Chemical Kinetics Question 259

Question: For the reaction

$ N_2O_5(g)\xrightarrow{{}}2NO_2(g)+1/2,O_2(g) $ the value of rate of disappearance of $ N_2O_5 $ is given as $ 6.25\times {10^{-3}}mol,{L^{-1}}{s^{-1}} $ . The rate of formation of $ NO_2 $ and $ O_2 $ is given respectively as:

Options:

A) $ 6.25\times {10^{-3}}mol{L^{-1}}{s^{-1}} $ and $ 6.25\times {10^{-3}}mol{L^{-1}}{s^{-1}} $

B) $ 1.25\times {10^{-2}}mol{L^{-1}}{s^{-1}} $ and $ 3.125\times {10^{-3}}mol{L^{-1}}{s^{-1}} $

C) $ 6.25\times {10^{-3}}mol{L^{-1}}{s^{-1}} $ and $ 3.125\times {10^{-3}}mol{L^{-1}}{s^{-1}} $

D) $ 1.125\times {10^{-2}}mol{L^{-1}}{s^{-1}} $ and $ 6.25\times {10^{-3}}mol{L^{-1}}{s^{-1}} $

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Answer:

Correct Answer: B

Solution:

[b] $ N_2O_5( g )\xrightarrow{{}}2NO_2( g )+1/2O_2( g ) $

$ -\frac{d}{dt}[N_2O_5]=+\frac{1}{2}\frac{d}{dt}[NO_2]=2\frac{d}{dt}[O_2] $

$ \frac{d}{dt}[NO_2]=1.25\times {10^{-2}}mol{L^{-1}}{s^{-1}} $ and $ \frac{d}{dt}[O_2]=3.125\times {10^{-3}}mol,{L^{-1}}{s^{-1}} $



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