Answer :
Certainly! Let's calculate the enthalpy of combustion of 1 mole of acetylene ([tex]\(C_2H_2\)[/tex]) for the reaction:
[tex]\[2 \, C_2H_2 + 5 \, O_2 \rightarrow 4 \, CO_2 + 2 \, H_2O\][/tex]
To calculate the enthalpy of combustion, we will use the enthalpy of formation ([tex]\(\Delta H_f\)[/tex]) values from the table provided:
1. List the [tex]\(\Delta H_f\)[/tex] values from the table:
- [tex]\(\Delta H_f\)[/tex] of acetylene ([tex]\(C_2H_2\)[/tex]) = 226.8 kJ/mol
- [tex]\(\Delta H_f\)[/tex] of carbon dioxide ([tex]\(CO_2\)[/tex]) = -393.5 kJ/mol
- [tex]\(\Delta H_f\)[/tex] of water ([tex]\(H_2O\)[/tex]) = -241.8 kJ/mol
2. Write the balanced chemical equation:
[tex]\[2 \, C_2H_2 + 5 \, O_2 \rightarrow 4 \, CO_2 + 2 \, H_2O\][/tex]
3. Calculate the total enthalpy of products:
- For [tex]\(4 \, CO_2\)[/tex]:
[tex]\[ \text{Enthalpy of formation for 4 \, CO_2} = 4 \times \Delta H_f \, CO_2 = 4 \times (-393.5) = -1574.0 \, \text{kJ} \][/tex]
- For [tex]\(2 \, H_2O\)[/tex]:
[tex]\[ \text{Enthalpy of formation for 2 \, H_2O} = 2 \times \Delta H_f \, H_2O = 2 \times (-241.8) = -483.6 \, \text{kJ} \][/tex]
- Therefore, the total enthalpy of the products is:
[tex]\[ \text{Total enthalpy of products} = -1574.0 \, \text{kJ} + (-483.6 \, \text{kJ}) = -2057.6 \, \text{kJ} \][/tex]
4. Calculate the total enthalpy of reactants:
- For [tex]\(2 \, C_2H_2\)[/tex]:
[tex]\[ \text{Enthalpy of formation for 2 \, C_2H_2} = 2 \times \Delta H_f \, C_2H_2 = 2 \times 226.8 = 453.6 \, \text{kJ} \][/tex]
- For [tex]\(5 \, O_2\)[/tex]:
[tex]\[ \Delta H_f \, O_2 = 0 \, \text{kJ/mol} \text{ (since O}_2\text{ is in its standard state)} \][/tex]
Therefore,
[tex]\[ \text{Enthalpy of formation for 5 \, O}_2 = 5 \times 0 = 0 \, \text{kJ} \][/tex]
- Therefore, the total enthalpy of the reactants is:
[tex]\[ \text{Total enthalpy of reactants} = 453.6 \, \text{kJ} + 0 \, \text{kJ} = 453.6 \, \text{kJ} \][/tex]
5. Calculate the enthalpy of combustion (∆H_comb) for 2 moles of [tex]\(C_2H_2\)[/tex]:
- Enthalpy of combustion (∆H_comb) is given by the enthalpy of products minus the enthalpy of reactants:
[tex]\[ \Delta H_{comb} = \text{Total enthalpy of products} - \text{Total enthalpy of reactants} \][/tex]
[tex]\[ \Delta H_{comb} = -2057.6 \, \text{kJ} - 453.6 \, \text{kJ} = -2511.2 \, \text{kJ} \][/tex]
6. Calculate the enthalpy of combustion for 1 mole of [tex]\(C_2H_2\)[/tex]:
- Since the above value is for 2 moles of [tex]\(C_2H_2\)[/tex], we need to divide it by 2 to find the enthalpy of combustion for 1 mole of [tex]\(C_2H_2\)[/tex]:
[tex]\[ \Delta H_{comb \, \text{(per mol)}} = \frac{\Delta H_{comb}}{2} = \frac{-2511.2 \, \text{kJ}}{2} = -1255.6 \, \text{kJ/mol} \][/tex]
Thus, the enthalpy of combustion of 1 mole of acetylene ([tex]\(C_2H_2\)[/tex]) in the given reaction is [tex]\(-1255.6 \, \text{kJ/mol}\)[/tex].
