Question

99% of a first order reaction was completed in 32 min. when will 99.9% of the reaction complete?
1) 50min.
2) 46 min
3) 49 min
4) 48 min

Answer 1

The reaction will complete in 46 min

Answer 2

Time required to complete 99.9% reaction is option 4) 48 min.

Rate of first order reaction is calculated as follows:

$K=\left(\frac{2.303}{t}\right) \log \left(\frac{A}{A}-X\right)$

Here, K is the rate constant, t is time, A is amount of product formed, (A - X) is amount of reactant left in the reaction after time t.

In the given problem,

When the reaction is 99 % completed

$\begin{array}{c}{A=100} \\ {A-X=100-99=1} \\ {t=32 \min =32 \times 60 \sec }\end{array}$

Now putting these values in rate of reaction of first order equation, we get:

$\begin{array}{l}{K=\left(\frac{2.303}{32} \times 60\right) \log \left(\frac{100}{1}\right)} \\ {=2.39 \times 10^{-3} \text { per second }}\end{array}$

In any reaction he rate constant or K remains same.

Therefore, when the reaction is 99.9% completed

$\begin{array}{c}{A=100} \\ {A-X=100-99.9=0.1}\end{array}$

Putting the values in the rate of reaction equation, we get:

$2.39 \times 10^{-3}=\left(\frac{2.303}{t}\right) \log \left(\frac{100}{0.1}\right)$

Therefore, t = 2890 sec  

$\begin{array}{l}{=\frac{2890}{60}=48.16\ \mathrm{min}} \\ {\cong 48\ \mathrm{min}}\end{array}$.