The equilibrium constant for the reaction
\( \text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g) \)
is \( K_C = 4.9 \times 10^{-2} \). The value of \( K_C \) for the reaction given below is
\( 2 \text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{SO}_3 \, (g) \) is
\( \text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g) \)
is \( K_C = 4.9 \times 10^{-2} \). The value of \( K_C \) for the reaction given below is
\( 2 \text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{SO}_3 \, (g) \) is
- 4.9
- 41.6
- 49
- 416
The Correct Option is D
Approach Solution - 1
Given the equilibrium constant for the reaction:
\[\text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g)\]
is \( K_c = 4.9 \times 10^{-2} \).
For the reaction:
\[2\text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2\text{SO}_3 \, (g)\]
we need the equilibrium constant \( K_c' \).
Since the second reaction is the reverse and doubled version of the original reaction, we calculate \( K_c' \) as:
\[K_c' = \left( \frac{1}{K_c} \right)^2 = \left( \frac{1}{4.9 \times 10^{-2}} \right)^2\]
\[K_c' = 416.49\]
Approach Solution -2
Given equilibrium reaction (1):
\[ \text{SO}_3 (g) \rightleftharpoons \text{SO}_2 (g) + \tfrac{1}{2} \text{O}_2 (g) \] and its equilibrium constant is \( K_{C1} = 4.9 \times 10^{-2}. \)
We need to find the equilibrium constant for reaction (2):
\[ 2\text{SO}_2 (g) + \text{O}_2 (g) \rightleftharpoons 2\text{SO}_3 (g) \] Let its equilibrium constant be \( K_{C2}. \)
Step 2: Relate the two reactions
Reaction (2) is the reverse of reaction (1) multiplied by 2.
To see this clearly, reverse reaction (1):
\[ \text{SO}_2 (g) + \tfrac{1}{2}\text{O}_2 (g) \rightleftharpoons \text{SO}_3 (g) \] For the reverse of (1), the equilibrium constant becomes the reciprocal: \[ K_{\text{reverse}} = \frac{1}{K_{C1}} = \frac{1}{4.9 \times 10^{-2}}. \] Now, multiply this reversed reaction by 2 to get exactly reaction (2):
\[ 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3. \] When we multiply a balanced reaction by a factor of 2, the new equilibrium constant becomes the square of the old one: \[ K_{C2} = (K_{\text{reverse}})^2 = \left( \frac{1}{K_{C1}} \right)^2. \]
Step 3: Substitute and calculate
\[ K_{C2} = \left( \frac{1}{4.9 \times 10^{-2}} \right)^2. \] Compute step by step: \[ \frac{1}{4.9 \times 10^{-2}} = \frac{1}{0.049} \approx 20.41. \] Now square this value: \[ K_{C2} = (20.41)^2 \approx 416.6. \] Hence, \[ K_{C2} \approx 416. \]
Step 4: Verify logical consistency
Since reaction (2) forms SO₃ from SO₂ and O₂ (the reverse of the decomposition), it is expected to have a large \(K_C\) value because the decomposition equilibrium constant was small. Therefore, \(K_C = 416\) is consistent with the expected direction of spontaneity.
Final answer
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