The standard heat of formation of carbon disulphide (l) given that standard heat of combustion of carbon (s), sulphur (s) and carbon disulphide (l) are -393.3, -293.72 and `-1108.76 KJ mol^(-1)` respectively is

Options

A. `-128.02" kJ mol"^(-1)`
B. `+12.802" kJ mol"^(-1)`
C. `+128.02" kJ mol"^(-1)`
D. `-12.802" kJ mol"^(-1)`

Answer

To find the standard heat of formation of carbon disulfide (CS2), we can use the standard heats of combustion provided for carbon, sulfur, and carbon disulfide. The reaction for the formation of carbon disulfide from its elements can be represented as:

\[ \text{C (s)} + \text{S}_2 \text{(g)} \rightarrow \text{CS}_2 \text{(l)} \]

### Step-by-Step Solution:

1. **Identify Given Values**:
- Standard heat of combustion of Carbon (C): \( \Delta H_c^\circ (\text{C}) = -393.5 \, \text{kJ/mol} \)
- Standard heat of combustion of Sulfur (S): \( \Delta H_c^\circ (\text{S}) = -293.72 \, \text{kJ/mol} \)
- Standard heat of combustion of Carbon Disulfide (CS2): \( \Delta H_c^\circ (\text{CS}_2) = -1108.76 \, \text{kJ/mol} \)

2. **Write the Formula for Heat of Formation**:
The standard heat of formation (\( \Delta H_f^\circ \)) for a compound can be calculated using the heats of combustion of the reactants and products:
\[ \Delta H_f^\circ (\text{CS}_2) = \Delta H_c^\circ (\text{C}) + \Delta H_c^\circ (\text{S}_2) - \Delta H_c^\circ (\text{CS}_2) \]

3. **Substitute the Values**:
Since sulfur (S) is in its elemental form as \( S_2 \), we will use the heat of combustion for sulfur as given:
\[ \Delta H_f^\circ (\text{CS}_2) = (-393.5 \, \text{kJ/mol}) + (-293.72 \, \text{kJ/mol}) - (-1108.76 \, \text{kJ/mol}) \]

4. **Calculate the Result**:
\[ \Delta H_f^\circ (\text{CS}_2) = -393.5 - 293.72 + 1108.76 \]
\[ \Delta H_f^\circ (\text{CS}_2) = -393.5 - 293.72 + 1108.76 = -128.02 \, \text{kJ/mol} \]

5. **Final Answer**:
The standard heat of formation of carbon disulfide (CS2) is:
\[ \Delta H_f^\circ (\text{CS}_2) = -128.02 \, \text{kJ/mol} \]