Chemical Kinetics Question 295

Question: The differential rate law for the reaction $ H_2(g)+I_2(g)\to 2HI(g) $ is

Options:

A) $ -\frac{d[H_2]}{dt}=-\frac{d[I_2]}{dt}=-\frac{d[HI]}{dt} $

B) $ \frac{d[H_2]}{dt}=\frac{d[I_2]}{dt}=\frac{1}{2}\frac{d[HI]}{dt} $

C) $ \frac{1}{2}\frac{d[H_2]}{dt}=\frac{1}{2}\frac{d[I_2]}{dt}=-\frac{d[HI]}{dt} $

D) $ -2\frac{d[H_2]}{dt}=-2\frac{d[I_2]}{dt}=\frac{d[HI]}{dt} $

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Answer:

Correct Answer: D

Solution:

[d] rate of appearance of $ HI=\frac{1}{2}\frac{d[HI]}{dt} $

rate of disappearance of $ H_2=\frac{-d[H_2]}{dt} $

rate of disappearance of $ I_2=\frac{-d[I_2]}{dt} $

hence $ \frac{-d[H_2]}{dt}=-\frac{d[I_2]}{dt}=\frac{1}{2}\frac{d[HI]}{dt} $ or $ -\frac{2d[H_2]}{dt}=-\frac{2d[I_2]}{dt}=\frac{d[HI]}{dt} $



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