### Chemical Kinetics Question 5

#### Question 5 - 25 January - Shift 2

A first order reaction has the rate constant, $k=4.6 \times 10^{-3} s^{-1}$. The number of correct statement/s from the following is/are

Given : $\log 3=0.48$

A. Reaction completes in $1000 s$.

B. The reaction has a half-life of $500 s$.

C. The time required for $10 \%$ completion is 25 times the time required for $90 \%$ completion.

D. The degree of dissociation is equal to $(1-e^{-kt})$.

E. The rate and the rate constant have the same unit.

## Show Answer

#### Answer: (2)

#### Solution:

$t _{10 \%}=\frac{1}{K} \ln (\frac{a}{a-x})=\frac{1}{K} \ln (\frac{100}{90})$

$t _{10 \%}=\frac{2.303}{K}(\log 10-\log 9)$

$t _{10 \%}=\frac{2.093}{K} \times(0.04)$

Similarly

$ \begin{aligned} & t _{90 \%}=\frac{1}{K} \ln (\frac{100}{10}) \\ & t _{90 \%}=\frac{2.303}{K} \\ & \frac{t _{90 \%}}{t _{10 \%}}=\frac{1}{0.04}=25 \\ & e^{kt}=\frac{a}{a-x} \\ & \frac{a-x}{a}=e^{-kt} \\ & \frac{1-\frac{x}{a}}{ma}=e^{-kt} \\ & x=a(1-e^{-kt}) \text{ matt } \\ & \alpha=\frac{x}{a}=(1-e^{-kt}) \end{aligned} $