Question

The time required for $10\% $ completion of a first order reaction at $298{\text{ K}}$ is equal to that required for its $25\% $ completion at $308{\text{ K}}$. If the pre-exponential factor for the reaction is $3.56 \times {10^9}{\text{ se}}{{\text{c}}^{ - 1}}$, calculate its rate constant at $318{\text{ K}}$ and also the energy of activation.

Answer 1

Hint: To solve this question we must know the equation for the rate constant of first order reaction. Using the equation calculate the rate constant at $10\% $ completion and $25\% $ completion. The minimum amount of energy that the reacting species must possess to undergo a specific reaction is known as the activation energy. The relation between the temperature, rate constant and activation energy is given by the Arrhenius equation.

Formulae Used:
1. $k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}$
2. $\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{{2.303 \times R}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
3. $\log k = \log A - \dfrac{{{E_a}}}{{2.303 \times R \times T}}$

Complete step by step solution:
We know the equation for the rate constant of a first order reaction is,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}$
where $k$ is the rate constant of a first order reaction,
             $t$ is time,
             ${\left[ a \right]^0}$ is the initial concentration of the reactant,
             $\left[ a \right]$ is the final concentration of the reactant.
The reaction is $10\% $ complete. Thus, the initial concentration is ${\text{100}}$ and the final concentration is $100 - 10 = 90$. Thus,
${k_{{\text{298 K}}}} = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{90}}$
The reaction is $25\% $ complete. Thus, the initial concentration is ${\text{100}}$ and the final concentration is $100 - 25 = 75$. Thus,
${k_{{\text{308 K}}}} = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{75}}$
Thus,
$\dfrac{{{k_{{\text{308 K}}}}}}{{{k_{{\text{298 K}}}}}} = \dfrac{{\dfrac{{2.303}}{t}\log \dfrac{{100}}{{75}}}}{{\dfrac{{2.303}}{t}\log \dfrac{{100}}{{90}}}}$
$\dfrac{{{k_{{\text{308 K}}}}}}{{{k_{{\text{298 K}}}}}} = 2.73$
We know the Arrhenius equation,
$\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{{2.303 \times R}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
where ${k_2}{\text{ and }}{k_1}$ are the constants for the reaction,
              ${E_a}$ is the energy of activation,
              $R$ is the universal gas constant,
              ${T_1}{\text{ and }}{T_2}$ are the temperatures.
Thus, for the given first order reaction, the Arrhenius equation is,
$\log \dfrac{{{k_{308{\text{ K}}}}}}{{{k_{298{\text{ K}}}}}} = \dfrac{{{E_a}}}{{2.303 \times R}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
Substitute $\dfrac{{{k_{308{\text{ K}}}}}}{{{k_{298{\text{ K}}}}}} = 2.73$, $8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ for the universal gas constant, ${T_1} = 298{\text{ K}}$ and ${T_2} = 308{\text{ K}}$. Thus,
$\log 2.73 = \dfrac{{{E_a}}}{{2.303 \times 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}}}\left( {\dfrac{1}{{298{\text{ K}}}} - \dfrac{1}{{308{\text{ K}}}}} \right)$
${E_a} = 76623{\text{ J mo}}{{\text{l}}^{ - 1}}$
Thus, the energy of activation is $76623{\text{ J mo}}{{\text{l}}^{ - 1}}$.
Calculate the rate constant at $318{\text{ K}}$ using the Arrhenius equation,
$\log k = \log A - \dfrac{{{E_a}}}{{2.303 \times R \times T}}$
where $k$ is the rate constant of a first order reaction,
              $A$ is the pre-exponential factor,
              ${E_a}$ is the energy of activation,
              $R$ is the universal gas constant,
              $T$ is the temperature.
Thus,
$\log k = \log \left( {3.56 \times {{10}^9}{\text{ se}}{{\text{c}}^{ - 1}}} \right) - \dfrac{{76623{\text{ J mo}}{{\text{l}}^{ - 1}}}}{{2.303 \times 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \times 318{\text{ K}}}}$
$\log k = - 3.0328$
$k = 9.27 \times {10^{ - 4}}{\text{ se}}{{\text{c}}^{ - 1}}$

Thus, the rate constant at $318{\text{ K}}$ is $9.27 \times {10^{ - 4}}{\text{ se}}{{\text{c}}^{ - 1}}$.

Note:
The unit of rate constant for first order reaction is ${\text{se}}{{\text{c}}^{ - 1}}$. The units do not contain concentration terms. Thus, we can say that the rate constant of a first order reaction is independent of the concentration of the reactant. The rate of any chemical reaction is inversely proportional to the activation energy. Higher the activation energy, slower is the chemical reaction.