Question

For a reaction rate constant is $0.693{s^{ - 1}}$ at temperature ${27^ \circ }C$ . If the activation energy of this reaction is $2.303 \times 600R$ , the calculate half life (approx.) of this reaction at temperature ${327^ \circ }C$ ( Assume no change in concentration due to change in temperature).

Answer 1

Hint:
In order to calculate the half life of rate constant at temperature ${T_2}$ we will first calculate the rate constant at temperature ${T_2}$ using Arrhenius equation. The Arrhenius equation is used to calculate the rate constant. It is helpful in understanding the effect of temperature on the rate of reaction.

Complete step by step answer:
Arrhenius equation is given as follows:
$k = A{e^{\dfrac{{ - Ea}}{{RT}}}}$
Where,$k = $ rate constant
$A = $ Frequency factor
${E_a} = $ activation energy
$R = $ universal gas constant
$T = $ Temperature
The alternative form of Arrhenius equation at two different temperatures is as follows:
$\ln \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = - \left( {\dfrac{{{E_a}}}{R}} \right)\left( {\dfrac{1}{{{T_2}}}} \right) - \left( {\dfrac{1}{{{T_1}}}} \right)$
Where, ${k_1} = $rate constant at the given temperature
${k_2} = $rate constant when the temperature changes
$R = $universal gas constant
${T_1} = $temperature
${T_2} = $increase in temperature.
${E_a} = $activation energy
Using this formula we will find the rate constant at temperature ${T_2}$.
Given:
${k_1} = 0.693{s^{ - 1}}$
${E_a} = 2.303 \times 600R$
${T_1} = {27^ \circ }C$
$\therefore {T_1} = 273 + 27$
$\therefore {T_1} = 300K$
${T_2} = {327^ \circ }C$
$\therefore {T_2} = 273 + 327$
$\therefore {T_2} = 600K$
To find: ${k_2} = ?$
Formula to be used: $\ln \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = - \left( {\dfrac{{{E_a}}}{R}} \right)\left( {\dfrac{1}{{{T_2}}}} \right) - \left( {\dfrac{1}{{{T_1}}}} \right)$
Soln:
$\ln \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = - \left( {\dfrac{{{E_a}}}{R}} \right)\left( {\dfrac{1}{{{T_2}}}} \right) - \left( {\dfrac{1}{{{T_1}}}} \right)$
Substituting the values we get,
$ \Rightarrow \ln \left( {\dfrac{{{k_2}}}{{0.693}}} \right) = - \left( {\dfrac{{2.303 \times 600R}}{R}} \right)\left( {\dfrac{1}{{600}}} \right) - \left( {\dfrac{1}{{300}}} \right)$

$ \Rightarrow \ln \left( {\dfrac{{{k_2}}}{{0.693}}} \right) = - 1381.8 \times \dfrac{{1 - 2}}{{600}}$
On further solving, we get
$ \Rightarrow \ln \left( {\dfrac{{{k_2}}}{{0.693}}} \right) = - 1381.8 \times \dfrac{{ - 1}}{{600}}$
$ \Rightarrow \ln \left( {\dfrac{{{k_2}}}{{0.693}}} \right) = 2.303$
on converting natural log to log to the base 10
$ \Rightarrow 2.303\log \left( {\dfrac{{{k_2}}}{{0.693}}} \right) = 2.303$
$ \Rightarrow \log \left( {\dfrac{{{k_2}}}{{0.693}}} \right) = 1$
$ \Rightarrow \left( {\dfrac{{{k_2}}}{{0.693}}} \right) = 10$
On further solving, we get the value
$ \Rightarrow {k_2} = 10 \times 0.693$
Now we will calculate the half life when the temperature is $600K$ .
Half life is defined as the amount of time required for a specific constant to decrease by half compared to its initial concentration.
For zero order reaction, half life is given by the formula : ${t_{\dfrac{1}{2}}} = \dfrac{{\left[ {{R_0}} \right]}}{{2k}}$
For first order reaction, half life is given by the formula: ${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$
For second order reaction, half life is given by the formula: \[{t_{\dfrac{1}{2}}} = \dfrac{1}{{\left[ {{R_0}} \right]k}}\]
Where, \[\left[ {{R_0}} \right] = \]initial reactant concentration, $k = $ rate constant, ${t_{\dfrac{1}{2}}} = $ half life of the reaction.
Since the concentration does not change at temperature ${T_2}$ , we will assume the reaction to be a first order reaction.
Using the half life formula for the first order reaction we will calculate the half life of the reaction.
Soln:
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$
Substituting the value of ${k_2}$ in the above formula we get,
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{10 \times 0.693}}$
${t_{\dfrac{1}{2}}} = \dfrac{1}{{10}}$
${t_{\dfrac{1}{2}}} = 0.1\sec $
Therefore the half life of the reaction at temperature ${327^ \circ }C$ is $0.1\sec $ .

Note:The rate of reaction increases with an increase in temperature whereas use of catalyst increases the rate of reaction and also alters the rate of reaction. The half life concept is highly used in the administration of drugs into the target.