Write the equilibrium constant expression (K_c) for the reaction.

The given reaction is:

$$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$$

The equilibrium constant expression for this reaction is:

$$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$

The given equilibrium concentrations are:
$[\text{N}_2] = 0.5 \text{ mol/litre}$
$[\text{H}_2] = 0.5 \text{ mol/litre}$
$[\text{NH}_3] = 0.15 \text{ mol/litre}$

Substitute these values into the $K_c$ expression:

$$K_c = \frac{(0.15 \text{ mol/litre})^2}{(0.5 \text{ mol/litre})(0.5 \text{ mol/litre})^3}$$

$$K_c = \frac{0.0225 \text{ (mol/litre)}^2}{(0.5 \text{ mol/litre})(0.125 \text{ (mol/litre)}^3)}$$ $$K_c = \frac{0.0225 \text{ (mol/litre)}^2}{0.0625 \text{ (mol/litre)}^4}$$ $$K_c = 0.36 \text{ (mol/litre)}^{-2}$$ $$K_c = 0.36 \text{ L}^2/\text{mol}^2$$

The value of $K_c$ is $\boxed{\text{0.36 L}^2/\text{mol}^2}$.