Reaction stoichiometry, kinetics and thermodynamics ltbgt Nitrosyl chloride `(NOCl)`, is a yellow gas that is most commonly encountered as a decomposition product of aqua regia. It is toxic and irritating to the lungs. On heating `NOCl` decomposes as
`2NOClrarr2NO+Cl_(2)`.
The enthalpy change `(DeltaH)` for the formation of `1` mole of `Cl_(2)` by the decomposition of `NOCl` is `75.3KJ` between `100.K "to" 600K`. The standard entropies `(S^(@)_(298K))` of different species are as given below:
`{:("Substance",NOCl,NO,Cl_(2)),(S_(298K)^(@),264,211,223):}`
Calculate the temperature at which `K_(p)` will be double the value at `298K`.

Answer

To solve the problem, we need to find the temperature at which the equilibrium constant \( K_p \) for the decomposition of nitrosyl chloride \( (NOCl) \) will be double its value at \( 298K \). We will use the Van't Hoff equation for this purpose.

### Step-by-Step Solution:

1. **Write the Reaction:**
The decomposition of nitrosyl chloride is given by:
\[ 2 NOCl \rightleftharpoons 2 NO + Cl_2 \]

2. **Identify Given Data:**
- Enthalpy change \( \Delta H = 75.3 \, \text{kJ/mol} \) (which is \( 75300 \, \text{J/mol} \))
- Standard entropies at \( 298K \):
- \( S_{NOCl} = 264 \, \text{J/(mol·K)} \)
- \( S_{NO} = 211 \, \text{J/(mol·K)} \)
- \( S_{Cl_2} = 223 \, \text{J/(mol·K)} \)

3. **Calculate the Change in Entropy \( \Delta S \):**
The change in entropy for the reaction can be calculated using the formula:
\[ \Delta S = \sum S_{products} - \sum S_{reactants} \]
For the reaction:
\[ \Delta S = [2 \times S_{NO} + S_{Cl_2}] - [2 \times S_{NOCl}] \]
Substituting the values:
\[ \Delta S = [2 \times 211 + 223] - [2 \times 264] \]
\[ \Delta S = [422 + 223] - [528] = 645 - 528 = 117 \, \text{J/(mol·K)} \]

4. **Use the Van't Hoff Equation:**
The Van't Hoff equation relates the change in the equilibrium constant \( K_p \) with temperature:
\[ \ln \left( \frac{K_{p2}}{K_{p1}} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Here, we know \( K_{p2} = 2 K_{p1} \), so:
\[ \ln(2) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{298} \right) \]

5. **Substituting Values:**
- \( R = 8.314 \, \text{J/(mol·K)} \)
- \( \Delta H = 75300 \, \text{J/mol} \)
- \( \ln(2) \approx 0.693 \)

The equation becomes:
\[ 0.693 = -\frac{75300}{8.314} \left( \frac{1}{T_2} - \frac{1}{298} \right) \]

6. **Rearranging the Equation:**
\[ \frac{1}{T_2} - \frac{1}{298} = -\frac{0.693 \times 8.314}{75300} \]
Calculate the right side:
\[ -\frac{0.693 \times 8.314}{75300} \approx -0.000068 \]
Therefore,
\[ \frac{1}{T_2} = -0.000068 + \frac{1}{298} \]
Calculate \( \frac{1}{298} \):
\[ \frac{1}{298} \approx 0.003356 \]
Thus,
\[ \frac{1}{T_2} \approx 0.003356 - 0.000068 \approx 0.003288 \]

7. **Calculate \( T_2 \):**
\[ T_2 \approx \frac{1}{0.003288} \approx 304.5 \, K \]

### Final Answer:
The temperature at which \( K_p \) will be double the value at \( 298K \) is approximately \( 305 \, K \).