Question

Which of the following values would you expect for the ratio of half-lives for a reaction with starting concentrations of 0.05M and 0.01M, t1/2(0.05M) / t1/2 (0.01M), if a reaction is known to be zero order?

Answer 1

Answer:

The expected ratio of half-lives for a reaction will be 5:1.

Explanation:

Integrated rate law  for zero order kinetics is given as:

$k=\frac{1}{t}([A_o]-[A])$

$[A_o]$ = initial concentration

[A]=concentration at time t

k = rate constant

if, $[A]=\frac{1}{2}[A_o]$

$t=t_{\frac{1}{2}}$, the equation (1) becomes:

$t_{\frac{1}{2}}=\frac{[A_o]}{2k}$

Half life when concentration was 0.05 M=$t_{\frac{1}{2}}$

Half life when concentration was 0.01 M=$t_{\frac{1}{2}}'$

Ratio of half-lives will be:

$\frac{t_{\frac{1}{2}}}{t_{\frac{1}{2}}'}=\frac{\frac{[0.05 M]}{2k}}{\frac{[0.01 M]}{2k}}=\frac{5}{1}$

The expected ratio of half-lives for a reaction will be 5:1.