JEE MAIN - Chemistry (2024 - 6th April Evening Shift - No. 25)
Consider the two different first order reactions given below
$$\begin{aligned} & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} \text { (Reaction 1) } \\ & \mathrm{P} \rightarrow \mathrm{Q} \text { (Reaction 2) } \end{aligned}$$
The ratio of the half life of Reaction 1 : Reaction 2 is $$5: 2$$ If $$t_1$$ and $$t_2$$ represent the time taken to complete $$2 / 3^{\text {rd }}$$ and $$4 / 5^{\text {th }}$$ of Reaction 1 and Reaction 2 , respectively, then the value of the ratio $$t_1: t_2$$ is _________ $$\times 10^{-1}$$ (nearest integer). [Given : $$\log _{10}(3)=0.477$$ and $$\log _{10}(5)=0.699$$]
Explanation
$$\mathrm{A+B \rightarrow C}$$ Reaction .... (1)
$$\mathrm{P} \rightarrow \mathrm{Q} \quad$$ Reaction .... (2)
$$\frac{\left(t_{\frac{1}{2}}\right)_1}{\left(t_{\frac{1}{2}}\right)_2}=\frac{k}{k_1}= \frac{5}{2}$$
$$\begin{aligned} & \frac{\frac{t_2}{3}}{t_{\frac{4}{5}}^5}=\frac{k_2}{k_1} \frac{\log 3}{\log 5} \\ & =\left(\frac{5}{2}\right) \frac{0.477}{0.699}=1.7 \\ & =17 \times 10^{-1} \\ & =x=17 \end{aligned}$$
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