The reaction L rightarrow M starts with 10 , | vedclass

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log(\frac{[A]_0}{[A]_t})$.At $t_1 = 30 \, min$,$[A]_0 = 10 \, g

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(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log(\frac{[A]_0}{[A]_t})$.At $t_1 = 30 \, min$,$[A]_0 = 10 \, gL^{-1}$ and $[A]_t = 5 \, gL^{-1}$.$k = \frac{2.303}{30} \log(\frac{10}{5}) = \frac{2.303}{30} \log(2)$.At $t_2 = 90 \, min$,$[A]_0 = 10 \, gL^{-1}$ and $[A]_t = 1.25 \, gL^{-1}$.$k = \frac{2.303}{90} \log(\frac{10}{1.25}) = \frac{2.303}{90} \log(8) = \frac{2.303}{90} \log(2^3) = \frac{2.303 	imes 3}{90} \log(2) = \frac{2.303}{30} \log(2)$.Since the rate constant $k$ is the same in both cases,the reaction is of the first order.

The reaction $L \rightarrow M$ starts with $10 \, gL^{-1}$. After $30$ and $90$ minutes,$5 \, gL^{-1}$ and $1.25 \, gL^{-1}$ remain,respectively. The order of the reaction is:

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