For which of the reaction, the ratio `(K_(P))/(K_(C))` is maximum and minimum respectively.
(a) `X(g)hArr2Y(g)` (b) `2X(g)hArrY(g)`
(c) `X(g)hArr2Y(g)+Z(g)` (d) `X(s)hArrY(g)`

Options

A. c,b
B. a,b
C. c,d
D. a,c

Explanation

To determine which reaction has the maximum and minimum ratio of \( \frac{K_p}{K_c} \), we can use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c \cdot R^T \cdot \Delta n \]

Where:
- \( R \) is the gas constant,
- \( T \) is the temperature,
- \( \Delta n \) is the change in moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

Thus, we can express the ratio \( \frac{K_p}{K_c} \) as:

\[ \frac{K_p}{K_c} = R^T \cdot \Delta n \]

Since \( R \) and \( T \) are constants, the ratio \( \frac{K_p}{K_c} \) depends solely on \( \Delta n \). Therefore, we need to calculate \( \Delta n \) for each of the given reactions.

### Step 1: Calculate \( \Delta n \) for each reaction

1. **Reaction (a)**: \( X(g) \rightleftharpoons 2Y(g) \)
- Moles of products = 2 (from \( 2Y \))
- Moles of reactants = 1 (from \( X \))
- \( \Delta n = 2 - 1 = 1 \)

2. **Reaction (b)**: \( 2X(g) \rightleftharpoons Y(g) \)
- Moles of products = 1 (from \( Y \))
- Moles of reactants = 2 (from \( 2X \))
- \( \Delta n = 1 - 2 = -1 \)

3. **Reaction (c)**: \( X(g) \rightleftharpoons 2Y(g) + Z(g) \)
- Moles of products = 3 (from \( 2Y + Z \))
- Moles of reactants = 1 (from \( X \))
- \( \Delta n = 3 - 1 = 2 \)

4. **Reaction (d)**: \( X(s) \rightleftharpoons Y(g) \)
- Moles of products = 1 (from \( Y \))
- Moles of reactants = 0 (since \( X \) is a solid)
- \( \Delta n = 1 - 0 = 1 \)

### Step 2: Compare \( \Delta n \) values

- For Reaction (a): \( \Delta n = 1 \)
- For Reaction (b): \( \Delta n = -1 \)
- For Reaction (c): \( \Delta n = 2 \)
- For Reaction (d): \( \Delta n = 1 \)

### Step 3: Determine maximum and minimum \( \frac{K_p}{K_c} \)

- The maximum \( \frac{K_p}{K_c} \) occurs when \( \Delta n \) is highest, which is for Reaction (c) with \( \Delta n = 2 \).
- The minimum \( \frac{K_p}{K_c} \) occurs when \( \Delta n \) is lowest, which is for Reaction (b) with \( \Delta n = -1 \).

### Conclusion

- Maximum \( \frac{K_p}{K_c} \): Reaction (c) \( X(g) \rightleftharpoons 2Y(g) + Z(g) \)
- Minimum \( \frac{K_p}{K_c} \): Reaction (b) \( 2X(g) \rightleftharpoons Y(g) \)

### Final Answer

- Maximum: (c)
- Minimum: (b)

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