Chemical Kinetics Question 54
Question: In a catalytic conversion of $ N_2 $ to $ NH_3 $ by Haber’s process, the rate of reaction was expressed as change in the concentration of ammonia per time is $ 40\times {10^{-3}},mol,litr{e^{-1}}{s^{-1}} $ . If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is (in mol $ litr{e^{-1}}{s^{-1}} $ )
Options:
A) $ 60\times {10^{-3}} $
B) $ 20\times {10^{-3}} $
C) 1.200
D) $ 10.3\times {10^{-3}} $
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Answer:
Correct Answer: A
Solution:
$ \frac{-,d(N_2)}{dt}=-\frac{1}{3}\frac{d(H_2)}{dt}=\frac{1}{2}\frac{d(NH_3)}{dt} $ = $ \frac{3}{2}\times 40\times {10^{-3}} $
$ =60\times {10^{-3}}. $
BETA
