Chemical Kinetics Question 260

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}} $

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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}} $



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