Answer :
To determine the effect of increasing pressure on the chemical equilibrium, we need to consider Le Chatelier's Principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of the equilibrium moves to counteract the change.
Here is the given equilibrium reaction:
[tex]\[ 2 H_2 (g) + O_2 (g) \Leftrightarrow 2 H_2O (g) \][/tex]
Let's analyze the number of gas molecules on both sides of the reaction:
- On the reactant side: [tex]\(2\)[/tex] molecules of [tex]\(H_2 (g)\)[/tex] + [tex]\(1\)[/tex] molecule of [tex]\(O_2 (g)\)[/tex] = [tex]\(3\)[/tex] molecules in total.
- On the product side: [tex]\(2\)[/tex] molecules of [tex]\(H_2O (g)\)[/tex].
When we increase the pressure of a system in equilibrium, the system will shift in the direction that reduces the pressure. This generally means it will favor the side with fewer gas molecules because fewer gas molecules result in lower pressure.
In this reaction:
- The reactant side has [tex]\(3\)[/tex] gas molecules.
- The product side has [tex]\(2\)[/tex] gas molecules.
Since the forward reaction converts [tex]\(3\)[/tex] gas molecules into [tex]\(2\)[/tex] gas molecules, increasing pressure will favor the forward reaction, which produces fewer gas molecules.
Therefore, the correct answer is:
"The forward reaction will be favored."
Here is the given equilibrium reaction:
[tex]\[ 2 H_2 (g) + O_2 (g) \Leftrightarrow 2 H_2O (g) \][/tex]
Let's analyze the number of gas molecules on both sides of the reaction:
- On the reactant side: [tex]\(2\)[/tex] molecules of [tex]\(H_2 (g)\)[/tex] + [tex]\(1\)[/tex] molecule of [tex]\(O_2 (g)\)[/tex] = [tex]\(3\)[/tex] molecules in total.
- On the product side: [tex]\(2\)[/tex] molecules of [tex]\(H_2O (g)\)[/tex].
When we increase the pressure of a system in equilibrium, the system will shift in the direction that reduces the pressure. This generally means it will favor the side with fewer gas molecules because fewer gas molecules result in lower pressure.
In this reaction:
- The reactant side has [tex]\(3\)[/tex] gas molecules.
- The product side has [tex]\(2\)[/tex] gas molecules.
Since the forward reaction converts [tex]\(3\)[/tex] gas molecules into [tex]\(2\)[/tex] gas molecules, increasing pressure will favor the forward reaction, which produces fewer gas molecules.
Therefore, the correct answer is:
"The forward reaction will be favored."