Question

What is the half-life of a first-order reaction if it takes 4.4 x 102 seconds for the concentration to decrease from 0.50 M to 0.20 M? 2.5 x 102 s 3.3 x 102s 1.6s 21 s 27 s

Answer 1

Answer: The half-life of a first-order reaction is, $3.3\times 10^2s$

Explanation:

All the radioactive reactions follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,  

k = rate constant = ?

t = time taken = 440 s

$[A_o]$ = initial amount of the reactant = 0.50 M

[A] = left amount =  0.20 M

Putting values in above equation, we get:

$k=\frac{2.303}{440s}\log\frac{0.50}{0.20}$

$k=2.083\times 10^{-3}s^{-1}$

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

Putting values in this equation, we get:

$t_{1/2}=\frac{0.693}{2.083\times 10^{-3}s^{-1}}=332.69s=3.3\times 10^2s$

Therefore, the half-life of a first-order reaction is, $3.3\times 10^2s$