Chemical Kinetics Question 278
Question: In a reaction, $ 2A\to $ products, the concentration of A decreases from 0.50 M to 0.38 M in 10 min. What is the rate of reaction (in $ M{s^{-1}} $ ) during this interval-
Options:
A) 0.012
B) 0.024
C) $ 2\times {10^{-3}} $
D) $ 2\times {10^{-4}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Rate of reaction $ =\frac{d[A]}{dt} $ Given, $ {{[A]}_{initial}}=0.50,M $
$ {{[A]}_{final}}=0.38,M $
$ dt=10min=600sec $
$ d[ A ]=0.12 $
$ Rate=\frac{0.12}{600}=2\times {10^{-4}}M,{s^{-1}} $
