Question

The decomposition of A to B is a first-order reaction with a half-life of 14.2 min: A → 2B If the initial concentration of A is 0.304 M, how long will it take for the concentration of A to decrease by 43.0 %?

Answer 1

Answer:

It take 16.9 min for the concentration of A to decrease by 43.0 %

Explanation:

The first-order reaction law is $Ln [A] = -k.t + Ln [A]_{0}$

And the half-life time for the first-order reaction is $t_{1/2} =\frac{Ln 2}{k}$

- [A] is the concentration of the reactant at any time (t) of the reaction

- [A]0 is the concentration of the reactant at the beginning of the reaction

- k is the rate constant.  

- t1/2 is the half-life time

Thus, for this reaction k is

$k = \frac{Ln 2}{t_{1/2}}  = \frac{Ln 2}{14.2min} = 0.05 min^{-1}$

Decreasing the concentration of the reactant by 43.0% means the concentration of A (reactant) at the end of the reaction has to be:  

$[A] = \frac{43.0}{100}x[A]_{0}$

Replacing [A] in the equation of the law  

$Ln \frac{43.0}{100} [A]_{0} = -k.t + Ln [A]_{0}$

Clearing the t  

$(Ln \frac{43.0}{100} [A]_{0} - Ln [A]_{0} ) / -k = t$

Replaicing with [A]0 = 0.304M and k = 0.05min-1

$t = (Ln \frac{43.0}{100}x0.304M - Ln 0.304M ) / -0.05 min^{-1} = 16.9 min$