`C(s) +H_2O(g) hArr CO(g) +H_2(g) , DeltaH lt 0`
the above equilibrium will proceed in forward direction when

Options

A. it is subjected to high pressure
B. it is subjected to high temperature
C. Inert gas (Argon) is added at constant pressure
D. Carbon (solid) is added

Answer

To determine the conditions under which the equilibrium reaction

\[ C(s) + H_2O(g) \rightleftharpoons CO(g) + H_2(g) \quad (\Delta H < 0) \]

will proceed in the forward direction, we will analyze each option provided.

### Step 1: Understand the Reaction
The reaction involves solid carbon and water vapor producing carbon monoxide and hydrogen gas. The enthalpy change (\(\Delta H\)) is negative, indicating that the reaction is exothermic (it releases heat).

### Step 2: Apply Le Chatelier's Principle
Le Chatelier's principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to counteract that change and restore a new equilibrium.

### Step 3: Analyze Each Option

**Option 1: Subjected to High Pressure**
- In the reaction, there is 1 mole of gaseous reactant (H2O) and 2 moles of gaseous products (CO and H2).
- Increasing the pressure favors the side with fewer moles of gas.
- Since the reactants have fewer moles of gas compared to the products, increasing pressure will shift the equilibrium to the left (backward direction).
- **Conclusion:** This option is incorrect.

**Option 2: Subjected to High Temperature**
- Since the reaction is exothermic, increasing the temperature will favor the endothermic direction (the reverse reaction).
- Therefore, increasing the temperature will shift the equilibrium to the left (backward direction).
- **Conclusion:** This option is incorrect.

**Option 3: Inert Gas Argon is Added at Constant Pressure**
- Adding an inert gas at constant pressure increases the total volume of the system without changing the partial pressures of the reactants and products.
- According to Le Chatelier's principle, the system will shift towards the side with more moles of gas to counteract the increase in volume.
- The product side has 2 moles of gas (CO and H2), while the reactant side has only 1 mole of gas (H2O).
- **Conclusion:** This option is correct, as it shifts the equilibrium to the right (forward direction).

**Option 4: Solid is Added**
- The addition of a solid does not affect the equilibrium position because the concentration of solids does not change.
- Therefore, adding more solid carbon will have no effect on the equilibrium.
- **Conclusion:** This option is incorrect.

### Final Answer
The equilibrium will proceed in the forward direction when **inert gas argon is added at constant pressure** (Option 3).

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