Chemical Kinetics Question 290
Question: In the reaction of formation of sulphur trioxide by contact process $ 2SO_2+O_2\rightarrow 2SO_3 $ the rate of reaction was measured as $ \frac{d[O_2]}{dt}=-2.5\times {10^{-4}}mol{L^{-1}}{s^{-1}} $ . The rate of reaction is terms of $ [SO_2] $ in $ mol{L^{-1}}{s^{-1}} $ will be:
Options:
A) $ -1.25\times {10^{-4}} $
B) $ -2.50\times {10^{-4}} $
C) $ -3.75\times {10^{-4}} $
D) $ -5.00\times {10^{-4}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] From rate law $ -\frac{1}{2}\frac{dSO_2}{dt}=-\frac{dO_2}{dt}=\frac{1}{2}\frac{dSO_3}{dt} $
$ \therefore -\frac{dSO_2}{dt}=-2\times \frac{dO_2}{dt} $
$ =-2\times 2.5\times {10^{-4}} $
$ =-5\times {10^{-4}}mol,{L^{-1}}{s^{-1}} $
