Question

1. the reaction a → products is first order in a. if 1.60 g of a is allowed to decompose for 45 min, the mass of a remaining undecomposed is found to be 0.40 g. what is the half-life, t₁/₂, of this reaction?

Explanation:

Step1: Recall first - order reaction formula

For a first - order reaction, the integrated rate law is $\ln\frac{[A]_0}{[A]_t}=kt$, where $[A]_0$ is the initial concentration (or in this case, since mass is proportional to concentration for the same substance, we can use mass directly), $[A]_t$ is the concentration (mass) at time $t$, $k$ is the rate constant, and $t$ is the time.

We know that $m_0 = 1.60\ g$, $m_t=0.40\ g$, and $t = 45\ min$.

First, calculate the ratio $\frac{m_0}{m_t}=\frac{1.60\ g}{0.40\ g}=4$.

Step2: Solve for the rate constant $k$

Substitute into the first - order rate law: $\ln(4)=k\times45\ min$.

We know that $\ln(4)=\ln(2^2) = 2\ln(2)\approx2\times0.693 = 1.386$.

So, $k=\frac{\ln(4)}{45\ min}=\frac{1.386}{45\ min}\approx0.0308\ min^{-1}$.

Step3: Recall the formula for half - life of a first - order reaction

The formula for the half - life $t_{1/2}$ of a first - order reaction is $t_{1/2}=\frac{\ln(2)}{k}$.

We know that $\ln(2)\approx0.693$.

Substitute $k = 0.0308\ min^{-1}$ into the formula: $t_{1/2}=\frac{0.693}{0.0308\ min^{-1}}\approx22.5\ min$.

We can also solve it in a more straightforward way. Since $\frac{m_0}{m_t}=4 = 2^2$, this means that the number of half - lives $n$ that have passed is 2 (because after 1 half - life, the mass is $\frac{m_0}{2}$, after 2 half - lives, the mass is $\frac{m_0}{2^2}=\frac{m_0}{4}$).

If $n = 2$ half - lives have passed in $t = 45\ min$, then $t_{1/2}=\frac{t}{n}=\frac{45\ min}{2}=22.5\ min$.

Answer:

The half - life of the reaction is $\boxed{22.5\ min}$