To solve this problem, we need to apply Le Chatelier's principle to the given equilibrium reaction:$\\ N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g) +$Heat$\\$Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions,$\\$the position of equilibrium shifts to counteract the change.$\\$Let's analyze each option:$\\$Option 1: Increasing the concentration of $NH_3 (g) \\$• Adding more $NH_3$ will shift the equilibrium to the left, favoring the formation of $N_2$ and $H_2. \\$• This does not shift the equilibrium in the forward direction.$\\$Option 2: Decreasing the pressure$\\$• Decreasing the pressure will favor the side with more moles of gas.$\\$• The left side has 4 moles (1 $N_2$ and 3 $H_2)$ while the right side has 2 moles of $NH_3. \\$• Therefore, decreasing pressure shifts the equilibrium to the left.$\\$Option 3: Decreasing the concentrations of $N_2 (g)$ and $H_2 (g) \\$• Reducing the reactants will shift the equilibrium to the left to produce more $N_2$ and $H_2. \\$• This does not favor the forward reaction.$\\$Option 4: Increasing pressure and decreasing temperature$\\$• Increasing pressure favors the side with fewer moles of gas, which is the right side (2 moles of $NH_3). \\$• Decreasing temperature favors the exothermic reaction, which is the forward reaction in this case.$\\$• Both changes shift the equilibrium in the forward direction.$\\$Therefore, the correct option is 4: Increasing pressure and decreasing temperature.