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.

Mark the conditions that favour the maximum product formation in the given reaction.
 $A_2$ $\left(g\right)$ $$ $+$ $B_2$ $\left(g\right)$ $$ $\leftrightharpoons$ $X_2$ $(g),$ ${∆}_{r}H$ $=$ $-X$ $k$ $J$ $$ 

1. Low temperature and High pressure
2. Low temperature and Low pressure
3. High temperature and High pressure
4. High temperature and Low pressure