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

Explanation:

The integrated first law is given by :

$[A]=[A]_o\times e^{-k\times t}$

Where:

$[A]_o$ = initial concentration of reactant

[A] = concentration of reactant after t time

k = rate constant

a)

$[A_o]=x$

$[A]=\frac{x}{2}$

t = 1000 s

$\frac{x}{2}=x\times e^{-k\times 1000 s}$

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

$0.53 mol/L=0.67 mol/L\times e^{-k\times 25s}$

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:

$\frac{2}{1}\times 0.034 mol= 0.068 mol$ 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

$0.085mol/L=0.53 mol/L\times e^{-k\times 115 s}$

Solving for k:

$k=0.01592 s^{-1}$

The rate constant for this reaction is $0.01592 s^{-1}$.