Question

In the reaction, a → products, a plot of ln[a] vs. Time is linear and the slope is equal to -0.860 s-1. How long will it take one-half of the initial amount of a to react?

Answer 1

The reaction is given as follows:

$a\rightarrow product$

The plot of ln[a] verses t is linear with slope $-0.860 s^{-1}$.

The plot for ln[a] verses t is linear for first order reaction. The integrated rate law equation is as follows:

$[A]=[A_{0}]e^{-kt}$

Here, k is rate constant, t is time of the reaction, $[A]$ is concentration of reactant at time t and $[A_{0}]$ is initial concentration of reactant.

Taking ln both sides,

$ln[A]=ln[A_{0}]-kt$...... (1)

Comparing with equation for linear graph, y=mx+c

Thus, slope =-k

Or, $k=0.860 s^{-1}$

Now, if final concentration become half of its initial value,$[A]=\frac{[A_{0}]}{2}$ and t becomes $t_{1/2}$

Putting the values in equation (1)

$ln\frac{[A_{0}]}{2}=ln[A_{0}]-(0.860 s^{-1})t_{1/2}$

On rearranging,

$t_{1/2}=\frac{ln2}{0.860 s^{-1}}=\frac{0.6932}{0.860 s^{-1}}=0.806 sec$

Therefore, it takes 0.806 sec to react half of the reactant a.