`F_(2)C=CF-CF = CF_(2) rarr F_(2)underset(FC=)underset(|)(C)-underset(CF)underset(|)(CF_(2))`
For this reaction (ring closure), `DeltaH =- 49 kJ mol^(-1), DeltaS =- 40.2 J K^(-1) mol^(-1).Up` to what temperature is the forward reaction spontaneous?

Options

A. `1492^(@)C`
B. `1219^(@)C`
C. `946^(@)C`
D. `1080^(@)C`

Answer

To determine the temperature up to which the forward reaction is spontaneous, we will use the Gibbs free energy equation:

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

For the reaction to be spontaneous, \(\Delta G\) must be less than 0:

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

Rearranging this inequality gives:

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

This can be rewritten as:

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

### Step 1: Convert \(\Delta H\) to the correct units
Given:
\[ \Delta H = -49 \text{ kJ/mol} \]
Convert \(\Delta H\) to joules:
\[ \Delta H = -49 \times 10^3 \text{ J/mol} = -49000 \text{ J/mol} \]

### Step 2: Use the given \(\Delta S\)
Given:
\[ \Delta S = -40.2 \text{ J K}^{-1} \text{ mol}^{-1} \]

### Step 3: Substitute values into the equation
Now, substitute \(\Delta H\) and \(\Delta S\) into the equation to find the temperature \(T\):

\[ T > \frac{-49000 \text{ J/mol}}{-40.2 \text{ J K}^{-1} \text{ mol}^{-1}} \]

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

\[ T > \frac{49000}{40.2} \approx 1219.4 \text{ K} \]

### Step 5: Conclusion
The forward reaction is spontaneous at temperatures less than approximately 1219.4 K. To convert this temperature to Celsius:

\[ T_{Celsius} = T_{Kelvin} - 273.15 \approx 1219.4 - 273.15 \approx 946.25 \text{ °C} \]

Thus, the reaction is spontaneous up to approximately 946 °C.

### Final Answer
The temperature up to which the forward reaction is spontaneous is approximately **946 °C**.

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