JEE PYQ: Chemical Equilibrium Question 1
Question 1 - 2021 (16 Mar Shift 1)
For the reaction $\text{A}(g) \rightleftharpoons \text{B}(g)$ at 495 K.
$\Delta_r G^\circ = -9.478 \text{ kJ mol}^{-1}$
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____ millimoles. (Round off to the Nearest Integer).
$[R = 8.314 \text{ J mol}^{-1} \text{K}^{-1};; \ell n10 = 2.303]$
Show Answer
Answer: 20
Solution
$\Delta G^\circ = -RT \ln K_{eq}$. Given $\Delta G^\circ = -9.478 \text{ kJ/mol}$, $T = 495$ K, $R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}$. $\ln K_{eq} = 2.303$, so $K_{eq} = 10$. ICE: $K_{eq} = \frac{x}{22-x} = 10 \Rightarrow x = 20$.
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