JEE PYQ: Chemical Kinetics Question 37
Question 37 - 2019 (10 Jan Shift 2)
For an elementary chemical reaction, $\text{A}2 \underset{k{-1}}{\overset{k_1}{\rightleftharpoons}} 2\text{A}$,
the expression for $\dfrac{d[\text{A}]}{dt}$ is:
(1) $k_1[\text{A}2] - k{-1}[\text{A}]^2$
(2) $2k_1[\text{A}2] - 2k{-1}[\text{A}]^2$
(3) $k_1[\text{A}2] + k{-1}[\text{A}]^2$
(4) $2k_1[\text{A}2] - 2k{-1}[\text{A}]^2$
Show Answer
Answer: (4)
Solution
Given: $\text{A}2 \underset{k{-1}}{\overset{k_1}{\rightleftharpoons}} 2\text{A}$. Now, $\frac{-1}{1}\frac{d[\text{A}2]}{dt} = \frac{1}{2}\frac{d[\text{A}]}{dt}$. $\frac{d[\text{A}]}{dt} = -2k{-1}[\text{A}]^2 + 2k_1[\text{A}_2]$. So $\frac{d[\text{A}]}{dt} = 2k_1[\text{A}2] - 2k{-1}[\text{A}]^2$.
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