Question

A first order reaction takes $40$ minutes for $30\%$ decomposition. Calculate ${t_{\dfrac{1}{2}}}$ value (log$7{\text{ }} = {\text{ }}0.845$)

Answer 1

We need to know that chemical kinetics is one of the topics used to study the kinetics nature of the chemical reaction. It is used to optimize the chemical reaction for industrial purposes. The reactant to the product so many parameters are required. All are optimised by using this chemical kinetics. Chemical kinetics is used as a mechanism of reactant to product in the chemical reaction.
Formula used:
The formula for first order reaction is
$t = \dfrac{{2.303}}{k}log\dfrac{{\left[ {{R_0}} \right]}}{{\left[ R \right]}}{\text{ }}$
k is the rate constant of the chemical reaction.
t is the time for the chemical reaction.
The initial concentration of the chemical reaction is${R_0}$.
The final concentration of the chemical reaction is$R$.
The time period calculation,
${{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}}}{{\text{k}}}$
Here, the time period is t.

Complete answer:
The given data is
A first order reaction takes $40$ minutes for $30\%$ decomposition. Calculate t1/2 value (log$7{\text{ }} = {\text{ }}0.845$)
The time for the chemical reaction t is $40$ minutes.
The initial concentration of the chemical reaction ${R_0}$ is $100\%$
The final concentration of the chemical reaction $R$ is $70\%$
The rate constant of the chemical reaction is calculated as,
$t = \dfrac{{2.303}}{k}log\dfrac{{\left[ {{R_0}} \right]}}{{\left[ R \right]}}$
We change the formula for our concern.
$k = \dfrac{{2.303}}{t}log\dfrac{{\left[ {{R_0}} \right]}}{{\left[ R \right]}}$
We substitute the known values in the formula,
$k = \dfrac{{2.303}}{{40}}log\dfrac{{100}}{{70}}$
$k = \dfrac{{2.303}}{{40}}log100 - log70$
$k = \dfrac{{2.303}}{{40}}(1 - 0.845)$
$k = 8.918 \times {10^{ - 3}}{\min ^{ - 1}}$
The rate constant of the chemical reaction is $8.918 \times {10^{ - 3}}{\min ^{ - 1}}$.
The half-life period for the chemical reaction is calculate as,
${{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}}}{{\text{k}}}$
$k = 8.918 \times {10^{ - 3}}{\min ^{ - 1}}$
${{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}}}{{8.918 \times {{10}^{ - 3}}{{\min }^{ - 1}}}}$
${{\text{t}}_{\dfrac{1}{2}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.6932}}\min }}{{8.918 \times {{10}^{ - 3}}}}$
On simplification we get,
${{\text{t}}_{\dfrac{1}{2}}}{\text{ = 77}}{\text{.7}}\min$
The half-life period for the chemical reaction is ${\text{77}}{\text{.7}}\min$.
According to the above discussion and calculation, we conclude a first order reaction takes $40$ minutes for $30\%$ decomposition and ${t_{1/2}}$value is ${\text{77}}{\text{.7}}\min$.

Note:
Rate of the reaction is an important factor for the study of reaction. The rate of reaction is an important concept for the chemical kinetics. Rate of the reaction depends on the concentration of the reactant. The rate of reaction is also calculated by using the concentration of the product in the chemical reaction. Depending on the concentration, the sign of the rate will change.
The rate of the reaction is directly proportional to the concentration of reactant and inversely proportional to the time of the reaction.
The rate of the reaction is equal to the product of the concentration of the reactant with respect to that order of the reaction.
$rate = k{[A]^m}{[B]^n}$
Here, k is proportionality constant, known as rate constant.
A and B are reactants of the reaction.
m and n are the order of the reaction of A and B respectively.