result Question 48 — JEE PYQ Chapterwise
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- Chemical and Ionic Equilibrium - Result Question 10
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- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
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- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 63-1
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise
- Chemical and Ionic Equilibrium - Result Question 110
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- Chemical and Ionic Equilibrium - Result Question 137
Chemical and Ionic Equilibrium - Result Question 13
The standard Gibbs energy change at $300 K$ for the reaction, $2 A \rightleftharpoons B+C$ is $2494.2 J$. At a given time, the composition of the reaction mixture is $[A]=\frac{1}{2},[B]=2$ and $[C]=\frac{1}{2}$. The reaction proceeds in the direction of lower Gibbs energy
$(R=8.314 $ $J K / mol, e=2.718)$
(2015, Main)
(a) forward direction because $Q < K_c$
(b) reverse direction because $Q > K _c$
(c) forward direction because $Q < K_c$
(d) reverse direction because $Q < K _c$
Show Answer
Answer:
Correct Answer: 13. (b)
Solution:
- Given, $\Delta G^{\circ}=2494.2\ \text{J}$
$ Q=\frac{[B][C]}{[A]^{2}}=\frac{2 \times \frac{1}{2}}{\left(\frac{1}{2}\right)^{2}}=4 $
$\therefore$ We know, $\Delta G=\Delta G^{\circ}+R T \ln Q$
$ \begin{alignedat} & =2494.2+8.314 \times 300 \ln 4 \\ & =28747.27\ \text{J} (+ \text{ve value}) \end{aligned} $
Also, we have $\quad \Delta G=R T \ln \frac{Q}{K}$
If $\Delta G$ is positive, $Q<K$
Therefore, reaction shifts in reverse direction.





