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Equilibrium Constant and Standard Free Energy Change

Calculate the...

Question

Calculate the equilibrium constant $(K_{c})$ for the reaction given below, if at equilibrium mixture contains 5.0 mole of $A_{2}$ , 3 mole of $B_{2}$ and 2 moles of $A B_{2}$ at 8.21 atm and 300K
$A_{2} (g) + 2 B_{2} (g) . \to . 2 A B_{2} (g) + H e a t$

A

1.333

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B

2.66

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C

20

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D

None of these

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Solution

The correct option is B 2.66
$K_{c} = \frac{n_{A B_{2}}^{2}}{n_{A_{2}} . n_{B_{2}}^{2}} \times V = \frac{n_{A B_{2}}^{2}}{n_{A_{2}} . n_{B_{2}}^{2}} \times (\frac{n_{t o t a l} R T}{P})$
$= \frac{2^{2}}{5 \times 3^{2}} \times (\frac{10 \times 0.0821 \times 300}{8.21}) = 2.66$

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