Chemical reactions are invariably associated with the transfer of energy either in the form of hear or light. In the laboratory, heat changes in physical and chemical processes are measured with an instrument called calorimeter. Heat change in the process is calculated as: q= ms `Delta T`, s= Specific heat = `c Delta T`= Heat capacity. Heat of reaction at constant pressure is measured using simple or water calorimeter. `Q_(v)= Delta U`= Internal energy change, `Q_(P) = DeltaH, Q_(P) = Q_(V) + P Delta V and DeltaH = Delta U+ Delta nRT`. The amount of energy released during a chemical change depends on the physical state of reactants and products, the condition of pressure, temperature and volume at which the reaction is carried out. The variation of heat of reaction with temperature and pressure is given by Kirchoff's equation: `(DeltaH_(2) - DeltaH_(1))/(T_(2)-T_(1))= Delta C_(P)` (At constant pressure), `(DeltaU_(2) - DeltaU_(1))/(T_(2)-T_(1)) = DeltaC_(V)` (At constant volume)
The specific heat of `I_(2)` in vapoour and solid state are 0.031 and 0.055 cal/g respectively. The heat of sublimation of iodine at `200^(@)C` is 6.096 kcal `mol^(-1)`. The heat of sublimation of iodine at `250^(0)C` will be

Options

A. `3.8 "k cal" mol^(-1)`
B. `4.8 "k cal" mol^(-1)`
C. `2.28 "k cal" mol^(-1)`
D. `5.8 "k cal" mol^(-1)`

Answer

To find the heat of sublimation of iodine at 250°C using Kirchhoff's equation, we can follow these steps:

### Step 1: Identify the known values
- **T1** = 200°C
- **T2** = 250°C
- **ΔH (at T1)** = 6.096 kcal/mol
- **Specific heat of solid iodine (s1)** = 0.055 cal/g
- **Specific heat of vapor iodine (s2)** = 0.031 cal/g

### Step 2: Calculate the change in specific heat (ΔCp)
Using the specific heats of iodine in solid and vapor states, we can calculate ΔCp:
\[ \Delta C_p = C_{p, vapor} - C_{p, solid} = 0.031 \, \text{kcal/g} - 0.055 \, \text{kcal/g} = -0.024 \, \text{kcal/g} \]

### Step 3: Apply Kirchhoff's equation
According to Kirchhoff's equation:
\[ \frac{\Delta H_2 - \Delta H_1}{T_2 - T_1} = \Delta C_p \]
Substituting the known values:
\[ \frac{\Delta H_{250} - 6.096}{250 - 200} = -0.024 \]

### Step 4: Rearrange and solve for ΔH at 250°C
Rearranging the equation gives:
\[ \Delta H_{250} - 6.096 = -0.024 \times (250 - 200) \]
\[ \Delta H_{250} - 6.096 = -0.024 \times 50 \]
\[ \Delta H_{250} - 6.096 = -1.2 \]
\[ \Delta H_{250} = 6.096 - 1.2 \]
\[ \Delta H_{250} = 4.896 \, \text{kcal/mol} \]

### Step 5: Round to the nearest option
The nearest option to 4.896 kcal/mol is **4.8 kcal/mol**.

### Final Answer:
The heat of sublimation of iodine at 250°C is approximately **4.8 kcal/mol**.

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