At equilibrium for the reaction A_2(g)+B_2(g) leftharpoons 2 AB(g) the concentrations of A_2, ~B_2 and AB respectively are 1.5 × 10^-3 M, 2.1 × 10^-3 M and 1.4 × 10^-3 M in a sealed vessel at 800 K . What will be K_p for the decomposition of AB at same temperature?

Correct answer: (B) 1.6. A_2(~g)+B_2(~g) leftharpoons 2 AB(~g) K_c=([AB]^2)/([~A_2 ] [B_2 ]) K_c=frac1.4 × 10^-3 × 1.4 × 10^-31.5 × 10^-3 × 2.1 × 10^-3 ~K_c=0.622 ~K_p=K_c x(RT)^Δ n Δ n=0 ~K_p=K_c=0.622 K_p for the decomposition of AB (~K^′ ) =frac1~K_p =1/0.622 ~K^′ =1.60

Q11 / 160 · AP EAMCET 2024 (22 May Shift 1) · Chemistry

At equilibrium for the reaction $\mathrm{A}_2(g)+\mathrm{B}_2(g) \rightleftharpoons 2 \mathrm{AB}(g)$ the concentrations of $\mathrm{A}_2, \mathrm{~B}_2$ and AB respectively are $1.5 \times 10^{-3} \mathrm{M}, 2.1 \times 10^{-3} \mathrm{M}$ and $1.4 \times 10^{-3} \mathrm{M}$ in a sealed vessel at 800 K . What will be $\mathrm{K}_{\mathrm{p}}$ for the decomposition of AB at same temperature?

A
0.62
Correct answer
B
1.6
C
0.44
D
2.27

Solution

$\begin{aligned} & \text { } \mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{~g}) \\ & \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{AB}]^2}{\left[\mathrm{~A}_2\right]\left[\mathrm{B}_2\right]} \\ & \mathrm{K}_{\mathrm{c}}=\frac{1.4 \times 10^{-3} \times 1.4 \times 10^{-3}}{1.5 \times 10^{-3} \times 2.1 \times 10^{-3}} \\ & \mathrm{~K}_{\mathrm{c}}=0.622 \\ & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \mathrm{x}(\mathrm{RT})^{\Delta \mathrm{n}} \\ & \Delta \mathrm{n}=0 \\ & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}=0.622 \end{aligned}$ $\begin{aligned} \mathrm{K}_{\mathrm{p}} \text { for the decomposition of } \mathrm{AB}\left(\mathrm{~K}^{\prime}\right) & =\frac{1}{\mathrm{~K}_{\mathrm{p}}} \\ & =\frac{1}{0.622} \\ \mathrm{~K}^{\prime} & =1.60 \end{aligned}$

Hence, the correct answer is option: 'B'.

Solution

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