For the reaction at 298 K
`2A + B to C`
`DeltaH = 40 kJ "mol"^(-1)` and `DeltaS = 0.2 kJ K^(-1) "mol"^(-1)`. At what temperature will the reaction beccomes sponatneous considering `DeltaH` and `DeltaS` to be constant over the temperature range.

Explanation

To determine the temperature at which the reaction becomes spontaneous, we need to analyze the Gibbs free energy change (ΔG) for the reaction. The relationship between ΔG, ΔH (enthalpy change), and ΔS (entropy change) is given by the equation:

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

Where:
- ΔG = Gibbs free energy change
- ΔH = Enthalpy change (given as 40 kJ/mol)
- ΔS = Entropy change (given as 0.2 kJ/K·mol)
- T = Temperature in Kelvin

### Step 1: Set ΔG to Zero for Spontaneity
For a reaction to be spontaneous, ΔG must be less than or equal to zero. At the point of spontaneity, we can set ΔG to zero:

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

### Step 2: Rearrange the Equation
Rearranging the equation gives us:

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

### Step 3: Solve for Temperature (T)
Now, we can solve for T:

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

### Step 4: Substitute the Given Values
Substituting the values for ΔH and ΔS:

\[ T = \frac{40 \text{ kJ/mol}}{0.2 \text{ kJ/K·mol}} \]

### Step 5: Calculate the Temperature
Calculating the above expression:

\[ T = \frac{40}{0.2} = 200 \text{ K} \]

### Conclusion
The reaction will become spontaneous at temperatures greater than 200 K.

### Summary
- The reaction becomes spontaneous when the temperature exceeds 200 K.