For a given reaction, `DeltaH = 35.5 kJ mol^(-1)` and `Delta S = 83.6 JK^(-1) mol^(-1)`. The reaction is spontaneous at (Assume that `DeltaH and DeltaS` do not vary with temperature)

Options

A. T lt 425 K
B. Tgt 425 K
C. All temperatures
D. Tgt 298 K

Explanation

To determine the temperature at which the given reaction is spontaneous, we can use the Gibbs free energy equation:

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

For a reaction to be spontaneous, \(\Delta G\) must be less than or equal to zero. Therefore, we can set up the inequality:

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

Rearranging this gives:

\[ T \Delta S \geq \Delta H \]

From this, we can express the temperature \(T\) as:

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

### Step-by-Step Solution

1. **Identify the given values**:
- \(\Delta H = 35.5 \, \text{kJ/mol}\)
- \(\Delta S = 83.6 \, \text{J/K/mol}\)

2. **Convert \(\Delta H\) from kJ to J**:
- Since \(1 \, \text{kJ} = 1000 \, \text{J}\),
\[ \Delta H = 35.5 \, \text{kJ/mol} = 35.5 \times 1000 \, \text{J/mol} = 35500 \, \text{J/mol} \]

3. **Substitute the values into the temperature equation**:
\[ T \geq \frac{\Delta H}{\Delta S} = \frac{35500 \, \text{J/mol}}{83.6 \, \text{J/K/mol}} \]

4. **Calculate the temperature**:
\[ T \geq \frac{35500}{83.6} \approx 424.4 \, \text{K} \]

5. **Conclusion**:
The reaction is spontaneous at temperatures greater than or equal to \(424.4 \, \text{K}\).

### Final Answer:
The reaction is spontaneous at temperatures \(T \geq 424.4 \, \text{K}\).