Question

For the equilibrium $X(g) \rightleftharpoons Y(g)$, $\Delta H$ is -100 kJ/mol. If the ratio of activation energy of forward and backward is $\frac{3}{4}$. The value of $Ea_f + Ea_b$ is _______ kJ/mol.

Answer 1

700 kJ/mol

Answer 2

The problem involves calculating the sum of forward and backward activation energies ($Ea_f + Ea_b$) for a reversible reaction, given the enthalpy change ($\Delta H$) and the ratio of activation energies.

1. Relate $\Delta H$ to Activation Energies:

For a reversible reaction $X(g) \rightleftharpoons Y(g)$, the enthalpy change ($\Delta H$) is related to the activation energy of the forward reaction ($Ea_f$) and the activation energy of the backward reaction ($Ea_b$) by the following equation:

$\Delta H = Ea_f - Ea_b$

2. Set up Equations from Given Information:

We are given:

• $\Delta H = -100$ kJ/mol
• Ratio of activation energies: $\frac{Ea_f}{Ea_b} = \frac{3}{4}$

From the ratio, we can express $Ea_f$ in terms of $Ea_b$:

$Ea_f = \frac{3}{4} Ea_b$ (Equation 1)

Substitute the given $\Delta H$ and Equation 1 into the relationship from step 1:

$-100 = \frac{3}{4} Ea_b - Ea_b$

3. Solve for $Ea_b$:

Combine the terms involving $Ea_b$:

$-100 = \left(\frac{3}{4} - 1\right) Ea_b$

$-100 = \left(-\frac{1}{4}\right) Ea_b$

Multiply both sides by -4 to solve for $Ea_b$:

$Ea_b = -100 \times (-4)$

$Ea_b = 400$ kJ/mol

4. Solve for $Ea_f$:

Substitute the value of $Ea_b$ back into Equation 1:

$Ea_f = \frac{3}{4} \times 400$

$Ea_f = 3 \times 100$

$Ea_f = 300$ kJ/mol

5. Calculate $Ea_f + Ea_b$:

The question asks for the sum of the activation energies:

$Ea_f + Ea_b = 300 \text{ kJ/mol} + 400 \text{ kJ/mol}$

$Ea_f + Ea_b = 700 \text{ kJ/mol}$