Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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].
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
