Chemical Kinetics Question 241
Question: A gaseous hypothetical chemical equation $ 2A $ ⇌ $ 4B+C $ is carried out in a closed vessel. The concentration of B is found to increase by $ 5\times {10^{-3}}mol{l^{-1}} $ in 10 second. The rate of appearance of B is [AFMC 2001]
Options:
A) $ 5\times {10^{-4}}mol{l^{-1}}se{c^{-1}} $
B) $ 5\times {10^{-5}}mol{l^{-1}}se{c^{-1}} $
C) $ 6\times {10^{-5}}mol{l^{-1}}{{\sec }^{-1}} $
D) $ 4\times {10^{-4}}mol,{l^{-1}}{{\sec }^{-1}} $
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Answer:
Correct Answer: A
Solution:
Increase in concentration of $ B=5\times {10^{-3}}mol{l^{-1}} $
$ Time=10\sec $ Rate of appearance of $ B= $
$ \frac{Increase,of,conc\text{. B}}{Timetaken} $
$ =\frac{5\times {10^{-3}}mol{l^{-1}}}{10\sec }=5\times {10^{-4}}mol{l^{-1}}Se{c^{-1}} $
