To determine the enthalpy of combustion of 1 mole of acetylene (C[tex]\(_2\)[/tex]H[tex]\(_2\)[/tex]), we need to consider the given chemical reaction and the provided enthalpy values for each compound involved. The reaction is:
[tex]\[ 2 \, \text{C}_2\text{H}_2 + 5 \, \text{O}_2 \rightarrow 4 \, \text{CO}_2 + 2 \, \text{H}_2\text{O} \][/tex]
The enthalpy of combustion will be the difference between the total enthalpy of the products and the total enthalpy of the reactants.
Given values:
- Enthalpy of formation of acetylene (C[tex]\(_2\)[/tex]H[tex]\(_2\)[/tex]): [tex]\( \Delta H_f^\circ = 226.7 \, \text{kJ/mol} \)[/tex]
- Enthalpy of formation of oxygen (O[tex]\(_2\)[/tex]): [tex]\( \Delta H_f^\circ = 0 \, \text{kJ/mol} \)[/tex] (since O[tex]\(_2\)[/tex] is in its standard state)
- Enthalpy of formation of carbon dioxide (CO[tex]\(_2\)[/tex]): [tex]\( \Delta H_f^\circ = -393.5 \, \text{kJ/mol} \)[/tex]
- Enthalpy of formation of water (H[tex]\(_2\)[/tex]O): [tex]\( \Delta H_f^\circ = -241.8 \, \text{kJ/mol} \)[/tex]
Step-by-Step Solution:
### 1. Calculate the total enthalpy of the reactants:
The reactants are 2 moles of C[tex]\(_2\)[/tex]H[tex]\(_2\)[/tex] and 5 moles of O[tex]\(_2\)[/tex].
[tex]\[
\text{Enthalpy of reactants} = 2 \times \Delta H_f^\circ (\text{C}_2\text{H}_2) + 5 \times \Delta H_f^\circ (\text{O}_2)
\][/tex]
Substituting the given values:
[tex]\[
\text{Enthalpy of reactants} = 2 \times 226.7 \, \text{kJ} + 5 \times 0 \, \text{kJ} = 453.4 \, \text{kJ}
\][/tex]
### 2. Calculate the total enthalpy of the products:
The products are 4 moles of CO[tex]\(_2\)[/tex] and 2 moles of H[tex]\(_2\)[/tex]O.
[tex]\[
\text{Enthalpy of products} = 4 \times \Delta H_f^\circ (\text{CO}_2) + 2 \times \Delta H_f^\circ (\text{H}_2\text{O})
\][/tex]
Substituting the given values:
[tex]\[
\text{Enthalpy of products} = 4 \times (-393.5 \, \text{kJ}) + 2 \times (-241.8 \, \text{kJ}) = -1574 \, \text{kJ} + (-483.6 \, \text{kJ}) = -2057.6 \, \text{kJ}
\][/tex]
### 3. Calculate the enthalpy of combustion:
The enthalpy of combustion is the total enthalpy of the products minus the total enthalpy of the reactants:
[tex]\[
\Delta H_{\text{combustion}} = \text{Enthalpy of products} - \text{Enthalpy of reactants}
\][/tex]
[tex]\[
\Delta H_{\text{combustion}} = -2057.6 \, \text{kJ} - 453.4 \, \text{kJ} = -2511.0 \, \text{kJ}
\][/tex]
Since the reaction involves 2 moles of acetylene and we want the enthalpy of combustion per mole of acetylene, we divide by 2:
[tex]\[
\Delta H_{\text{combustion per mole}} = \frac{-2511.0 \, \text{kJ}}{2} = -1255.5 \, \text{kJ/mol}
\][/tex]
Hence, the enthalpy of combustion of 1 mole of acetylene is:
[tex]\[
\boxed{-1255.5 \, \text{kJ/mol}}
\][/tex]
