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Activation Energy

For the equil...

Question

For the equilibrim, $A (g) ⇌ B (g)$ , $Δ H = - 40 k J / m o l$ . If the ratio of the activation energies of the forward $(E_{f})$ and reverse $(E_{b})$ reaction is $\frac{2}{3}$ then:

A

E_{f} = 30 k J / m o l; E_{b} = 70 k J / m o l

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B

E_{f} = 70 k J / m o l; E_{b} = 30 k J / m o l

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C

E_{f} = 80 k J / m o l; E_{b} = 120 k J / m o l

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D

E_{f} = 60 k J / m o l; E_{b} = 100 k J / m o l

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Solution

The correct option is A $E_{f} = 80 k J / m o l; E_{b} = 120 k J / m o l$
(A) $E_{f} = 30 k J / m o l; E_{b} = 70 k J / m o l$ $\frac{E_{f}}{E_{b}} = \frac{30 k J / m o l}{70 k J / m o l} = \frac{3}{7}$

(B) $E_{f} = 70 k J / m o l; E_{b} = 30 k J / m o l$ $\frac{E_{f}}{E_{b}} = \frac{70 k J / m o l}{30 k J / m o l} = \frac{7}{3}$

(C) $E_{f} = 80 k J / m o l; E_{b} = 120 k J / m o l$ $\frac{E_{f}}{E_{b}} = \frac{80 k J / m o l}{120 k J / m o l} = \frac{2}{3}$

(D) $E_{f} = 60 k J / m o l; E_{b} = 100 k J / m o l$ $\frac{E_{f}}{E_{b}} = \frac{60 k J / m o l}{100 k J / m o l} = \frac{3}{5}$

But the given ratio is $\frac{E_{f}}{E_{b}} = \frac{2}{3}$

Hence, the option $C$ is the correct answer.

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