Question
Consider system at equilibrium:
2CO(g) + O2(g) -> 2CO2(g)
How will decreasing the volume of the container shift the equilibrium?
1. to the left
2. No effect
3. to the right
2CO(g) + O2(g) -> 2CO2(g)
How will decreasing the volume of the container shift the equilibrium?
1. to the left
2. No effect
3. to the right
Answers
GPT-4o mini
To determine how a decrease in the volume of the container will affect the equilibrium of the reaction:
\[ 2CO(g) + O2(g) \rightleftharpoons 2CO2(g) \]
we can apply Le Chatelier's Principle, which states that if a system at equilibrium is subjected to a change (such as a change in volume), the system will adjust to counteract that change and restore a new equilibrium.
1. **Initial State**: The reaction has 3 moles of gas on the reactant side (2 moles of \( CO \) and 1 mole of \( O2 \)) and 2 moles of gas on the product side (2 moles of \( CO2 \)).
2. **Decrease in Volume**: Reducing the volume of the container increases the pressure of the system. According to Le Chatelier’s Principle, the equilibrium will shift toward the side with fewer moles of gas in order to decrease the pressure.
In this case, since the reactants side has 3 moles of gas and the products side has 2 moles of gas, the equilibrium will shift to the right, toward the formation of \( CO2 \).
Therefore, the answer is:
**3. to the right**.
\[ 2CO(g) + O2(g) \rightleftharpoons 2CO2(g) \]
we can apply Le Chatelier's Principle, which states that if a system at equilibrium is subjected to a change (such as a change in volume), the system will adjust to counteract that change and restore a new equilibrium.
1. **Initial State**: The reaction has 3 moles of gas on the reactant side (2 moles of \( CO \) and 1 mole of \( O2 \)) and 2 moles of gas on the product side (2 moles of \( CO2 \)).
2. **Decrease in Volume**: Reducing the volume of the container increases the pressure of the system. According to Le Chatelier’s Principle, the equilibrium will shift toward the side with fewer moles of gas in order to decrease the pressure.
In this case, since the reactants side has 3 moles of gas and the products side has 2 moles of gas, the equilibrium will shift to the right, toward the formation of \( CO2 \).
Therefore, the answer is:
**3. to the right**.