For the reversible reaction:
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) + heat 

The equilibrium shifts in a forward direction:

1. by increasing the concentration of $N{H}_3\left(g\right)$
2. by decreasing the pressure.
3. by decreasing the concentration of $N_2\left(g\right)$ $and$ $H_2\left(g\right)$
4. by increasing pressure and decreasing temperature.

The value of $∆\mathrm{H}$ for the reaction is less than zero. Formation of $X_2\left(g\right)$ $+$ $4$ $Y_2\left(g\right)$ $\leftrightharpoons 2X{Y}_4\left(g\right)$ will be favoured at: