Question

what is the half - life for a particular reaction if the rate law is rate = (1171 min^(-1))a?

Explanation:

Step1: Identify the reaction order

The rate - law is of the form rate = k[A], which indicates a first - order reaction. For a first - order reaction, the rate constant k is given as $k = 1171\ min^{-1}$.

Step2: Use the half - life formula for first - order reactions

The half - life formula for a first - order reaction is $t_{1/2}=\frac{\ln(2)}{k}$.
Substitute $k = 1171\ min^{-1}$ into the formula:
$t_{1/2}=\frac{\ln(2)}{1171\ min^{-1}}$.
We know that $\ln(2)\approx0.693$.
So, $t_{1/2}=\frac{0.693}{1171\ min^{-1}}\approx6\times10^{-4}\ min$.

Answer:

$6\times 10^{-4}\ min$