What is the effect of reducing the volume on the system described below? 2C(s) + O2(g)= 2CO(g)

Explanation

To analyze the effect of reducing the volume on the given system, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ 2C(s) + O_2(g) \rightleftharpoons 2CO(g) \]

In this reaction:
- **Reactants**: 2 moles of solid carbon (C) and 1 mole of gaseous oxygen (O₂).
- **Products**: 2 moles of gaseous carbon monoxide (CO).

### Step 2: Identify the Moles of Gases
Next, we need to count the number of moles of gases on both sides of the reaction:
- **Reactants**: 1 mole of O₂ (gaseous).
- **Products**: 2 moles of CO (gaseous).

### Step 3: Apply Le Chatelier's Principle
Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in conditions (such as pressure, temperature, or volume), the system will adjust itself to counteract that change and restore a new equilibrium.

### Step 4: Effect of Reducing Volume
When we reduce the volume of the system:
- The pressure of the gaseous components increases.
- According to Le Chatelier's Principle, the system will shift in the direction that reduces the pressure.

### Step 5: Determine the Direction of Shift
In our case:
- The reactants side has 1 mole of gas (O₂).
- The products side has 2 moles of gas (CO).

Since there are fewer moles of gas on the reactants side (1 mole) compared to the products side (2 moles), reducing the volume will shift the equilibrium towards the side with fewer moles of gas to decrease the pressure.

### Conclusion
Thus, reducing the volume of the system will shift the equilibrium to the left, favoring the formation of the reactants (O₂ and C).

### Summary of Steps:
1. Identify the reaction and its components.
2. Count the moles of gas on both sides.
3. Apply Le Chatelier's Principle.
4. Analyze the effect of reducing volume on pressure.
5. Determine the direction of the equilibrium shift.