[tex]\[2 \, C_2H_2 + 5 \, O_2 \rightarrow 4 \, CO_2 + 2 \, H_2O\][/tex]
To calculate the enthalpy of combustion, we will use the enthalpy of formation ([tex]\(\Delta H_f\)[/tex]) values from the table provided:
1. List the [tex]\(\Delta H_f\)[/tex] values from the table:
- [tex]\(\Delta H_f\)[/tex] of acetylene ([tex]\(C_2H_2\)[/tex]) = 226.8 kJ/mol
- [tex]\(\Delta H_f\)[/tex] of carbon dioxide ([tex]\(CO_2\)[/tex]) = -393.5 kJ/mol
- [tex]\(\Delta H_f\)[/tex] of water ([tex]\(H_2O\)[/tex]) = -241.8 kJ/mol
2. Write the balanced chemical equation:
[tex]\[2 \, C_2H_2 + 5 \, O_2 \rightarrow 4 \, CO_2 + 2 \, H_2O\][/tex]
3. Calculate the total enthalpy of products:
- For [tex]\(4 \, CO_2\)[/tex]:
[tex]\[ \text{Enthalpy of formation for 4 \, CO_2} = 4 \times \Delta H_f \, CO_2 = 4 \times (-393.5) = -1574.0 \, \text{kJ} \][/tex]
- For [tex]\(2 \, H_2O\)[/tex]:
[tex]\[ \text{Enthalpy of formation for 2 \, H_2O} = 2 \times \Delta H_f \, H_2O = 2 \times (-241.8) = -483.6 \, \text{kJ} \][/tex]
- Therefore, the total enthalpy of the products is:
[tex]\[ \text{Total enthalpy of products} = -1574.0 \, \text{kJ} + (-483.6 \, \text{kJ}) = -2057.6 \, \text{kJ} \][/tex]
4. Calculate the total enthalpy of reactants:
- For [tex]\(2 \, C_2H_2\)[/tex]:
[tex]\[ \text{Enthalpy of formation for 2 \, C_2H_2} = 2 \times \Delta H_f \, C_2H_2 = 2 \times 226.8 = 453.6 \, \text{kJ} \][/tex]
- For [tex]\(5 \, O_2\)[/tex]:
[tex]\[ \Delta H_f \, O_2 = 0 \, \text{kJ/mol} \text{ (since O}_2\text{ is in its standard state)} \][/tex]
Therefore,
[tex]\[ \text{Enthalpy of formation for 5 \, O}_2 = 5 \times 0 = 0 \, \text{kJ} \][/tex]
- Therefore, the total enthalpy of the reactants is:
[tex]\[ \text{Total enthalpy of reactants} = 453.6 \, \text{kJ} + 0 \, \text{kJ} = 453.6 \, \text{kJ} \][/tex]
5. Calculate the enthalpy of combustion (∆H_comb) for 2 moles of [tex]\(C_2H_2\)[/tex]:
- Enthalpy of combustion (∆H_comb) is given by the enthalpy of products minus the enthalpy of reactants:
[tex]\[ \Delta H_{comb} = \text{Total enthalpy of products} - \text{Total enthalpy of reactants} \][/tex]
[tex]\[ \Delta H_{comb} = -2057.6 \, \text{kJ} - 453.6 \, \text{kJ} = -2511.2 \, \text{kJ} \][/tex]
6. Calculate the enthalpy of combustion for 1 mole of [tex]\(C_2H_2\)[/tex]:
- Since the above value is for 2 moles of [tex]\(C_2H_2\)[/tex], we need to divide it by 2 to find the enthalpy of combustion for 1 mole of [tex]\(C_2H_2\)[/tex]:
[tex]\[ \Delta H_{comb \, \text{(per mol)}} = \frac{\Delta H_{comb}}{2} = \frac{-2511.2 \, \text{kJ}}{2} = -1255.6 \, \text{kJ/mol} \][/tex]
Thus, the enthalpy of combustion of 1 mole of acetylene ([tex]\(C_2H_2\)[/tex]) in the given reaction is [tex]\(-1255.6 \, \text{kJ/mol}\)[/tex].
