The enthalpy and entropy changes for the reaction C(s) diamond `+ O_(2) rarr CO_(2)(g)` at `25^(@)C` and 1 atm. Are -393.4 kJ `mol^(-1)` and 0.006 `kJ mol^(-1)K^(-1)` respectively . Is the conversion of diamond to `CO_(2)` at room temperature a spontaneous process ?

Explanation

To determine if the conversion of diamond (C(s), diamond) to carbon dioxide (CO₂(g)) at room temperature is a spontaneous process, we can use the Gibbs free energy change (ΔG) for the reaction. The relationship between ΔG, enthalpy change (ΔH), and entropy change (ΔS) is given by the equation:

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

Where:
- ΔG = Gibbs free energy change
- ΔH = Enthalpy change
- T = Temperature in Kelvin
- ΔS = Entropy change

### Step 1: Convert the temperature to Kelvin
The given temperature is 25°C. To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]
\[ T = 25 + 273.15 = 298.15 \, K \]

### Step 2: Substitute the values into the Gibbs free energy equation
We have:
- ΔH = -393.4 kJ/mol
- ΔS = 0.006 kJ/mol·K

Now, we can calculate \(T \Delta S\):

\[ T \Delta S = 298.15 \, K \times 0.006 \, \text{kJ/mol·K} = 1.789 \, \text{kJ/mol} \]

### Step 3: Calculate ΔG
Now we substitute the values into the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]
\[ \Delta G = -393.4 \, \text{kJ/mol} - 1.789 \, \text{kJ/mol} \]
\[ \Delta G = -395.189 \, \text{kJ/mol} \]

### Step 4: Determine spontaneity
A process is spontaneous if ΔG is negative. Since we calculated:

\[ \Delta G = -395.189 \, \text{kJ/mol} < 0 \]

This indicates that the conversion of diamond to CO₂ at room temperature is indeed a spontaneous process.

### Final Answer
Yes, the conversion of diamond to CO₂ at room temperature is a spontaneous process.

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