Question

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to 4.0 min at 25°C. Calculate the order of the reaction and the rate constant.

Answer 1

2

`0.4167\ \text{M}^(-1)\text{min}^(-1)`

Step-by-step explanation:

Half-life

`{t_(1/2)}A=2\ \text{min}`

`{t_(1/2)}B=4\ \text{min}`

Concentration

`{[A]_0}_A=1.2\ \text{M}`

`{[A]_0}_B=0.6\ \text{M}`

We have the relation

`t_(1/2)\propto (1)/([A]_0^(n-1))`

So

`\frac{{t_(1/2)}_A}{{t_(1/2)}_B}=\left(\frac{{[A]_0}_B}{{[A]_0}_A}\right)^(n-1)\\\Rightarrow (2)/(4)=\left((0.6)/(1.2)\right)^(n-1)\\\Rightarrow (1)/(2)=\left((1)/(2)\right)^(n-1)`

Comparing the exponents we get

`1=n-1\\\Rightarrow n=2`

The order of the reaction is 2.

`t_(1/2)=(1)/(k[A]_0^(n-1))\\\Rightarrow k=(1)/(t_(1/2)[A]_0^(n-1))\\\Rightarrow k=(1)/(2* 1.2^(2-1))\\\Rightarrow k=0.4167\ \text{M}^(-1)\text{min}^(-1)`

The rate constant is
`0.4167\ \text{M}^(-1)\text{min}^(-1)`