Calculate `Delta_(r)G^(@)` for the reaction : `CO(g) + 1/2O_(2)(g) to CO_(2)(g) DeltaH^(@) = -282.8 kJ`
Standard entropies : `CO_(2)(g) = 213.6, CO(g) = 197.6` and `O_(2)(g) = 205.0` (all in J `mol^(-1)`).Predict whether the reaction is spontaneous or not at standard state.

Explanation

To calculate the standard Gibbs free energy change (Δ_rG^(@)) for the reaction:

\[ \text{CO(g)} + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) \]

we will follow these steps:

### Step 1: Write down the given data
- ΔH^(@) = -282.8 kJ (convert to J: -282.8 kJ = -282800 J)
- Standard entropies:
- S^(@)(CO2) = 213.6 J/mol·K
- S^(@)(CO) = 197.6 J/mol·K
- S^(@)(O2) = 205.0 J/mol·K

### Step 2: Calculate the change in entropy (ΔS^(@))
The change in entropy for the reaction can be calculated using the formula:

\[ \Delta S^(@) = S^(@)(\text{products}) - S^(@)(\text{reactants}) \]

For our reaction:

\[ \Delta S^(@) = S^(@)(\text{CO}_2) - \left( S^(@)(\text{CO}) + \frac{1}{2} S^(@)(\text{O}_2) \right) \]

Substituting the values:

\[ \Delta S^(@) = 213.6 - \left( 197.6 + \frac{1}{2} \times 205.0 \right) \]

Calculating the entropy of the reactants:

\[ \Delta S^(@) = 213.6 - \left( 197.6 + 102.5 \right) = 213.6 - 300.1 = -86.5 \text{ J/mol·K} \]

### Step 3: Calculate Δ_rG^(@) using the Gibbs free energy equation
The Gibbs free energy change can be calculated using the formula:

\[ \Delta_rG^(@) = \Delta H^(@) - T \Delta S^(@) \]

Where:
- T = 298 K (standard temperature)

Substituting the values:

\[ \Delta_rG^(@) = -282800 \text{ J} - (298 \text{ K} \times -86.5 \text{ J/mol·K}) \]

Calculating the second term:

\[ 298 \times -86.5 = -25707 \text{ J} \]

Now substituting back into the Gibbs free energy equation:

\[ \Delta_rG^(@) = -282800 \text{ J} + 25707 \text{ J} = -257093 \text{ J} \]

### Step 4: Convert Δ_rG^(@) to kJ
\[ \Delta_rG^(@) = -257.1 \text{ kJ/mol} \]

### Step 5: Determine spontaneity
Since Δ_rG^(@) is negative, the reaction is spontaneous under standard conditions.

### Final Answer
\[ \Delta_rG^(@) = -257.1 \text{ kJ/mol} \quad \text{(Spontaneous reaction)} \]

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