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

Options

A. `T lt 425 K`
B. `T gt 425 K`
C. All temperature
D. `T gt 298 K`

Answer

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:

\[ \Delta G < 0 \]

Substituting the Gibbs free energy equation into this inequality gives:

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

Rearranging this inequality, we find:

\[ T \Delta S > \Delta H \]

or

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

Now, we can substitute the given values for \(\Delta H\) and \(\Delta S\):

- \(\Delta H = 35.5 \, \text{kJ mol}^{-1}\)
- \(\Delta S = 83.6 \, \text{JK}^{-1} \, \text{mol}^{-1}\)

First, we need to convert \(\Delta H\) from kJ to J:

\[ \Delta H = 35.5 \, \text{kJ mol}^{-1} \times 1000 \, \text{J/kJ} = 35500 \, \text{J mol}^{-1} \]

Now, we can substitute the values into the inequality:

\[ T > \frac{35500 \, \text{J mol}^{-1}}{83.6 \, \text{JK}^{-1} \, \text{mol}^{-1}} \]

Calculating the right side:

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

Thus, the reaction is spontaneous at temperatures greater than approximately 425.2 K.

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