A reaction has `DeltaH=-33kJ and DeltaS=-58J//K`. This reaction would be:

Options

A. Spontaneous at all temperature
B. non-spontaneous at all temperatures
C. spontaneous above a certain temperature only
D. spontaneous below a certain temperature only

Explanation

To determine the nature of the reaction based on the given values of ΔH and ΔS, we can 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
- \(\Delta H = -33 \, \text{kJ} = -33000 \, \text{J}\) (since we need to convert kJ to J)
- \(\Delta S = -58 \, \text{J/K}\)

### Step 2: Substitute the values into the Gibbs free energy equation
We can rewrite the equation as:

\[ \Delta G = -33000 \, \text{J} - T(-58 \, \text{J/K}) \]

This simplifies to:

\[ \Delta G = -33000 \, \text{J} + 58T \]

### Step 3: Determine the conditions for spontaneity
For the reaction to be spontaneous, \(\Delta G\) must be less than zero:

\[ -33000 + 58T < 0 \]

### Step 4: Solve for temperature \(T\)
Rearranging the inequality gives:

\[ 58T < 33000 \]

Dividing both sides by 58:

\[ T < \frac{33000}{58} \approx 569.0 \, \text{K} \]

### Step 5: Analyze the results
- If \(T < 569.0 \, \text{K}\), then \(\Delta G < 0\) and the reaction is spontaneous.
- If \(T > 569.0 \, \text{K}\), then \(\Delta G > 0\) and the reaction is non-spontaneous.

### Conclusion
The reaction is spontaneous below a certain temperature (569.0 K) and non-spontaneous above that temperature. Therefore, the correct answer is that the reaction is spontaneous below a certain temperature.