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Sagot :
Let's analyze the given chemical equilibrium:
[tex]\[ 2 \text{H}_2(g) + \text{O}_2(g) \Leftrightarrow 2 \text{H}_2\text{O}(g) \][/tex]
To determine the effect of increasing pressure on the system, we should consider Le Chatelier's principle. This principle states that if a stress is applied to a system at equilibrium, the system will respond in a way that counteracts the stress.
In this scenario, the stress applied is an increase in pressure. We must consider the number of moles of gas on each side of the equilibrium equation to predict the direction in which the reaction will shift.
On the reactant side of the equilibrium equation, we have:
[tex]\[ 2 \text{H}_2(g) + 1 \text{O}_2(g) \Rightarrow 2 + 1 = 3 \text{ moles of gas} \][/tex]
On the product side of the equilibrium equation, we have:
[tex]\[ 2 \text{H}_2\text{O}(g) \Rightarrow 2 \text{ moles of gas} \][/tex]
According to Le Chatelier's principle, if we increase the pressure of the system, the equilibrium will shift toward the side with fewer moles of gas to reduce the pressure.
Here, the product side has fewer moles of gas (2 moles) compared to the reactant side (3 moles). Therefore, increasing the pressure will favor the forward reaction, where the system produces fewer gas molecules.
Hence, the correct answer to the question is:
The forward reaction will be favored.
[tex]\[ 2 \text{H}_2(g) + \text{O}_2(g) \Leftrightarrow 2 \text{H}_2\text{O}(g) \][/tex]
To determine the effect of increasing pressure on the system, we should consider Le Chatelier's principle. This principle states that if a stress is applied to a system at equilibrium, the system will respond in a way that counteracts the stress.
In this scenario, the stress applied is an increase in pressure. We must consider the number of moles of gas on each side of the equilibrium equation to predict the direction in which the reaction will shift.
On the reactant side of the equilibrium equation, we have:
[tex]\[ 2 \text{H}_2(g) + 1 \text{O}_2(g) \Rightarrow 2 + 1 = 3 \text{ moles of gas} \][/tex]
On the product side of the equilibrium equation, we have:
[tex]\[ 2 \text{H}_2\text{O}(g) \Rightarrow 2 \text{ moles of gas} \][/tex]
According to Le Chatelier's principle, if we increase the pressure of the system, the equilibrium will shift toward the side with fewer moles of gas to reduce the pressure.
Here, the product side has fewer moles of gas (2 moles) compared to the reactant side (3 moles). Therefore, increasing the pressure will favor the forward reaction, where the system produces fewer gas molecules.
Hence, the correct answer to the question is:
The forward reaction will be favored.
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