JEE PYQ: Chemical Equilibrium Question 25
Question 25 - 2019 (10 Jan Shift 2)
(5). (1 g $\text{NH}_4\text{SH}$ is introduced in 3.0 L evacuated flask at 327$^\circ$C, 30% of the solid decomposed to $\text{NH}_3$ and $\text{H}_2\text{S}$ as gases. The $K_p$ of the reaction at 327$^\circ$C is ($R = 0.082$ L atm mol$^{-1}$ K$^{-1}$, molar mass of S = 32, N = 14))
(1) $0.242 \times 10^{-4}$ atm$^2$ (2) $1 \times 10^{-4}$ atm$^2$ (3) $4.9 \times 10^{-3}$ atm$^2$ (4) $0.242$ atm$^2$
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Answer: (4) $0.242$ atm$^2$
Solution
$K_c = (0.03/3)^2 = 10^{-4}$. $K_p = K_c(RT)^2 = 10^{-4} \times (0.082 \times 600)^2 = 0.242$ atm$^2$.
BETA
