Chemical Kinetics Question 47
Question: In the formation of sulphur trioxide by the contact process, $ 2SO_2+O_2\rightarrow 2SO_3 $ , the rate of reaction was measured as $ \frac{d[O_2]}{dt}=3.0\times {10^{-4}}mol,{L^{-1}}{s^{-1}} $ . The rate of reaction expressed in terms of $ SO_3 $ will be
Options:
A) $ 3.0\times {10^{-4}}mol,{L^{-1}}{s^{-1}} $
B) $ 6.0\times {10^{-4}}mol,{L^{-1}}{s^{-1}} $
C) $ 1.5\times {10^{-4}}mol,{L^{-1}}{s^{-1}} $
D) $ 4.5\times {10^{-4}}mol,{L^{-1}}{s^{-1}} $
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Answer:
Correct Answer: B
Solution:
[b] $ -\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[SO_3]}{dt} $
$ 3\times {10^{-4}}=\frac{1}{2}\frac{d[SO_3]}{dt} $
$ \frac{d[SO_3]}{dt}=6\times {10^{-4}}M/sec. $
