Reaction stoichiometry, kinetics and thermodynamics that of 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 above which the reaction will become non-spontaneous.

Answer

To determine the temperature above which the decomposition of nitrosyl chloride (NOCl) becomes non-spontaneous, we will use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- \(\Delta G\) = Gibbs free energy change
- \(\Delta H\) = Enthalpy change
- \(T\) = Temperature in Kelvin
- \(\Delta S\) = Entropy change

### Step 1: Identify the given values
- \(\Delta H = 75.3 \, \text{kJ} = 75300 \, \text{J}\) (since we need to convert kJ to J)
- Standard entropies at 298 K:
- \(S^\circ_{NOCl} = 264 \, \text{J/K}\)
- \(S^\circ_{NO} = 211 \, \text{J/K}\)
- \(S^\circ_{Cl_2} = 223 \, \text{J/K}\)

### Step 2: Calculate the entropy change (\(\Delta S\))
Using the formula for the entropy change of the reaction:

\[ \Delta S^\circ = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]

For the reaction:

\[ 2 \, \text{NOCl} \rightarrow 2 \, \text{NO} + \text{Cl}_2 \]

The entropy change can be calculated as:

\[ \Delta S^\circ = [2 \times S^\circ_{NO} + S^\circ_{Cl_2}] - [2 \times S^\circ_{NOCl}] \]

Substituting the values:

\[ \Delta S^\circ = [2 \times 211 + 223] - [2 \times 264] \]
\[ = [422 + 223] - [528] \]
\[ = 645 - 528 = 117 \, \text{J/K} \]

### Step 3: Set up the Gibbs free energy equation for non-spontaneity
For the reaction to become non-spontaneous, \(\Delta G\) must be equal to zero:

\[ 0 = \Delta H - T \Delta S \]

Rearranging gives:

\[ T \Delta S = \Delta H \]

Thus:

\[ T = \frac{\Delta H}{\Delta S} \]

### Step 4: Substitute the values to find the temperature
Substituting the known values:

\[ T = \frac{75300 \, \text{J}}{117 \, \text{J/K}} \]
\[ T \approx 643.59 \, \text{K} \]

### Conclusion
The temperature above which the reaction will become non-spontaneous is approximately **643.6 K**.

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