Alright, let's analyze the effect of increasing pressure on the reaction:
Given the chemical equation:
[tex]\[ 3 H_2 + N_2 \rightleftarrows 2 NH_3 \][/tex]
First, we need to count the moles of gas on each side of the reaction:
- On the reactants side, you have 3 moles of [tex]\( H_2 \)[/tex] and 1 mole of [tex]\( N_2 \)[/tex]. So, the total is [tex]\( 3 + 1 = 4 \)[/tex] moles of gas.
- On the products side, you have 2 moles of [tex]\( NH_3 \)[/tex].
Next, we consider Le Chatelier's Principle, which states that if a system at equilibrium is subjected to a change in pressure, the system will adjust to minimize that change. In the context of gas reactions, increasing the pressure will shift the equilibrium toward the side with fewer moles of gas.
Since the reactants side has 4 moles of gas and the products side has 2 moles of gas:
- Increasing the pressure will shift the equilibrium toward the side with fewer moles of gas, which in this case is the products side (2 moles).
Therefore, increasing the pressure will favor the reaction that produces [tex]\( NH_3 \)[/tex], because it results in fewer moles of gas.
Thus, the correct answer is:
D. [tex]\( H_2 \)[/tex] and [tex]\( N_2 \)[/tex] would react to produce more [tex]\( NH_3 \)[/tex].
