Discuss the effect of increase of pressure on the following reactions
`CH_4(g) +2O_2(g) hArr CO_2(g) +2H_2O(g)`

Answer

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

\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightleftharpoons \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \]

we will follow these steps:

### Step 1: Identify the reaction and its components
The reaction involves:
- Reactants: 1 mole of methane (CH₄) and 2 moles of oxygen (O₂)
- Products: 1 mole of carbon dioxide (CO₂) and 2 moles of water (H₂O)

### Step 2: Count the total number of moles on each side
- Total moles of reactants = 1 (CH₄) + 2 (O₂) = 3 moles
- Total moles of products = 1 (CO₂) + 2 (H₂O) = 3 moles

### Step 3: 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 4: Analyze the effect of pressure increase
When pressure is increased, the equilibrium will shift towards the side with fewer moles of gas. In this case:
- Reactants: 3 moles of gas
- Products: 3 moles of gas

Since the number of moles of gas is the same on both sides of the reaction (3 moles on the reactant side and 3 moles on the product side), there is no side with fewer moles.

### Step 5: Conclusion
Therefore, increasing the pressure will have no effect on the position of equilibrium for this reaction, as the number of moles of gas is equal on both sides.

### Summary
In conclusion, for the reaction:

\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightleftharpoons \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \]

an increase in pressure will not affect the equilibrium position because the total number of moles of gas is the same on both sides of the reaction.

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