Question

In the first order reaction A → products, the initial concentration of A is 0.1 108M, and 44 s later, 0.0554M. What is the initial rate of this reaction? (Initial rate-k[A]). (t %-0.693/k)

Answer 1

Answer : The initial rate of the reaction is, $1.739\times 10^{-3}s^{-1}$

Explanation :

First we have to calculate the rate constant of the reaction.

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant  = ?

t = time taken for the process  = 44 s

$[A_o]$ = initial amount or concentration of the reactant  = 0.1108 M

$[A]$ = amount or concentration left time 44 s = 0.0554 M

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

$k=\frac{2.303}{44}\log\frac{0.1108}{0.0554}$

$k=0.0157$

Now we have to calculate the initial rate of the reaction.

Initial rate = K [A]

At t = 0, $[A]=[A_o]$

Initial rate = 0.0157 × 0.1108 = $1.739\times 10^{-3}s^{-1}$

Therefore, the initial rate of the reaction is, $1.739\times 10^{-3}s^{-1}$