For a given reaction, `DeltaH=35.5 KJ "mol"^(-1)` and `DeltaS=83.6 JK^(-1) "mol"^(-1)`. The reaction is spontaneous at: (Assume that `DeltaH and deltaS` so not vary with temperature)

Options

A. `T lt 425 K`
B. `T gt 425 K`
C. All temperatures
D. `T gt 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 zero (\(\Delta G < 0\)). This leads us to the inequality:

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

Rearranging this gives us:

\[ \Delta H < T \Delta S \]

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

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

### Step 1: Convert \(\Delta H\) to the correct units
Given:
\[ \Delta H = 35.5 \, \text{kJ/mol} \]
We need to convert this to Joules:
\[ \Delta H = 35.5 \times 10^3 \, \text{J/mol} = 35500 \, \text{J/mol} \]

### Step 2: Use the given \(\Delta S\)
Given:
\[ \Delta S = 83.6 \, \text{J/K/mol} \]

### Step 3: Substitute values into the temperature inequality
Now we substitute \(\Delta H\) and \(\Delta S\) into the inequality:

\[ T > \frac{35500 \, \text{J/mol}}{83.6 \, \text{J/K/mol}} \]

### Step 4: Calculate the temperature
Calculating the right-hand side gives:

\[ T > \frac{35500}{83.6} \approx 425.2 \, \text{K} \]

### Conclusion
Thus, the reaction is spontaneous at temperatures greater than approximately 425 K.

### Final Answer
The reaction is spontaneous at temperatures greater than 425 K.

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