JEE PYQ: Chemical Kinetics Question 33
Question 33 - 2019 (12 Apr Shift 1)
In the following reaction: $x\text{A} \to y\text{B}$
$\log_{10}\left[-\dfrac{d[\text{A}]}{dt}\right] = \log_{10}\left[\dfrac{d[\text{B}]}{dt}\right] + 0.3010$
‘A’ and ‘B’ respectively can be:
(1) $n$-Butane and Iso-butane
(2) $\text{C}_2\text{H}_2$ and $\text{C}_6\text{H}_6$
(3) $\text{C}_2\text{H}_4$ and $\text{C}_4\text{H}_8$
(4) $\text{N}_2\text{O}_4$ and $\text{NO}_2$
Show Answer
Answer: (3)
Solution
$x\text{A} \to y\text{B}$. $\frac{-dA}{x,dt} = \frac{1}{y}\frac{dB}{dt}$. $\log\left[-\frac{dA}{dt}\right] = \log\left[\frac{dB}{dt}\right] + \log\left(\frac{x}{y}\right)$. Comparing with the given equation: $\log\frac{x}{y} = 0.3010 = \log 2$. So $\frac{x}{y} = 2$. The reaction is of type $2\text{A} \to \text{B}$. Hence, option (3) is correct: $2\text{C}_2\text{H}_4 \to \text{C}_4\text{H}_8$.
BETA
