Question

The rate coefficient $(k)$ for a particular reaction is $1.3 \times {10^{ - 4}}{M^{ - 1}}{s^{ - 1}}$ at ${100^ \circ }C$and $1.3 \times {10^{ - 3}}{M^{ - 1}}{s^{ - 1}}$ at ${150^ \circ }C$ . What is the energy of activation $({\text{in }}kJ)$ for this reaction?
(A)$16$
(B)$60$
(C)$99$
(D)$132$

Answer 1

Hint : The Arrhenius equation gives the relation between the rate of the reaction and temperature for many chemical and physical reactions. The equation is temperature dependent. The Arrhenius equation helps in solving problems related to the rate of reactions. The Arrhenius equation helps in calculating the activation energy if rate constants are known.
Formula used:
$\ln \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$

Complete Step By Step Answer:
Given in question that:
${k_1} = 1.3 \times {10^{ - 4}}{M^{ - 1}}{s^{ - 1}}$
${k_2} = 1.3 \times {10^{ - 3}}{M^{ - 1}}{s^{ - 1}}$
${T_1} = 373K$
${T_2} = 423K$
We know that value of $R$at standard atmospheric pressure is $R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$
By substituting all these values in Arrhenius equation we get:
$\ln \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
$\ln \dfrac{{1.3 \times {{10}^{ - 3}}}}{{1.3 \times {{10}^{ - 4}}}} = \dfrac{{{E_a}}}{{8.314}}\left( {\dfrac{1}{{373}} - \dfrac{1}{{423}}} \right)$
On solving further we get,
${E_a} = 60kJ$

Therefore, option B is correct.

Note :
In the Arrhenius equation, $k$is the rate constant, $A$is called the Arrhenius constant which is also called a frequency factor, ${E_a}$is the activation energy and $R$is the universal gas constant. We also have $T$which is the temperature in kelvin. The Arrhenius constant takes into account the frequency of collisions with the correct orientations for the reaction to occur. The activation energy can be calculated from the value of rate constants at two different temperatures. An increase in temperature, generally speeds up the reaction because the rate constant increases according to the Arrhenius equation.