If pressure increases then its effect on given equilibrium `C_((s)) + H_(2)O_((g))hArr CO_((g)) + H_(2(g))` it is satisfied in

Options

A. Forward direction
B. Backward direction
C. No effect
D. None of these

Explanation

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

\[ C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_2_{(g)} \]

we will follow these steps:

### Step 1: Identify the states of the reactants and products
- The reactants are:
- Carbon (C) in solid state (s)
- Water (H2O) in gaseous state (g)
- The products are:
- Carbon monoxide (CO) in gaseous state (g)
- Hydrogen (H2) in gaseous state (g)

### Step 2: Count the number of moles of gaseous reactants and products
- Gaseous reactants:
- 1 mole of H2O (g)
- Gaseous products:
- 1 mole of CO (g) + 1 mole of H2 (g) = 2 moles of gas

### Step 3: Calculate the change in the number of moles of gas (Δn)
- Δn = (moles of gaseous products) - (moles of gaseous reactants)
- Δn = 2 (from products) - 1 (from reactants) = 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.
- Since Δn = 1 (which is greater than 0), this indicates that there are more moles of gas on the product side than on the reactant side.

### Step 5: Determine the direction of the shift
- Because the reaction has more moles of gas on the product side, an increase in pressure will cause the equilibrium to shift to the left (toward the reactants) to reduce the number of gaseous moles.

### Conclusion
- Therefore, if the pressure increases, the equilibrium will shift to the left, favoring the formation of reactants.