Question

Consider the following initial rate data (at 273 K) for the decomposition of a substrate (substrate 1) which decomposes to product 1 and product 2: [Substrate 1] (M) Initial Rate (M/s) 0.4 0.183 0.8 0.183 2 0.183 Determine the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M.

Answer 1

Answer:

15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.

Explanation:

A → B + C

The rate law of the reaction will be :

$R=k[A]^x$

Initial rate of the reaction when concentration of the substrate was 0.4 M:

$0.183 M/s=k[0.4 M]^x$..[1]

Initial rate of the reaction when concentration of the substrate was 0.8 M:

$0.183 M/s=k[0.8 M]^x$...[2]

[1] ÷ [2] :

$\frac{0.183 M/s}{0.183 M/s}=\frac{k[0.4 M]^x}{k[0.8 M]^x}$

x = 0

The order of the reaction is zero.

For the value of rate constant ,k:

$0.183 M/s=k[0.4 M]^x$..[1]

x = 0

$0.183 M/s=k[0.4 M]^0$

k= 0.183 M/s

The half life of the zero order kinetics is given by :

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

Where:

$[A_o]$ = Initial concentration of A

k = Rate constant of the reaction

So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:

$t_{1/2}=\frac{2.77 M}{0.183 M/s}=15.1 s$

15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.