Question

The slope of Arrhenius Plot (ln k v/s \(\frac{1}{T}\)) of first order reaction is −5×10³ K. The value of Ea of the reaction is. Choose the correct option for your answer. [Given R=8.314 JK⁻¹mol⁻¹ ]

• A. −83 kJ mol⁻¹
• B. 41.5 kJ mol⁻¹
• C. 83.0 kJ mol⁻¹
• D. 166 kJ mol⁻¹

Answer 1

The Correct Option is B

To find the activation energy \(E_a\) for a reaction from an Arrhenius plot, we use the Arrhenius equation in its logarithmic form:

\(\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}\) 

where:

• \(k\) is the rate constant.
• \(A\) is the pre-exponential factor, a constant.
• \(E_a\) is the activation energy.
• \(R\) is the universal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\)).
• \(T\) is the temperature in Kelvin.

The slope of the Arrhenius plot (\(\ln k\) vs. \(\frac{1}{T}\)) is given by \(-\frac{E_a}{R}\).

From the problem, we know that the slope is \(-5 \times 10^3 \text{ K}\).

Hence, we can set up the equation:

\(-\frac{E_a}{R} = -5 \times 10^3\)

This implies:

\(\frac{E_a}{R} = 5 \times 10^3\)

Now solve for \(E_a\):

\(E_a = (5 \times 10^3) \times R \text{ J/mol}\)

Substitute the given value of R:

\(E_a = (5 \times 10^3) \times 8.314 \text{ J/mol} = 41570 \text{ J/mol}\)

Convert Joules to kilojoules (since \(1\text{ kJ} = 1000\text{ J}\)):

\(E_a = \frac{41570}{1000} \text{ kJ/mol} = 41.57 \text{ kJ/mol}\)

Therefore, the activation energy \(E_a\) is approximately 41.5 kJ/mol.

Thus, the correct option is 41.5 kJ/mol.