For a reversible gaseous reaction A + B \rightleftharpoons C + D, the potential energy of … — JAMB UTME

For a reversible gaseous reaction $A + B \rightleftharpoons C + D$ , the potential energy of the reactants is $120\text{ kJ mol}^{-1}$ and the potential energy of the products is $45\text{ kJ mol}^{-1}$ . If the activation energy for the reverse reaction is $140\text{ kJ mol}^{-1}$ , calculate the activation energy for the forward reaction in $\text{kJ mol}^{-1}$ .

Answer: 65 kJ mol^{-1}

Answer

The activation energy for the forward reaction is

65\text{ kJ mol}^{-1}

The forward activation energy (

E_{a,\text{forward}}

) is the energy barrier measured from the energy level of the reactants (

120\text{ kJ mol}^{-1}

) to the peak of the activated complex. Since the products lie at

45\text{ kJ mol}^{-1}

and require

140\text{ kJ mol}^{-1}

to reach the activated complex, the peak energy is

45 + 140 = 185\text{ kJ mol}^{-1}

. Subtracting the reactant energy gives

185 - 120 = 65\text{ kJ mol}^{-1}

. Alternatively, using

\Delta H = E_{a,\text{forward}} - E_{a,\text{reverse}}

, where

\Delta H = 45 - 120 = -75\text{ kJ mol}^{-1}

, gives

-75 = E_{a,\text{forward}} - 140 \implies E_{a,\text{forward}} = 65\text{ kJ mol}^{-1}

Step-by-Step Solution

Determine the potential energy of the activated complex (transition state).

Transition state energy

E_{\text{peak}} = H_{\text{products}} + E_{a,\text{reverse}} = 45 + 140 = 185\text{ kJ mol}^{-1}

The reverse activation energy is the energy required to overcome the barrier going from products up to the peak.

Calculate the forward activation energy.

E_{a,\text{forward}} = E_{\text{peak}} - H_{\text{reactants}} = 185 - 120 = 65\text{ kJ mol}^{-1}

The forward activation energy is the energy difference between the peak (activated complex) and the energy level of the reactants.

Key Concept

Activation energy and potential energy relationships in energy profile diagrams

Estimated Time:1m 30s

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