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

The given problem involves understanding the Arrhenius equation and its graphical representation. The Arrhenius equation is given by:

\( k = A e^{-\frac{E_a}{RT}} \)

where:

• \( k \) is the rate constant,
• \( A \) is the pre-exponential factor,
• \( E_a \) is the activation energy,
• \( R \) is the universal gas constant,
• \( T \) is the temperature in Kelvin.

To linearize the Arrhenius equation, we take the natural logarithm:

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

This can be compared to the equation of a straight line \( y = mx + c \), where:

• \( y = \ln k \),
• \( m = -\frac{E_a}{R} \) (slope),
• \( x = \frac{1}{T} \),
• \( c = \ln A \).

In the given problem, the slope of the Arrhenius plot (\(\ln k\) vs \(\frac{1}{T}\)) is provided as \( -5 \times 10^3 \) K. From the linearized equation, the slope (m) is related to the activation energy (\(E_a\)) by:

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

Solving for \( E_a \):

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

\(E_a = 5 \times 10^3 \times R\)

Substituting the given value of \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \):

\(E_a = 5 \times 10^3 \times 8.314\)

\(E_a = 41570 \, \text{J mol}^{-1}\)

Converting J/mol to kJ/mol (since 1 kJ = 1000 J):

\(E_a = \frac{41570}{1000} \, \text{kJ mol}^{-1} = 41.57 \, \text{kJ mol}^{-1}\)

Rounding to the appropriate number of significant figures, the activation energy \( E_a \) is approximately 41.5 kJ mol^-1.

Thus, the correct option is 41.5 kJ mol^-1.