Chemical Kinetics Question 88
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}} $
BETA
