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}} $



NCERT Chapter Video Solution

Dual Pane