Chemical EquilibriumMedium
Question
An amount of 4 moles of NH3 gas is introduced with a previously evacuated 1 L container in which it is partially dissociated at high temperature as 2NH3(g) $\rightleftharpoons$N2(g) + 3H2(g) At equilibrium, 2 mole of NH3(g) remained. The value of KC for the reaction is
Options
A.1.5M2
B.6.75M2
C.0.44M–2
D.2.25M2
Solution
2NH3 $\rightleftharpoons$ N2 + 3H2
4 mole 0 0
Equilibrium 4 – 2x x 3x
= 2 = 1 = 3
∴ x = 1
Now, $K_{C} = \frac{\left\lbrack N_{2} \right\rbrack\left\lbrack H_{2} \right\rbrack^{3}}{\left\lbrack NH_{3} \right\rbrack^{2}} = \frac{\frac{1}{1} \times \left( \frac{3}{1} \right)^{3}}{\left( \frac{2}{1} \right)^{2}} = 6.75\text{ }\text{M}^{2}$
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