Chemical Kinetics Question 240
Question: For the reaction $ N_2+3H_2\to 2NH_3 $ if $ \frac{\Delta [NH_3]}{\Delta t}=2\times {10^{-4}}mol,{l^{-1}}{s^{-1}}, $ the value of $ \frac{-\Delta [H_2]}{\Delta t} $ would be [MP PMT 2000]
Options:
A) $ 1\times {10^{-4}}mol{l^{-1}}{s^{-1}} $
B) $ 3\times {10^{-4}}mol{l^{-1}}{s^{-1}} $
C) $ 4\times {10^{-4}}mol{l^{-1}}{s^{-1}} $
D) $ 6\times {10^{-4}}mol,{l^{-1}}{s^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ N_2+3H_2 $ ⇌ $ 2NH_3 $
$ \frac{-\Delta [N_2]}{\Delta t}=-\frac{1}{3}\frac{\Delta [H_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [NH_3]}{\Delta t} $
$ \therefore \frac{\Delta [H_2]}{\Delta t}=\frac{3}{2}\times \frac{\Delta [NH_3]}{\Delta t}=\frac{3}{2}\times 2\times {10^{-4}} $
$ =3\times {10^{-4}}mollitr{e^{-1}}se{c^{-1}} $
