Question

The rate constant for a first order reaction becomes six times when the temperature is raised from $350{\text{ K}}$ to $400{\text{ K}}$. Calculate the activation energy for the reaction. $\left( {R = 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}} \right)$.

Answer 1

Hint: The minimum amount of energy that the reacting species must possess to undergo a specific reaction is known as the activation energy. We know that the relation between the temperature, rate constant and activation energy is given by the Arrhenius equation.

Formula Used: $\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{{2.303 \times R}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$

Complete step by step answer:
We know the Arrhenius equation,
$\Rightarrow\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{{2.303 \times R}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
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.
The rate constant for a first order reaction becomes six times when the temperature is raised from $350{\text{ K}}$ to $400{\text{ K}}$.

Consider ${T_1} = 350{\text{ K}}$ and ${T_2} = 400{\text{ K}}$. Thus,
$\Rightarrow {k_2} = 6 \times {k_1}$
$\Rightarrow \dfrac{{{k_2}}}{{{k_1}}} = 6$
Substitute $\dfrac{{{k_2}}}{{{k_1}}} = 6$, $8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ for the universal gas constant, ${T_1} = 350{\text{ K}}$ and ${T_2} = 400{\text{ K}}$. Thus,
$\Rightarrow \log 6 = \dfrac{{{E_a}}}{{2.303}}\left( {\dfrac{1}{{350{\text{ K}}}} - \dfrac{1}{{400{\text{ K}}}}} \right)$
$\Rightarrow \log 6 = \dfrac{{{E_a}}}{{2.303 \times 8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}}}\left( {\dfrac{1}{{350{\text{ K}}}} - \dfrac{1}{{400{\text{ K}}}}} \right)$
$\Rightarrow 0.77815 = \dfrac{{{E_a}}}{{19.147{\text{ J }}{{{{\text{K}}^{ - 1}}}}{\text{ mo}}{{\text{l}}^{ - 1}}}}\left( {\dfrac{{400 - 350}}{{350 \times 400}}{{\text{K}}}} \right)$
$\Rightarrow 0.77815 = \dfrac{{{E_a}}}{{19.147{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}}} \times 3.571 \times {10^{ - 4}}$
$\Rightarrow {E_a} = \dfrac{{0.77815 \times 19.147{\text{ J mo}}{{\text{l}}^{ - 1}}}}{{3.571 \times {{10}^{ - 4}}}}$
$\Rightarrow {E_a} = 4.17 \times {10^4}{\text{ J mo}}{{\text{l}}^{ - 1}}$

Thus, the activation energy for the reaction is $4.17 \times {10^4}{\text{ J mo}}{{\text{l}}^{ - 1}}$.

Note: The rate of any chemical reaction is inversely proportional to the activation energy. Higher the activation energy, slower is the chemical reaction. Reactions with very high activation energy do not proceed unless energy is supplied to them. For reactions that have a high activation energy, catalysts are used for the reaction to proceed. The catalyst reduces the activation energy and thus, the rate of the reaction increases. The catalyst forms a transition state. The transition state forms because energy is released.