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Activation Energy

For the equil...

Question

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

A

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

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B

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

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Solution

The correct option is D $E_{f} = 80 k J / m o l; E_{b} = 120 k J / m o l$
The given reaction is,
$A (g) ⇌ B (g)$
$△ H = - 40 K J / m o l$
& $\frac{E_{f}}{E_{b}} = \frac{2}{3} \Rightarrow E_{f} = \frac{2}{3} E_{b}$
From graph we have,
$△ H = E_{b} - E_{f}$
$\Rightarrow - 40 = E_{b} - \frac{2}{3} E_{b}$
$\Rightarrow - 40 = \frac{1}{3} E_{b} \Rightarrow E_{b} = - 120 K J / m o l$
& $E_{f} = E_{b} - △ H$
$= - 120 - (- 40)$
$- 80$ $K J / m o l$ .

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Similar questions

Q. For the equilibrim,

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