Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
For the reaction: H2 +I2 -> 2HI , the differential rate law is
Question Error Report
Question is incomplete/wrong
Question not belongs to this Chapter
Answer is wrong
Solution is wrong
Answer & Solution is not matching
Spelling mistake
Image missing
Website not working properly
Other (not listed above)
Error description
Thank you for reporting, we will resolve it shortly
Back to Question
Thank you for reporting, we will resolve it shortly
Q. For the reaction : $ \ce{H_2 +I_2 -> 2HI} $ , the differential rate law is
MHT CET
MHT CET 2008
A
$ -\frac{d\left[H_{2}\right]}{dt}=-\frac{d\left[I_{2}\right]}{dt}=2 \frac{d\left[HI\right]}{dt} $
B
$ -\frac{d\left[H_{2}\right]}{dt}=-2\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt} $
C
$ -\frac{d\left[H_{2}\right]}{dt}=-\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt} $
D
$ -\frac{d\left[H_{2}\right]}{dt}=\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt} $
Solution:
$H _{2}+ I _{2} \longrightarrow 2 HI$
Rate of reaction $=\frac{-d\left[ H _{2}\right]}{d t}$
$=\frac{-d\left[ I _{2}\right]}{d t}= \frac{1}{2} \frac{d[ HI ]}{d t}$
or $\frac{-2 d\left[ H _{2}\right]}{d t}=\frac{-2 d\left[ I _{2}\right]}{d t}$
$=\frac{d[ HI ]}{d t}$