NEET 2025, Chemistry Q28: Law of Chemical Equilibrium And Equilibrium Constant

NEET2025Chemistry

For the reaction A(g) $⇌$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of $2500$ , at $1000$ K.

[Given: $R = 0.0831$ L atm mol $^{- 1}$ K $^{- 1}$ ]

$K_{p}$ for the reaction at $1000$ K is

A
83.1
B
$2.077 \times 1 0^{5}$
C
0.033
D
0.021

Solution

$K_{C} = \frac{k _{f}}{k _{b}} = \frac{k _{f}}{2500 k _{f}} = \frac{1}{2500}$ .

$Δ n_{g} = 2 - 1 = 1$ , so $K_{p} = K_{C} (R T)^{Δ n_{g}} = \frac{1}{2500} \times (0.0831 \times 1000)^{1} = \frac{83.1}{2500} \approx 0.033$ .

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NEET 2025, Chemistry Q28: Law of Chemical Equilibrium And Equilibrium Constant

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