Question

For a unimolecular reaction A → B has rate constant K1 and for another unimolecular reaction C → D has rate constnat K2. If half life of first reaction is half of half life of second reaction then find Ea1 – Ea2 will be.

Answer 1

-RT ln 2

Answer 2

To solve this problem, we need to use the concepts of half-life for first-order reactions and the Arrhenius equation.

1. Assumption of First-Order Kinetics:
   For a unimolecular reaction, it is generally assumed to follow first-order kinetics unless stated otherwise.
   For a first-order reaction, the half-life ($t_{1/2}$) is related to the rate constant (K) by the formula: $t_{1/2} = \frac{\ln 2}{K}$

2. Relationship between Half-lives and Rate Constants:
   Given that the half-life of the first reaction ($t_{1/2, 1}$) is half of the half-life of the second reaction ($t_{1/2, 2}$): $t_{1/2, 1} = \frac{1}{2} t_{1/2, 2}$ Substituting the half-life formula for both reactions: $\frac{\ln 2}{K_1} = \frac{1}{2} \frac{\ln 2}{K_2}$ Simplifying the equation: $\frac{1}{K_1} = \frac{1}{2K_2}$ $K_1 = 2K_2$

3. Arrhenius Equation:
   The rate constant (K) is related to the activation energy ($E_a$) by the Arrhenius equation: $K = A e^{-E_a / RT}$ where A is the pre-exponential factor (frequency factor), R is the gas constant, and T is the absolute temperature.

4. Applying Arrhenius Equation to Both Reactions:
   For the first reaction (A → B): $K_1 = A_1 e^{-E_{a1} / RT}$ For the second reaction (C → D): $K_2 = A_2 e^{-E_{a2} / RT}$ Here, we assume that both reactions occur at the same temperature T.
   Also, in the absence of information to the contrary, it is a standard assumption in such problems that the pre-exponential factors (A) are the same for similar types of reactions or cancel out in the comparison. So, we assume $A_1 = A_2 = A$.

5. Solving for $E_{a1} - E_{a2}$:
   Substitute the Arrhenius expressions for $K_1$ and $K_2$ into the relationship $K_1 = 2K_2$: $A e^{-E_{a1} / RT} = 2 A e^{-E_{a2} / RT}$ Since A is common and non-zero, we can cancel it out: $e^{-E_{a1} / RT} = 2 e^{-E_{a2} / RT}$ Rearrange the terms: $\frac{e^{-E_{a1} / RT}}{e^{-E_{a2} / RT}} = 2$ Using the property $e^x / e^y = e^{x-y}$: $e^{(E_{a2} - E_{a1}) / RT} = 2$ Take the natural logarithm ($\ln$) on both sides: $\ln\left(e^{(E_{a2} - E_{a1}) / RT}\right) = \ln 2$ $\frac{E_{a2} - E_{a1}}{RT} = \ln 2$ Finally, solve for $E_{a1} - E_{a2}$: $E_{a2} - E_{a1} = RT \ln 2$ $E_{a1} - E_{a2} = -RT \ln 2$

The difference in activation energies, $E_{a1} - E_{a2}$, is $-RT \ln 2$.

Explanation of the solution:

1. Assume both unimolecular reactions are first-order, so $t_{1/2} = \frac{\ln 2}{K}$.
2. Given $t_{1/2, 1} = \frac{1}{2} t_{1/2, 2}$, substitute the half-life formula to get $K_1 = 2K_2$.
3. Use the Arrhenius equation $K = A e^{-E_a / RT}$. Assume pre-exponential factors (A) and temperature (T) are the same for both reactions.
4. Substitute Arrhenius expressions into $K_1 = 2K_2$: $A e^{-E_{a1} / RT} = 2 A e^{-E_{a2} / RT}$.
5. Simplify and take natural logarithm: $e^{(E_{a2} - E_{a1}) / RT} = 2 \implies \frac{E_{a2} - E_{a1}}{RT} = \ln 2$.
6. Solve for $E_{a1} - E_{a2} = -RT \ln 2$.