Question

For the following reaction,
\(\mathrm{A + B \rightleftharpoons X^{\ddagger} \rightarrow product}\)
If the value of \(\Delta^{\ddagger}U^{o} = \dfrac{5}{2}RT\), then the value of \(\Delta^{\ddagger}H^{o}\) is
(\(X^{\ddagger}\) is an activated complex; \(\Delta^{\ddagger}U^{o}\) is the standard change in internal energy; \(\Delta^{\ddagger}H^{o}\) is the standard enthalpy of activation; \(R\) is the gas constant; \(T\) is the temperature)

• A. \(\dfrac{1}{2}RT\)
• B. \(\dfrac{3}{2}RT\)
• C. \(\dfrac{7}{2}RT\)
• D. \(\dfrac{5}{2}RT\)

Hint:

Use $\Delta H^{\ddagger}=\Delta U^{\ddagger}+\Delta n^{\ddagger}RT$ with $\Delta n^{\ddagger}=1-2=-1$ since two molecules combine into one activated complex.

Answer 1

The Correct Option is B

A second way to reach the same result is to think of the pre-equilibrium step $A + B \rightleftharpoons X^{\ddagger}$ as an ordinary chemical equilibrium and use the standard relation between $K_p$ and $K_c$ (or equivalently between $\Delta H$ and $\Delta U$ for a reaction of gases).

1. For any reaction of ideal gases, $\Delta H = \Delta U + \Delta(PV)$, and since $PV = n_{gas}RT$ for each species, this becomes $\Delta H = \Delta U + (\Delta n_{gas})RT$, exactly the relation used to connect $K_p$ and $K_c$ through $K_p = K_c(RT)^{\Delta n_{gas}}$.
2. Count the moles of gas on each side of the activation step $A + B \rightarrow X^{\ddagger}$: 2 moles of gas go in ($A$ and $B$), 1 mole of gas comes out ($X^{\ddagger}$), so $\Delta n_{gas} = 1 - 2 = -1$.
3. Apply this to the activation quantities directly: $\Delta^{\ddagger}H^o = \Delta^{\ddagger}U^o + \Delta n^{\ddagger}RT = \frac{5}{2}RT + (-1)(RT)$.
4. Carrying out the subtraction: $\frac{5}{2}RT - \frac{2}{2}RT = \frac{3}{2}RT$.

This is the same numeric answer as before, reached by treating the activation step as a mole-counting problem on an equilibrium rather than a direct thermodynamic definition.

Let's summarize:

• Whenever the number of gas particles changes during a step, $\Delta H$ and $\Delta U$ for that step differ by $(\Delta n_{gas})RT$.
• Here two particles become one, so $\Delta n^{\ddagger}=-1$, and $\Delta^{\ddagger}H^o$ ends up $1RT$ smaller than $\Delta^{\ddagger}U^o$.

The correct option is (B), $\frac{3}{2}RT$.