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(B) For a reaction of order $n$,the half-life $t_{1/2}$ is related to the initial concentration $[A]_0$ as $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$.Giv

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proportional to the second power of concentration

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(B) For a reaction of order $n$,the half-life $t_{1/2}$ is related to the initial concentration $[A]_0$ as $t_{1/2} \propto \frac{1}{[A]_0^{n-1}}$.Given that the concentration decreases from $0.1 \, M$ to $0.01 \, M$,the concentration becomes $\frac{1}{10}$ of the initial value.The half-life increases by $10$ times,so $10 = (\frac{1}{10})^{-(n-1)}$.$10^1 = 10^{n-1}$.Comparing the exponents,$1 = n - 1$,which gives $n = 2$.Thus,the rate of reaction is proportional to the second power of the concentration.

In the reaction $A + B \to \text{Products}$,the initial concentration of both $A$ and $B$ is $0.1 \, M$. When the concentration decreases to $1.0 \times 10^{-2} \, M$,the half-life period increases by ten times. The rate of the reaction is:

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