In what manner will increases of pressure affect the following equation ?
`C(s) +H_2O(g) hArr CO(g) +H_2(g)`

Options

A. Shift in the forward direction
B. shift in the reverse reaction
C. increase in the yield of `H_2`
D. No effect

Answer

To analyze how an increase in pressure affects the equilibrium of the reaction:

**Given Reaction:**
\[ C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g) \]

**Step 1: Identify the states of the reactants and products.**
- Reactants:
- Carbon (C) is a solid (s).
- Water vapor (H₂O) is a gas (g).
- Products:
- Carbon monoxide (CO) is a gas (g).
- Hydrogen (H₂) is a gas (g).

**Step 2: Count the number of moles of gaseous reactants and products.**
- Gaseous reactants: 1 mole of H₂O
- Gaseous products: 1 mole of CO + 1 mole of H₂ = 2 moles

**Step 3: Calculate the change in the number of moles of gas (ΔN).**
\[ \Delta N = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
\[ \Delta N = 2 - 1 = 1 \]

**Step 4: Apply Le Chatelier's Principle.**
According to Le Chatelier's Principle, if the pressure of a system at equilibrium is increased, the equilibrium will shift in the direction that produces fewer moles of gas.

**Step 5: Determine the effect of increased pressure.**
Since we have:
- 1 mole of gaseous reactant (H₂O)
- 2 moles of gaseous products (CO and H₂)

The increase in pressure will favor the side with fewer moles of gas. In this case, the reactant side has 1 mole of gas, while the product side has 2 moles of gas.

**Conclusion:**
An increase in pressure will shift the equilibrium to the left (toward the reactants), which means it will favor the formation of H₂O and solid carbon (C).

**Final Answer:**
The increase in pressure will shift the equilibrium in the reverse direction (toward the reactants).

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