The standard entropies of `CO_(2(g)), C_((s)), and O_(2(g))` are 213.5, 5.740 and 205 `JK^-1` respectively. The standard entropy of formation of `CO_2` is _________.

Options

A. `1.86 JK^(-1)`
B. `1.96JK^(-1)`
C. `2.81JK^(-1)`
D. `2.86JK^(-1)`

Answer

To find the standard entropy of formation of \( CO_2 \), we will use the following reaction:

\[ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \]

The standard entropy of formation (\( \Delta S^\circ_f \)) can be calculated using the formula:

\[ \Delta S^\circ_f = S^\circ_{\text{products}} - S^\circ_{\text{reactants}} \]

Where:
- \( S^\circ_{\text{products}} \) is the standard entropy of the products.
- \( S^\circ_{\text{reactants}} \) is the standard entropy of the reactants.

### Step 1: Identify the standard entropies
From the question, we have:
- \( S^\circ_{CO_2(g)} = 213.5 \, \text{J K}^{-1} \)
- \( S^\circ_{C(s)} = 5.740 \, \text{J K}^{-1} \)
- \( S^\circ_{O_2(g)} = 205 \, \text{J K}^{-1} \)

### Step 2: Calculate the total standard entropy of the reactants
The total standard entropy of the reactants is the sum of the standard entropies of carbon and oxygen:

\[ S^\circ_{\text{reactants}} = S^\circ_{C(s)} + S^\circ_{O_2(g)} = 5.740 + 205 = 210.740 \, \text{J K}^{-1} \]

### Step 3: Calculate the standard entropy of formation
Now, we can substitute the values into the entropy of formation equation:

\[ \Delta S^\circ_f = S^\circ_{CO_2(g)} - S^\circ_{\text{reactants}} = 213.5 - 210.740 = 2.76 \, \text{J K}^{-1} \]

### Conclusion
The standard entropy of formation of \( CO_2 \) is approximately \( 2.76 \, \text{J K}^{-1} \).