Chemical Kinetics Question 133
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} $
BETA
