Q.In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively $-x\ \text{kJ mol}^{-1}$ and $y\ \text{kJ mol}^{-1}$. Therefore, the energy of activation in the backward direction is

Step 1. On a reaction's potential-energy diagram (Fig 7.5-style), the peak's energy equals reactant energy plus the forward activation energy, and equals product energy plus the backward activation energy: $E_{\text{reactant}}+E_{a,\text{fwd}}=E_{\text{product}}+E_{a,\text{bwd}}$.

Step 2. Rearranging: $E_{\text{product}}-E_{\text{reactant}}=E_{a,\text{fwd}}-E_{a,\text{bwd}}$, i.e. $\Delta H=E_{a,\text{fwd}}-E_{a,\text{bwd}}$. …

Use the potential-energy-diagram relation $\Delta H=E_{a,\text{forward}}-E_{a,\text{backward}}$, substitute the given $\Delta H=-x$ and $E_{a,\text{f …

• Missing that $(x+y)$ kJ mol$^{-1}$ and $(x+y)\times10^3$ J mol$^{-1}$ are the SAME physical quantity in different units, and rejecting (d) because it 'doesn't look like' the derived kJ answer. …