Consider the equilibrium:
\[
\text{CO(g)} + \text{3H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)}
\]
If the pressure applied over the system increases by two fold at constant temperature then:
Show Hint
- (A) and (B) only
- (A), (B) and (D) only
- (B) and (C) only
- (A), (B) and (C) only
The Correct Option is A
Approach Solution - 1
According to Le Chatelier's principle, increasing the pressure on a gaseous system will favor the side with fewer moles of gas. In the given equilibrium reaction, the left-hand side has 4 moles of gas (1 mole of CO and 3 moles of H\(_2\)) and the right-hand side has 2 moles of gas (1 mole of CH\(_4\) and 1 mole of H\(_2\)O). Therefore, increasing the pressure will shift the equilibrium towards the right (in the forward direction), increasing the concentration of products and decreasing the concentration of reactants.
- (A) The concentration of reactants and products increases because the equilibrium shifts toward the products side.
- (B) The equilibrium will shift in the forward direction to produce more CH\(_4\) and H\(_2\)O.
- (C) The equilibrium constant remains unchanged, as pressure does not affect the value of the equilibrium constant at constant temperature.
Therefore, the correct answer is (1) (A) and (B) only.
Approach Solution -2
\[ \text{CO(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)} \] This is a homogeneous gaseous equilibrium involving 1 mole of CO and 3 moles of H₂ on the left, and 1 mole of CH₄ and 1 mole of H₂O on the right.
Step 2: Calculate the change in the number of moles of gas (Δn).
\[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = (1 + 1) - (1 + 3) = 2 - 4 = -2 \] Hence, the number of moles decreases during the forward reaction.
Step 3: Apply Le Chatelier’s Principle.
When pressure is increased, the equilibrium shifts in the direction that reduces the total number of gaseous moles. Here, since Δn is negative (–2), the equilibrium will shift **toward the right** (the product side).
Therefore, increasing the pressure favors the formation of CH₄ and H₂O (the products).
Step 4: Analyze the result when pressure is doubled at constant temperature.
- The equilibrium shifts toward the side with fewer moles of gas (right side).
- The concentrations of CH₄ and H₂O increase.
- The concentrations of CO and H₂ decrease.
Step 5: Conclusion.
When the pressure on the system is increased twofold at constant temperature:
(A) The equilibrium shifts to the right.
(B) Formation of CH₄ and H₂O increases.
Final Answer:
(A) and (B) only.
\[ \boxed{\text{(A) and (B) only}} \]
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