MP Board · Class 12 · Chemistry · Chemical KineticsThe rate constant of a reaction increases from $1.5 \times 10^4 \text{ s}^{-1}$ at $300 \text{ K}$ to $4.5 \times 10^4 \text{ s}^{-1}$ at $320 \text{ K}$. Calculate the activation energy ($Ea$) and the Arrhenius factor ($A$). (Given: $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$)

Step-by-Step Numerical Solution

Given Data:

• Temperature $T_1 = 300 \text{ K}$, Rate constant $k_1 = 1.5 \times 10^4 \text{ s}^{-1}$
• Temperature $T_2 = 320 \text{ K}$, Rate constant $k_2 = 4.5 \times 10^4 \text{ s}^{-1}$
• Gas constant $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$

Part 1: Calculation of Activation Energy ($E_a$)

\nThe modified Arrhenius equation relating rate constants at two different temperatures is: $$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$$ \nSubstitute the given values into the equation: $$\log \left( \frac{4.5 \times 10^4}{1.5 \times 10^4} \right) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{320 - 300}{300 \times 320} \right]$$

$$\log(3) = \frac{E_a}{19.147} \left[ \frac{20}{96000} \right]$$ \nSince $\log(3) = 0.4771$: $$0.4771 = \frac{E_a}{19.147} \times 0.0002083$$ \nRearranging for $E_a$: $$E_a = \frac{0.4771 \times 19.147}{0.0002083}$$ $$E_a = \frac{9.135}{0.0002083} = 43855 \text{ J mol}^{-1} = 43.86 \text{ kJ mol}^{-1}$$

Part 2: Calculation of Arrhenius Factor ($A$)

\nUsing the Arrhenius equation at $T_1 = 300 \text{ K}$: $$k_1 = A e^{-E_a / RT_1}$$ $$\log k_1 = \log A - \frac{E_a}{2.303 R T_1}$$ \nSubstitute the known values: $$\log(1.5 \times 10^4) = \log A - \frac{43855}{2.303 \times 8.314 \times 300}$$

$$\log(1.5) + 4 = \log A - \frac{43855}{5744.1}$$ $$0.1761 + 4 = \log A - 7.6348$$ $$4.1761 = \log A - 7.6348$$

$$\log A = 4.1761 + 7.6348 = 11.8109$$ \nTaking antilog: $$A = \text{antilog}(11.8109) = 6.47 \times 10^{11} \text{ s}^{-1}$$

Answer:

• Activation energy ($E_a$) = $43.86 \text{ kJ mol}^{-1}$
• Arrhenius factor ($A$) = $6.47 \times 10^{11} \text{ s}^{-1}$