Answer :
Sure! Let's find the enthalpy of combustion of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex] (ethyne) per mole using the provided reaction:
[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]
To begin, the standard enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) for the substances involved are given as follows:
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{C}_2\text{H}_2(g) = 226.77\, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{CO}_2(g) = -393.5\, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{H}_2\text{O(g)} = -241.82\, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{O}_2(g) = 0\, \text{kJ/mol} \)[/tex] (since [tex]\( \text{O}_2 \)[/tex] is in its elemental form)
The balanced chemical equation for combustion is given as:
[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]
### Step 1: Calculate the total enthalpy of products
The enthalpy of the products can be calculated by summing the enthalpies of formation of the products, taking into account their coefficients in the balanced equation.
[tex]\[ \Delta H_{\text{products}} = (4 \times \Delta H_f \text{ of } \text{CO}_2) + (2 \times \Delta H_f \text{ of } \text{H}_2\text{O}) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{\text{products}} = (4 \times -393.5\, \text{kJ/mol}) + (2 \times -241.82\, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1574\, \text{kJ} + (-483.64\, \text{kJ}) = -2057.64\, \text{kJ} \][/tex]
### Step 2: Calculate the total enthalpy of reactants
The enthalpy of the reactants is calculated similarly by summing the enthalpies of formation of the reactants, also considering their coefficients in the balanced equation. Note that the enthalpy of formation of [tex]\( \text{O}_2 \)[/tex] is zero.
[tex]\[ \Delta H_{\text{reactants}} = (2 \times \Delta H_f \text{ of } \text{C}_2\text{H}_2) + (5 \times \Delta H_f \text{ of } \text{O}_2) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{\text{reactants}} = (2 \times 226.77\, \text{kJ/mol}) + (5 \times 0\, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = 453.54\, \text{kJ} \][/tex]
### Step 3: Calculate the enthalpy change of the reaction ([tex]\( \Delta H_{\text{rxn}} \)[/tex])
The enthalpy change of the reaction is the difference between the total enthalpy of products and the total enthalpy of reactants:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
Substituting the values:
[tex]\[ \Delta H_{\text{rxn}} = -2057.64\, \text{kJ} - 453.54\, \text{kJ} = -2511.18\, \text{kJ} \][/tex]
### Step 4: Find the enthalpy change per mole of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex]
Since the given balanced equation is for 2 moles of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex], we'll divide the enthalpy change of the reaction by 2 to get the enthalpy change per mole of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex]:
[tex]\[ \Delta H_{\text{rxn per mole}} = \frac{\Delta H_{\text{rxn}}}{2} \][/tex]
Substituting the value:
[tex]\[ \Delta H_{\text{rxn per mole}} = \frac{-2511.18\, \text{kJ}}{2} = -1255.59\, \text{kJ/mol} \][/tex]
Thus, the enthalpy of combustion of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex] per mole is [tex]\( -1255.59\, \text{kJ/mol} \)[/tex].
From the provided choices:
- [tex]\( -2511.2\, \text{kJ/mol} \)[/tex]
- [tex]\( -1255.6\, \text{kJ/mol} \)[/tex]
- [tex]\( -862.1\, \text{kJ/mol} \)[/tex]
- [tex]\( -431.0\, \text{kJ/mol} \)[/tex]
The correct answer is:
[tex]\[ -1255.6\, \text{kJ/mol} \][/tex]
[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]
To begin, the standard enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) for the substances involved are given as follows:
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{C}_2\text{H}_2(g) = 226.77\, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{CO}_2(g) = -393.5\, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{H}_2\text{O(g)} = -241.82\, \text{kJ/mol} \)[/tex]
- [tex]\( \Delta H_f \)[/tex] of [tex]\( \text{O}_2(g) = 0\, \text{kJ/mol} \)[/tex] (since [tex]\( \text{O}_2 \)[/tex] is in its elemental form)
The balanced chemical equation for combustion is given as:
[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]
### Step 1: Calculate the total enthalpy of products
The enthalpy of the products can be calculated by summing the enthalpies of formation of the products, taking into account their coefficients in the balanced equation.
[tex]\[ \Delta H_{\text{products}} = (4 \times \Delta H_f \text{ of } \text{CO}_2) + (2 \times \Delta H_f \text{ of } \text{H}_2\text{O}) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{\text{products}} = (4 \times -393.5\, \text{kJ/mol}) + (2 \times -241.82\, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1574\, \text{kJ} + (-483.64\, \text{kJ}) = -2057.64\, \text{kJ} \][/tex]
### Step 2: Calculate the total enthalpy of reactants
The enthalpy of the reactants is calculated similarly by summing the enthalpies of formation of the reactants, also considering their coefficients in the balanced equation. Note that the enthalpy of formation of [tex]\( \text{O}_2 \)[/tex] is zero.
[tex]\[ \Delta H_{\text{reactants}} = (2 \times \Delta H_f \text{ of } \text{C}_2\text{H}_2) + (5 \times \Delta H_f \text{ of } \text{O}_2) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{\text{reactants}} = (2 \times 226.77\, \text{kJ/mol}) + (5 \times 0\, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = 453.54\, \text{kJ} \][/tex]
### Step 3: Calculate the enthalpy change of the reaction ([tex]\( \Delta H_{\text{rxn}} \)[/tex])
The enthalpy change of the reaction is the difference between the total enthalpy of products and the total enthalpy of reactants:
[tex]\[ \Delta H_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
Substituting the values:
[tex]\[ \Delta H_{\text{rxn}} = -2057.64\, \text{kJ} - 453.54\, \text{kJ} = -2511.18\, \text{kJ} \][/tex]
### Step 4: Find the enthalpy change per mole of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex]
Since the given balanced equation is for 2 moles of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex], we'll divide the enthalpy change of the reaction by 2 to get the enthalpy change per mole of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex]:
[tex]\[ \Delta H_{\text{rxn per mole}} = \frac{\Delta H_{\text{rxn}}}{2} \][/tex]
Substituting the value:
[tex]\[ \Delta H_{\text{rxn per mole}} = \frac{-2511.18\, \text{kJ}}{2} = -1255.59\, \text{kJ/mol} \][/tex]
Thus, the enthalpy of combustion of [tex]\( \text{C}_2\text{H}_2(g) \)[/tex] per mole is [tex]\( -1255.59\, \text{kJ/mol} \)[/tex].
From the provided choices:
- [tex]\( -2511.2\, \text{kJ/mol} \)[/tex]
- [tex]\( -1255.6\, \text{kJ/mol} \)[/tex]
- [tex]\( -862.1\, \text{kJ/mol} \)[/tex]
- [tex]\( -431.0\, \text{kJ/mol} \)[/tex]
The correct answer is:
[tex]\[ -1255.6\, \text{kJ/mol} \][/tex]
