Answer :
To calculate the standard enthalpy of formation of carbon disulfide [tex]\( \left( \text{CS}_2 \right) \)[/tex] from its elements, we will follow a systematic approach using the given reactions and Hess's Law.
### Given Data:
1. [tex]\( \text{C (graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \)[/tex]
- [tex]\( \Delta H^\circ_{\text{rxn}} = -393.5 \, \text{kJ/mol} \)[/tex]
2. [tex]\( \text{S (rhombic)} + \text{O}_2(g) \rightarrow \text{SO}_2(g) \)[/tex]
- [tex]\( \Delta H^\circ_{\text{rxn}} = -296.4 \, \text{kJ/mol} \)[/tex]
3. [tex]\( \text{CS}_2(l) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \)[/tex]
- [tex]\( \Delta H^\circ_{\text{rxn}} = -1073.5 \, \text{kJ/mol} \)[/tex]
### Step-by-Step Solution:
1. Combustion Reactions:
- For the combustion of carbon:
[tex]\[ \text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5 \, \text{kJ/mol} \][/tex]
- For the combustion of sulfur (remembering there are two moles of sulfur in the final balanced reaction):
[tex]\[ 2 \times \left( \text{S(rhombic)} + \text{O}_2(g) \rightarrow \text{SO}_2(g) \right) \quad \Delta H = 2 \times (-296.4 \, \text{kJ/mol}) = -592.8 \, \text{kJ/mol} \][/tex]
2. Total Enthalpy Change for Combustion Reactions:
- Summing these reactions gives the overall reaction:
[tex]\[ \text{C(graphite)} + 2\text{S(rhombic)} + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \][/tex]
- Total enthalpy change for this combined reaction:
[tex]\[ \Delta H = -393.5 \, \text{kJ/mol} + (-592.8 \, \text{kJ/mol}) = -986.3 \, \text{kJ/mol} \][/tex]
3. Formation of CS[tex]\(_2\)[/tex] from Elements:
- Now let's consider the reverse of the given combustion of carbon disulfide:
[tex]\[ \text{CO}_2(g) + 2\text{SO}_2(g) \rightarrow \text{CS}_2(l) + 3\text{O}_2(g) \][/tex]
- The enthalpy change for the reverse reaction is the negative of the given value:
[tex]\[ \Delta H_{\text{reverse}} = +1073.5 \, \text{kJ/mol} \][/tex]
4. Combining the Reactions:
- Using Hess's Law, we combine the formation reactions and the reverse combustion of CS[tex]\(_2\)[/tex]:
[tex]\[ \text{C(graphite)} + 2\text{S(rhombic)} + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \quad (\Delta H = -986.3 \, \text{kJ/mol}) \][/tex]
[tex]\[ \text{CO}_2(g) + 2\text{SO}_2(g) \rightarrow \text{CS}_2(l) + 3\text{O}_2(g) \quad (\Delta H = +1073.5 \, \text{kJ/mol}) \][/tex]
5. Calculating the Enthalpy of Formation:
- To form CS[tex]\(_2\)[/tex] from its elements:
[tex]\[ \text{C(graphite)} + 2\text{S(rhombic)} \rightarrow \text{CS}_2(l) \][/tex]
- The overall enthalpy change for this reaction:
[tex]\[ \Delta H^\circ_{\text{formation}} = +1073.5 \, \text{kJ/mol} - 986.3 \, \text{kJ/mol} = +87.2 \, \text{kJ/mol} \][/tex]
### Conclusion:
The standard enthalpy of formation of carbon disulfide [tex]\(\text{CS}_2\)[/tex] from its elements (graphite and rhombic sulfur) is [tex]\(-87.2 \, \text{kJ/mol}\)[/tex].
### Given Data:
1. [tex]\( \text{C (graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \)[/tex]
- [tex]\( \Delta H^\circ_{\text{rxn}} = -393.5 \, \text{kJ/mol} \)[/tex]
2. [tex]\( \text{S (rhombic)} + \text{O}_2(g) \rightarrow \text{SO}_2(g) \)[/tex]
- [tex]\( \Delta H^\circ_{\text{rxn}} = -296.4 \, \text{kJ/mol} \)[/tex]
3. [tex]\( \text{CS}_2(l) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \)[/tex]
- [tex]\( \Delta H^\circ_{\text{rxn}} = -1073.5 \, \text{kJ/mol} \)[/tex]
### Step-by-Step Solution:
1. Combustion Reactions:
- For the combustion of carbon:
[tex]\[ \text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5 \, \text{kJ/mol} \][/tex]
- For the combustion of sulfur (remembering there are two moles of sulfur in the final balanced reaction):
[tex]\[ 2 \times \left( \text{S(rhombic)} + \text{O}_2(g) \rightarrow \text{SO}_2(g) \right) \quad \Delta H = 2 \times (-296.4 \, \text{kJ/mol}) = -592.8 \, \text{kJ/mol} \][/tex]
2. Total Enthalpy Change for Combustion Reactions:
- Summing these reactions gives the overall reaction:
[tex]\[ \text{C(graphite)} + 2\text{S(rhombic)} + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \][/tex]
- Total enthalpy change for this combined reaction:
[tex]\[ \Delta H = -393.5 \, \text{kJ/mol} + (-592.8 \, \text{kJ/mol}) = -986.3 \, \text{kJ/mol} \][/tex]
3. Formation of CS[tex]\(_2\)[/tex] from Elements:
- Now let's consider the reverse of the given combustion of carbon disulfide:
[tex]\[ \text{CO}_2(g) + 2\text{SO}_2(g) \rightarrow \text{CS}_2(l) + 3\text{O}_2(g) \][/tex]
- The enthalpy change for the reverse reaction is the negative of the given value:
[tex]\[ \Delta H_{\text{reverse}} = +1073.5 \, \text{kJ/mol} \][/tex]
4. Combining the Reactions:
- Using Hess's Law, we combine the formation reactions and the reverse combustion of CS[tex]\(_2\)[/tex]:
[tex]\[ \text{C(graphite)} + 2\text{S(rhombic)} + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g) \quad (\Delta H = -986.3 \, \text{kJ/mol}) \][/tex]
[tex]\[ \text{CO}_2(g) + 2\text{SO}_2(g) \rightarrow \text{CS}_2(l) + 3\text{O}_2(g) \quad (\Delta H = +1073.5 \, \text{kJ/mol}) \][/tex]
5. Calculating the Enthalpy of Formation:
- To form CS[tex]\(_2\)[/tex] from its elements:
[tex]\[ \text{C(graphite)} + 2\text{S(rhombic)} \rightarrow \text{CS}_2(l) \][/tex]
- The overall enthalpy change for this reaction:
[tex]\[ \Delta H^\circ_{\text{formation}} = +1073.5 \, \text{kJ/mol} - 986.3 \, \text{kJ/mol} = +87.2 \, \text{kJ/mol} \][/tex]
### Conclusion:
The standard enthalpy of formation of carbon disulfide [tex]\(\text{CS}_2\)[/tex] from its elements (graphite and rhombic sulfur) is [tex]\(-87.2 \, \text{kJ/mol}\)[/tex].
