The chemical reaction : A `rarr` P, `Delta H^(@)` = 2.8 kJ is spontaneous only above 400 K. Therefore `Delta S` of reaction must be at least `(JK^(-1))`.

Answer

To solve the problem, we need to determine the minimum value of the entropy change (ΔS) for the reaction A → P, given that the reaction is spontaneous only above 400 K and that the standard enthalpy change (ΔH) is 2.8 kJ.

### Step-by-Step Solution:

1. **Understand the Conditions**: The reaction is spontaneous at temperatures greater than 400 K. This means that at 400 K, the reaction is at equilibrium, and the Gibbs free energy change (ΔG) is zero.

2. **Use the Gibbs Free Energy Equation**: The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
At equilibrium (400 K), we set ΔG to 0:
\[ 0 = \Delta H - T \Delta S \]

3. **Convert ΔH to Joules**: The given ΔH is 2.8 kJ. We need to convert this into joules:
\[ \Delta H = 2.8 \, \text{kJ} = 2.8 \times 10^3 \, \text{J} \]

4. **Substitute Known Values into the Equation**: Now we can substitute ΔH and T into the equation:
\[ 0 = 2.8 \times 10^3 \, \text{J} - (400 \, \text{K}) \Delta S \]

5. **Rearrange to Solve for ΔS**: Rearranging the equation gives:
\[ (400 \, \text{K}) \Delta S = 2.8 \times 10^3 \, \text{J} \]
\[ \Delta S = \frac{2.8 \times 10^3 \, \text{J}}{400 \, \text{K}} \]

6. **Calculate ΔS**: Performing the division:
\[ \Delta S = \frac{2800 \, \text{J}}{400 \, \text{K}} = 7 \, \text{J/K} \]

7. **Conclusion**: Therefore, the minimum value of ΔS for the reaction is:
\[ \Delta S \geq 7 \, \text{J/K} \]

### Final Answer:
The value of ΔS must be at least **7 J/K**.