Write the expression for the equilibrium constant Kp for the given reaction.

The reaction is:

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

The expression for $\text{Kp}$ is:

$$\text{Kp} = \frac{(P_{\text{NH}_3})^2}{(P_{\text{N}_2})(P_{\text{H}_2})^3}$$

Given partial pressures at equilibrium:
$P_{\text{N}_2} = 1.49 \text{ atm}$
$P_{\text{H}_2} = 0.40 \text{ atm}$
$P_{\text{NH}_3} = 0.11 \text{ atm}$

$$\text{Kp} = \frac{(0.11 \text{ atm})^2}{(1.49 \text{ atm})(0.40 \text{ atm})^3}$$

$$\text{Kp} = \frac{0.0121}{(1.49)(0.064)}$$ $$\text{Kp} = \frac{0.0121}{0.09536}$$ $$\text{Kp} \approx 0.12688$$

The equilibrium constant, $\text{Kp}$, is $\boxed{\text{0.127}}$.