Q. For the reaction A(g) $\rightleftharpoons$ 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) $2.077\times10^5$
C) 0.033
D) 0.021
Explanation

[Solution by Claude - not from source] $K_c=\frac{k_f}{k_b}=\frac{1}{2500}=4\times10^{-4}$. $\Delta n=1$, so $K_P=K_c(RT)=4\times10^{-4}\times83.1=0.033$. Answer (3).

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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