Answer :
To determine the enthalpy of combustion (per mole) of ethyne [tex]\((\text{C}_2\text{H}_2(g))\)[/tex], we need to follow these steps:
1. Identify the given enthalpy values for reactants and products:
- Enthalpy of formation of ethyne ([tex]\(\text{C}_2\text{H}_2\)[/tex]): [tex]\( \Delta H_{\text{C}_2\text{H}_2} = 226.77\)[/tex] kJ/mol
- Enthalpy of formation of carbon dioxide ([tex]\(\text{CO}_2\)[/tex]): [tex]\( \Delta H_{\text{CO}_2} = -393.5\)[/tex] kJ/mol
- Enthalpy of formation of water ([tex]\(\text{H}_2\text{O(g)}}\)[/tex]): [tex]\( \Delta H_{\text{H}_2\text{O}} = -241.82\)[/tex] kJ/mol
2. Write down the balanced chemical equation:
[tex]\[ 2 \text{C}_2\text{H}_2 (g) + 5 \text{O}_2 (g) \rightarrow 4 \text{CO}_2 (g) + 2 \text{H}_2\text{O (g)} \][/tex]
3. Calculate the total enthalpy of products by summing the enthalpies of formation multiplied by their respective coefficients:
[tex]\[ \text{Enthalpy of products} = (4 \text{ mol} \times \Delta H_{\text{CO}_2}) + (2 \text{ mol} \times \Delta H_{\text{H}_2\text{O}}) \][/tex]
[tex]\[ \text{Enthalpy of products} = (4 \times -393.5\ \text{kJ/mol}) + (2 \times -241.82\ \text{kJ/mol}) \][/tex]
4. Calculate the total enthalpy of reactants by summing the enthalpies of formation multiplied by their respective coefficients:
[tex]\[ \text{Enthalpy of reactants} = 2 \text{ mol} \times \Delta H_{\text{C}_2\text{H}_2} \][/tex]
[tex]\[ \text{Enthalpy of reactants} = 2 \times 226.77\ \text{kJ/mol} \][/tex]
5. Apply the enthalpy change formula for the reaction:
[tex]\[ \Delta H_{\text{combustion}} = \text{Enthalpy of products} - \text{Enthalpy of reactants} \][/tex]
Using our values:
[tex]\[ \Delta H_{\text{combustion}} = (4 \times -393.5 + 2 \times -241.82) - (2 \times 226.77) \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = (-1574 - 483.64) - 453.54 \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = -2057.64 - 453.54 \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = -2511.18\ \text{kJ} \][/tex]
6. Find the enthalpy of combustion per mole of [tex]\(\text{C}_2\text{H}_2\)[/tex]:
Since the combustion equation is for 2 moles of [tex]\(\text{C}_2\text{H}_2\)[/tex], we need to divide the total enthalpy change by 2 to find it per mole:
[tex]\[ \Delta H_{\text{combustion per mole}} = \frac{\Delta H_{\text{combustion total}}}{2} \][/tex]
[tex]\[ \Delta H_{\text{combustion per mole}} = \frac{-2511.18\ \text{kJ}}{2} \][/tex]
[tex]\[ \Delta H_{\text{combustion per mole}} = -1255.59\ \text{kJ/mol} \][/tex]
Therefore, the enthalpy of combustion (per mole) of [tex]\(\text{C}_2\text{H}_2 (g)\)[/tex] is [tex]\(-1255.6\ \text{kJ/mol}\)[/tex]. The correct choice from the given options is:
[tex]\[ -1255.6\ \text{kJ/mol} \][/tex]
1. Identify the given enthalpy values for reactants and products:
- Enthalpy of formation of ethyne ([tex]\(\text{C}_2\text{H}_2\)[/tex]): [tex]\( \Delta H_{\text{C}_2\text{H}_2} = 226.77\)[/tex] kJ/mol
- Enthalpy of formation of carbon dioxide ([tex]\(\text{CO}_2\)[/tex]): [tex]\( \Delta H_{\text{CO}_2} = -393.5\)[/tex] kJ/mol
- Enthalpy of formation of water ([tex]\(\text{H}_2\text{O(g)}}\)[/tex]): [tex]\( \Delta H_{\text{H}_2\text{O}} = -241.82\)[/tex] kJ/mol
2. Write down the balanced chemical equation:
[tex]\[ 2 \text{C}_2\text{H}_2 (g) + 5 \text{O}_2 (g) \rightarrow 4 \text{CO}_2 (g) + 2 \text{H}_2\text{O (g)} \][/tex]
3. Calculate the total enthalpy of products by summing the enthalpies of formation multiplied by their respective coefficients:
[tex]\[ \text{Enthalpy of products} = (4 \text{ mol} \times \Delta H_{\text{CO}_2}) + (2 \text{ mol} \times \Delta H_{\text{H}_2\text{O}}) \][/tex]
[tex]\[ \text{Enthalpy of products} = (4 \times -393.5\ \text{kJ/mol}) + (2 \times -241.82\ \text{kJ/mol}) \][/tex]
4. Calculate the total enthalpy of reactants by summing the enthalpies of formation multiplied by their respective coefficients:
[tex]\[ \text{Enthalpy of reactants} = 2 \text{ mol} \times \Delta H_{\text{C}_2\text{H}_2} \][/tex]
[tex]\[ \text{Enthalpy of reactants} = 2 \times 226.77\ \text{kJ/mol} \][/tex]
5. Apply the enthalpy change formula for the reaction:
[tex]\[ \Delta H_{\text{combustion}} = \text{Enthalpy of products} - \text{Enthalpy of reactants} \][/tex]
Using our values:
[tex]\[ \Delta H_{\text{combustion}} = (4 \times -393.5 + 2 \times -241.82) - (2 \times 226.77) \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = (-1574 - 483.64) - 453.54 \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = -2057.64 - 453.54 \][/tex]
[tex]\[ \Delta H_{\text{combustion}} = -2511.18\ \text{kJ} \][/tex]
6. Find the enthalpy of combustion per mole of [tex]\(\text{C}_2\text{H}_2\)[/tex]:
Since the combustion equation is for 2 moles of [tex]\(\text{C}_2\text{H}_2\)[/tex], we need to divide the total enthalpy change by 2 to find it per mole:
[tex]\[ \Delta H_{\text{combustion per mole}} = \frac{\Delta H_{\text{combustion total}}}{2} \][/tex]
[tex]\[ \Delta H_{\text{combustion per mole}} = \frac{-2511.18\ \text{kJ}}{2} \][/tex]
[tex]\[ \Delta H_{\text{combustion per mole}} = -1255.59\ \text{kJ/mol} \][/tex]
Therefore, the enthalpy of combustion (per mole) of [tex]\(\text{C}_2\text{H}_2 (g)\)[/tex] is [tex]\(-1255.6\ \text{kJ/mol}\)[/tex]. The correct choice from the given options is:
[tex]\[ -1255.6\ \text{kJ/mol} \][/tex]
