Calculate the temperature above which the given reaction become spontaneous.
`C_((s)) + H_(2)O_((g)) rarr CO_((g)) + H_(2(g))`
`DeltaH^(@) = + 131.3 KJ//"mole"` ,
`DeltaS^(@) = + 0.1336 KJ//"mole" K`

Options

A. `98.8 K`
B. `709.8^(@)C`
C. `491.4 K`
D. `354.9^(@)C`

Explanation

To determine the temperature above which the given reaction becomes 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 or equal to zero. Therefore, we set \(\Delta G\) to zero:

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

Rearranging the equation gives us:

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

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

Given:
- \(\Delta H = +131.3 \, \text{kJ/mole}\)
- \(\Delta S = +0.1336 \, \text{kJ/mole K}\)

Since \(\Delta H\) is in kJ and \(\Delta S\) is also in kJ, we can directly use these values in our calculation.

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

Now, substitute the values of \(\Delta H\) and \(\Delta S\) into the equation for temperature:

\[ T = \frac{131.3 \, \text{kJ/mole}}{0.1336 \, \text{kJ/mole K}} \]

### Step 3: Calculate the temperature

Perform the division:

\[ T = \frac{131.3}{0.1336} \approx 982.8 \, \text{K} \]

### Step 4: Convert the temperature to Celsius

To convert from Kelvin to Celsius, use the formula:

\[ T(°C) = T(K) - 273.15 \]

Substituting the temperature we found:

\[ T(°C) = 982.8 \, \text{K} - 273.15 \approx 709.65 \, °C \]

### Final Answer

Thus, the temperature above which the reaction becomes spontaneous is approximately:

\[ T \approx 982.8 \, \text{K} \quad \text{or} \quad T \approx 709.65 \, °C \]