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Consider a reversible reaction taking place in a 10 litre closed vessel at 300K, $X (g) ⇌ 2 Z (g);$ Initially three moles of ‘X’ is present in the vessel. At equilibrium two moles of ‘Z’ is formed. $K_{c}$ for the reaction at 300 K will be

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$1 M$

$0.2 M$

$0.04 M$

$0.02 M$

answer is B.

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Detailed Solution

$stoichiometry 1 mole 2 moles X (g) ⇌ 2 Z (g); initial moles 3 - moles at eqbm (3 - 1) 2 eq . conc \frac{2}{10} \frac{2}{10} K_{c} = \frac{{[Z]}^{2}}{[X]} = \frac{{(0.2)}^{2}}{0.2} = 0.2 M$

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