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

Step 1: Understanding the Question.
In transition state theory, two reactant molecules \(A\) and \(B\) come together to form one activated complex \(X^{\ddagger}\), which then goes on to product. We are given the standard internal energy of activation, \(\Delta^{\ddagger}U^o = \frac{5}{2}RT\), and asked for the standard enthalpy of activation, \(\Delta^{\ddagger}H^o\).

Step 2: Key Formula.
For a process at constant pressure involving ideal gases, enthalpy and internal energy are related by \(H = U + PV\), so a change obeys \(\Delta H = \Delta U + \Delta(PV)\). For ideal gases, \(PV=nRT\), so at constant temperature \(\Delta(PV) = \Delta n_{gas}\,RT\), where \(\Delta n_{gas}\) is the change in the number of moles of gas going from reactants to the species being formed. Applied to activation,
\[ \Delta^{\ddagger}H^o = \Delta^{\ddagger}U^o + \Delta n^{\ddagger} RT \]

Step 3: Find \(\Delta n^{\ddagger}\).
The reaction is \(A + B \rightarrow X^{\ddagger}\): two moles of reactant gas combine into one mole of activated complex. So
\[ \Delta n^{\ddagger} = (\text{moles of } X^{\ddagger}) - (\text{moles of } A + B) = 1 - 2 = -1 \]

Step 4: Substitute and calculate.
\[ \Delta^{\ddagger}H^o = \frac{5}{2}RT + (-1)RT = \frac{5}{2}RT - RT = \frac{5-2}{2}RT = \frac{3}{2}RT \]

Final Answer:
The standard enthalpy of activation is \(\frac{3}{2}RT\), smaller than \(\Delta^{\ddagger}U^o\) because the number of gas moles decreases by 1 on forming the activated complex. \[ \boxed{\Delta^{\ddagger}H^o = \frac{3}{2}RT} \]