Step 1: Understanding the Concept:
The Arrhenius equation $k = A e^{-E_a/RT}$ relates the rate constant $k$ to the temperature $T$.
Taking the natural logarithm ($\ln$) on both sides gives:
$\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)$.
Comparing this to the straight-line equation $y = mx + c$:
$y = \ln k$
$x = 1/T$
$m (\text{slope}) = -E_a/R$
Step 2: Key Formula or Approach:
\[ E_a = -\text{slope} \times R \]
Note: Activation energy is always positive ($E_a>0$). The slope is negative, which cancels out the negative sign in the formula.
Step 3: Detailed Explanation:
Given values:
Slope = $-5 \times 10^3$ K
$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$
Substitute the values:
\[ E_a = -(-5 \times 10^3 \text{ K}) \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \]
\[ E_a = 5000 \times 8.314 \text{ J mol}^{-1} \]
\[ E_a = 41570 \text{ J mol}^{-1} \]
Convert the answer to kiloJoules per mole (kJ/mol):
\[ E_a = \frac{41570}{1000} \text{ kJ mol}^{-1} = 41.57 \text{ kJ mol}^{-1} \]
Rounding to the nearest provided option, we get 41.5 kJ/mol.
The physical significance of this value is that the reactants need 41.5 kJ of energy per mole to reach the transition state.
If the slope were based on $\log k$ (base 10), the formula would have been $slope = -E_a / (2.303 R)$. Careful reading of the plot axis is essential.
Step 4: Final Answer:
The activation energy $E_a$ is 41.5 kJ/mol.