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There are two competing first order reactions:
\begin{aligned} & \mathrm{A \stackrel{k_1}{\longrightarrow} P+Q} \\ & \mathrm{B \stackrel{k_2}{\longrightarrow} R+S} \end{aligned}

If half of the reaction of A was completed when 84 % of the reaction of B was completed, the ratio of their rate constants \mathrm{\left(\frac{k_2}{k_1}\right)} :
 

Option: 1

2.64
 


Option: 2

1.32
 


Option: 3

3.12
 


Option: 4

2.56
 


Answers (1)

best_answer

\begin{aligned} \mathrm{t=\frac{1}{k_1} \ln \left(\frac{100}{100-50}\right)} &\mathrm{ =\frac{1}{k_2} \ln \left(\frac{100}{100-84}\right) }\\ \mathrm{\frac{\ln 2}{k_1}} &\mathrm{ =\frac{\ln (100 / 16)}{k_2} }\\ \mathrm{\frac{k_2}{k_1} }&\mathrm{ =\frac{2.303(\log 100-\log 16)}{2.303 \log 2} }\\ \mathrm{\frac{k_2}{k_1}} &\mathrm{ =\frac{2-4 \times 0.3010}{0.3010}=2.64}\\ \end{aligned}

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