if a reaction is first order with a rate constant of 0.0450 s⁻¹, how much time is required for 45% of the…

if a reaction is first order with a rate constant of 0.0450 s⁻¹, how much time is required for 45% of the initial quantity of reactant to be consumed?

if a reaction is first order with a rate constant of 0.0450 s⁻¹, how much time is required for 45% of the initial quantity of reactant to be consumed?

Answer

Explanation:

Step1: Define the first - order reaction formula

The integrated rate law for a first - order reaction is $\ln\left(\frac{[A]_t}{[A]_0}\right)=-kt$, where $[A]_t$ is the concentration of the reactant at time $t$, $[A]_0$ is the initial concentration of the reactant, $k$ is the rate constant, and $t$ is the time. If 45% of the reactant is consumed, then the fraction of the reactant remaining is $[A]_t/[A]_0=(1 - 0.45)=0.55$.

Step2: Rearrange the formula to solve for $t$

We can rewrite the first - order rate law as $t =-\frac{\ln\left(\frac{[A]_t}{[A]_0}\right)}{k}$.

Step3: Substitute the values

Given $k = 0.0450\ s^{-1}$ and $\frac{[A]_t}{[A]_0}=0.55$, we substitute into the formula: $t=-\frac{\ln(0.55)}{0.0450\ s^{-1}}$. We know that $\ln(0.55)\approx - 0.5978$. Then $t=\frac{0.5978}{0.0450\ s^{-1}}\approx13.3\ s$.

Answer:

$13.3\ s$