For a given reaction `DeltaH="35.5 k Jmol"^(-1) and DeltaS="83.6 Jk"^(-1)"mol"^(-1)`. The reaction is spontaneous at
(Assume that `DeltaH and DeltaS)` do not vary with temperature)

Options

A. `T gt 425K`
B. All temperatures
C. `T gt 398 K`
D. `T lt 525K`

Answer

To determine the temperature at which the given reaction is spontaneous, we need to 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 0:

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

This can be rearranged to find the temperature:

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

Or,

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

### Step 1: Convert \(\Delta H\) to the same units as \(\Delta S\)

Given:
- \(\Delta H = 35.5 \, \text{kJ/mol} = 35.5 \times 10^3 \, \text{J/mol}\)
- \(\Delta S = 83.6 \, \text{J/(K mol)}\)

### Step 2: Substitute the values into the equation

Now, we can substitute the values into the equation:

\[ T > \frac{35.5 \times 10^3 \, \text{J/mol}}{83.6 \, \text{J/(K mol)}} \]

### Step 3: Calculate the temperature

Perform the calculation:

\[ T > \frac{35.5 \times 10^3}{83.6} \]

Calculating:

\[ T > 424.6 \, \text{K} \]

### Conclusion

Thus, the reaction is spontaneous at temperatures greater than \(424.6 \, \text{K}\).