Question

Determine the rate constant for each of the following fi rst-order reactions, in each case expressed for the rate of loss of A: (a) A S B, given that the concentration of A decreases to one-half its initial value in 1000. s; (b) A S B, given that the concentration of A decreases from 0.67 molL1 to 0.53 molL1 in 25 s; (c) 2 A S B C, given that [A]0 0.153 molL1 and that after 115 s the concentration of B rises to 0.034 molL1

Answer 1

Step-by-step explanation:

The integrated first law is given by :

`[A]=[A]_o* e^(-k* t)`

Where:

`[A]_o` = initial concentration of reactant

[A] = concentration of reactant after t time

k = rate constant

a)

`[A_o]=x`

`[A]=(x)/(2)`

t = 1000 s

`(x)/(2)=x* e^(-k* 1000 s)`

Solving for k:

`k=0.06934 s^(-1)`

The rate constant for this reaction is
`0.06934 s^(-1)`.

b)

`[A_o]=0.67 mol/L`

`[A]=0.53 mol/L`

t = 25 s

`0.53 mol/L=0.67 mol/L* e^(-k* 25s)`

Solving for k:

`k=0.009376 s^(-1)`

The rate constant for this reaction is
`0.009376 s^(-1)`.

c) 2 A → B +C

`[A_o]=0.153 mol/L`

`[A]=?`

`[B]=0.034 mol/L`

According to reaction, 1 mole of B is obtained from 2 moles of A.

Then 0.034 mole of B will be obtained from:

`(2)/(1)* 0.034 mol= 0.068 mol` of A

So, the concentration left after 115 seconds:

`[A]=0.153 mol/L-0.068 mol/L=0.085 mol/L`

t = 115 s

`0.085mol/L=0.53 mol/L* e^(-k* 115 s)`

Solving for k:

`k=0.01592 s^(-1)`

The rate constant for this reaction is
`0.01592 s^(-1)`.