Chemical Kinetics Question 91
Question: The reactions rate $ N_2( g )+3H_2( g )\to 2NH_3( g ) $ was measured $ \frac{d[NH_3]}{dt}=2\times {10^{-4}}mol,{{\sec }^{-1}} $ . The rates of reactions expressed in terms of $ N_2 $ and $ H_2 $ are
Options:
A) Rate in terms of $ N_2 $
$ (mol,{L^{-1}}se{c^{-1}}) $ Rate in terms of $ H_2 $
$ (mol,{L^{-1}}se{c^{-1}}) $ $ 2\times {10^{-4}} $
$ 2\times {10^{-4}} $
B) Rate in terms of $ N_2 $
$ (mol,{L^{-1}}se{c^{-1}}) $ Rate in terms of $ H_2 $
$ (mol,{L^{-1}}se{c^{-1}}) $ $ 3\times {10^{-4}} $
$ 1\times {10^{-4}} $
C) Rate in terms of $ N_2 $
$ (mol,{L^{-1}}se{c^{-1}}) $ Rate in terms of $ H_2 $
$ (mol,{L^{-1}}se{c^{-1}}) $ $ 1\times {10^{-4}} $
$ 3\times {10^{-4}} $
D) Rate in terms of $ N_2 $
$ (mol,{L^{-1}}se{c^{-1}}) $ Rate in terms of $ H_2 $
$ (mol,{L^{-1}}se{c^{-1}}) $ $ 2\times {10^{-1}} $
$ 2\times {10^{-3}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ N_2+3H_2\rightarrow 2NH_3 $ ; Rate is given by any of the expressions $ -\frac{d[ N_2 ]}{dt}=-\frac{1d[ H_2 ]}{3dt}\text{=}\frac{1}{2}\frac{d( NH_3 ]}{dt} $ Rate of disappearance of $ N_2=\frac{1}{2} $ the rate of formation of $ NH_3=1\times {10^{-4}} $ Rate of disappearance of $ H_2=3/2 $ the rate of formation of $ NH_3=3\times {10^{-4}} $
BETA
