Understanding Equilibrium Shift

The question concerns the reaction $2C(s) + O_2(g) \rightleftharpoons 2CO(g)$ and asks how to favor the forward direction (formation of $CO(g)$).

Applying Le Chatelier's Principle

Le Chatelier's principle helps predict how equilibrium changes in response to external conditions. It states that the system will shift to counteract the applied change.

Analyzing Pressure Impact

We need to consider the number of gas moles on each side of the equation:

• Reactant side: 1 mole of $O_2(g)$
• Product side: 2 moles of $CO(g)$

Note that solid carbon ($C(s)$) does not influence the equilibrium position because its concentration remains constant.

Decreasing the pressure exerts a stress on the system. To counteract this, the equilibrium will shift towards the side with a greater number of gas moles, as this increases the overall pressure within the system. In this case, the product side has more gas moles (2 moles of $CO$) compared to the reactant side (1 mole of $O_2$). Thus, decreasing the pressure favors the forward reaction.

Evaluating Other Factors

• Increasing carbon amount: Changing the amount of a pure solid reactant or product does not shift the equilibrium.
• Decreasing volume: This action increases the system's pressure. According to Le Chatelier's principle, increasing pressure shifts the equilibrium towards the side with fewer gas moles. Here, that is the reactant side (1 mole $O_2$ vs 2 moles $CO$), so it opposes the forward direction.
• Increasing temperature: The reaction $2C(s) + O_2(g) \rightarrow 2CO(g)$ is exothermic (releases heat). Increasing the temperature favors the endothermic (reverse) direction to absorb the added heat, thus shifting the equilibrium away from the forward direction.