Question

When the concentration of A in a reaction was changed from 1.60M to 0.80 M , the half life increased from 2.0 min to 4.0 min at 25C. Calculate the the order of the reaction and the rate constant .

Answer 1

The order of reaction is, 2 (second order reaction).

The value of rate constant is,
`0.31M^(-1)min^(-1)`

Explanation :

Half life : It is defined as the time in which the concentration of a reactant is reduced to half of its original value.

The general expression of half-life for nth order is:

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

or,

`(t_(1/2)_1)/(t_(1/2)_2)=([A_2]^(n-1))/([A_1]^(n-1))`

or,

`n=\left((\log((t_(1/2))_1)/((t_(1/2))_2))/(\log((A)_2)/((A)_1))\right )+1` .............(1)

where,

`t_(1/2)` = half-life of the reaction

n = order of reaction

[A] = concentration

As we are given:

Initial concentration of A = 1.60 M

Final concentration of A = 0.80 M

Initial half-life of the reaction = 2.0 min

Final half-life of the reaction = 4.0 min

Now put all the given values in the above formula 1, we get:

`n=\left ((\log (2.0)/(4.0))/(\log(0.80)/(1.60))\right )+1`

`n=2`

Thus, the order of reaction is, 2 (second order reaction).

Now we have to determine the rate constant.

To calculate the rate constant for second order the expression will be:

`t_(1/2)=(1)/(k* [A_o])`

When,

`t_(1/2)` = 2.0 min

`[A_o]` = 1.60 M

`2.0min=(1)/(k* 1.60M)`

`k=0.31M^(-1)min^(-1)`

Thus, the value of rate constant is,
`0.31M^(-1)min^(-1)`