For the equilibrium A(g)⇌B(g),ΔH is −40kJ/mol, If the ratio of activation energies of the forward Ef and reverse Eb is 2/3, then

A
$E_{f} = 60 kJ / mol; E_{b} = 100 kJ / mol$
B
$E_{f} = 30 kJ / mol; E_{b} = 70 kJ / mol$
C
$E_{f} = 80 kJ / mol; E_{b} = 120 kJ / mol$
D
$E_{f} = 70 kJ / mol : E_{b} = 30 kJ / mol$

Solution:

The given reaction is,

$\begin{array}{l} A (g) ⇌ B (g) \\ △ H = - 40 KJ / mol \\ & \frac{E_{f}}{E_{b}} = \frac{2}{3} \Rightarrow E_{f} = \frac{2}{3} E_{b} \end{array}$

From graph we have,

$\begin{array}{l} △ H = E_{b} - E_{f} \\ \Rightarrow - 40 = E_{b} - \frac{2}{3} E_{b} \\ \Rightarrow - 40 = \frac{1}{3} E_{b} \Rightarrow E_{b} = - 120 KJ / mol \\ & E_{f} = E_{b} - ΔH \\ = - 120 - (- 40) \\ - 80 KJ / mol \end{array}$

For the equilibrium $A (g) ⇌ B (g), Δ H is - 40 kJ / mol$ , If the ratio of activation energies of the forward $(E_{f})$ and reverse $(E_{b})$ is 2/3, then

Solution:

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