Enthalpy of sublimation of iodine is `"24 cal g"^(-1)" at " 200^(@)C`. If specific heat of `l_(2)(s) and l_(2)` (vap) are 0.055 and 0.031 respectively, then enthalpy of sublimation of iodine at `250^(@)C` in `"cal g"^(-1)` is:

Options

A. 5 . 7
B. 22.8
C. 11.4
D. 2.85

Explanation

To find the enthalpy of sublimation of iodine at 250°C, we can follow these steps:

### Step 1: Understand the Given Data
- Enthalpy of sublimation at 200°C (T1): \( \Delta H_1 = 24 \, \text{cal g}^{-1} \)
- Specific heat of solid iodine (\( C_{p, \text{solid}} \)): \( 0.055 \, \text{cal g}^{-1} \, \text{°C}^{-1} \)
- Specific heat of gaseous iodine (\( C_{p, \text{gas}} \)): \( 0.031 \, \text{cal g}^{-1} \, \text{°C}^{-1} \)
- Initial temperature (T1): \( 200 \, \text{°C} \)
- Final temperature (T2): \( 250 \, \text{°C} \)

### Step 2: Calculate the Change in Specific Heat Capacity (\( \Delta C_p \))
\[ \Delta C_p = C_{p, \text{gas}} - C_{p, \text{solid}} = 0.031 - 0.055 = -0.024 \, \text{cal g}^{-1} \, \text{°C}^{-1} \]

### Step 3: Convert Temperatures to Kelvin
- \( T_1 = 200 + 273 = 473 \, \text{K} \)
- \( T_2 = 250 + 273 = 523 \, \text{K} \)

### Step 4: Calculate the Change in Temperature (\( \Delta T \))
\[ \Delta T = T_2 - T_1 = 523 - 473 = 50 \, \text{K} \]

### Step 5: Calculate the Change in Enthalpy (\( \Delta H \))
Using the formula:
\[ \Delta H = \Delta C_p \times \Delta T \]
Substituting the values:
\[ \Delta H = -0.024 \, \text{cal g}^{-1} \, \text{°C}^{-1} \times 50 \, \text{°C} = -1.2 \, \text{cal g}^{-1} \]

### Step 6: Calculate the Enthalpy of Sublimation at 250°C (\( \Delta H_2 \))
Using the relationship:
\[ \Delta H_2 = \Delta H + \Delta H_1 \]
Substituting the values:
\[ \Delta H_2 = -1.2 + 24 = 22.8 \, \text{cal g}^{-1} \]

### Final Answer
The enthalpy of sublimation of iodine at 250°C is:
\[ \Delta H_2 = 22.8 \, \text{cal g}^{-1} \]