In what manner will the increase of pressure affect the following equation ? ` C(s) + H_(2) O(g) hArr CO(g) +H_(2)(g)`

Options

A. Shift in the forward direction
B. Shift in the reverse direction
C. increase in the yield of hydrogen
D. No effect

Explanation

To analyze the effect of an increase in pressure on the equilibrium of the reaction:

\[ C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g) \]

we will follow these steps:

### Step 1: Identify the gaseous reactants and products
In this reaction, we have:
- Gaseous reactants: \( H_2O(g) \)
- Gaseous products: \( CO(g) \) and \( H_2(g) \)

### Step 2: Calculate the change in the number of moles of gas (ΔN)
To determine how the equilibrium will shift with a change in pressure, we need to calculate ΔN, which is defined as the number of moles of gaseous products minus the number of moles of gaseous reactants.

- Moles of gaseous products:
- \( CO(g) \): 1 mole
- \( H_2(g) \): 1 mole
- Total = 1 + 1 = 2 moles

- Moles of gaseous reactants:
- \( H_2O(g) \): 1 mole
- Total = 1 mole

Now, calculate ΔN:
\[ \Delta N = \text{Moles of products} - \text{Moles of reactants} = 2 - 1 = 1 \]

### Step 3: Analyze the effect of pressure increase
According to Le Chatelier's principle, if the pressure of a system at equilibrium is increased, the system will respond by shifting the equilibrium position to the side with fewer moles of gas to counteract the change.

In our case:
- The reactant side has 1 mole of gas (1 mole of \( H_2O(g) \)).
- The product side has 2 moles of gas (1 mole of \( CO(g) \) + 1 mole of \( H_2(g) \)).

Since there are more moles of gas on the product side (2 moles) compared to the reactant side (1 mole), an increase in pressure will cause the equilibrium to shift towards the side with fewer moles of gas, which is the reactant side.

### Conclusion
Therefore, when the pressure is increased, the equilibrium will shift to the left, favoring the formation of \( C(s) + H_2O(g) \).

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