For a hypothetical reaction , `X rarr Y `, the enthalpy and entropy changes are 46.3 `kJ mol^(-1)` and `108.80 JK^(-1) mol^(-1)` respectively. Find the temperature. Find the temperature above which this reaction is spontaneous

Answer

To determine the temperature above which the reaction \( X \rightarrow Y \) 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 negative. This means we need to find the temperature at which \(\Delta G = 0\):

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

Rearranging this gives us:

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

### Step 1: Convert the enthalpy change from kJ to J
Given:
\[ \Delta H = 46.3 \, \text{kJ mol}^{-1} = 46.3 \times 1000 \, \text{J mol}^{-1} = 46300 \, \text{J mol}^{-1} \]

### Step 2: Use the given entropy change
Given:
\[ \Delta S = 108.80 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 3: Substitute the values into the temperature equation
Now we can substitute the values into the equation for temperature:

\[ T = \frac{\Delta H}{\Delta S} = \frac{46300 \, \text{J mol}^{-1}}{108.80 \, \text{J K}^{-1} \text{mol}^{-1}} \]

### Step 4: Calculate the temperature
Calculating the above expression:

\[ T = \frac{46300}{108.80} \approx 425.5 \, \text{K} \]

### Conclusion
The temperature above which the reaction \( X \rightarrow Y \) is spontaneous is approximately \( 425.5 \, \text{K} \).