For the given reversible reaction: $\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}$

Initial moles: $n_A = 1$ $n_B = 1$ $n_C = 0$ $n_D = 0$

Let $x$ be the number of moles of C formed at equilibrium. The moles of each species at equilibrium will be: $n_A = 1 - x$ $n_B = 1 - x$ $n_C = x$ $n_D = x$

We are given that $0.7\text{ mol}$ of C is formed at equilibrium, so $x = 0.7$. Since the volume of the container is $1\text{ L}$, the molar concentrations at equilibrium are equal to the number of moles : $[\text{A}] = 1 - 0.7 = 0.3\text{ M}$ $[\text{B}] = 1 - 0.7 = 0.3\text{ M}$ $[\text{C}] = 0.7\text{ M}$ $[\text{D}] = 0.7\text{ M}$

The equilibrium constant expression ($K_c$) is: $K_c = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

Substituting the equilibrium concentrations into the expression: $K_c = \frac{0.7 \times 0.7}{0.3 \times 0.3} = \frac{0.49}{0.09} = \frac{49}{9} \approx 5.44$

Therefore, the equilibrium constant ($K_c$) is approximately $5.4$.