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To calculate the enthalpy of combustion of 1 mole of acetylene ([tex]\( \text{C}_2\text{H}_2 \)[/tex]), we will use the given standard enthalpies of formation ([tex]\( \Delta H_f \)[/tex]) from the table. The reaction we are considering 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]
First, we need to determine the enthalpy change for the combustion of 2 moles of acetylene by finding the enthalpy change of the reactants and products.
### Step 1: Calculate the enthalpy of the reactants
The reactants are 2 moles of acetylene ([tex]\( \text{C}_2\text{H}_2 \)[/tex]) and 5 moles of oxygen ([tex]\( \text{O}_2 \)[/tex]). Since oxygen is in its standard state, its enthalpy of formation is zero:
[tex]\[ \Delta H_{\text{reactants}} = 2 \times \Delta H_f \left( \text{C}_2\text{H}_2 \right) + 5 \times \Delta H_f \left( \text{O}_2 \right) \][/tex]
From the table, [tex]\( \Delta H_f \left( \text{C}_2\text{H}_2 \right) = 226.8 \, \text{kJ/mol} \)[/tex] and [tex]\( \Delta H_f \left( \text{O}_2 \right) = 0 \, \text{kJ/mol} \)[/tex]:
[tex]\[ \Delta H_{\text{reactants}} = 2 \times 226.8 + 5 \times 0 = 453.6 \, \text{kJ} \][/tex]
### Step 2: Calculate the enthalpy of the products
The products are 4 moles of carbon dioxide ([tex]\( \text{CO}_2 \)[/tex]) and 2 moles of water ([tex]\( \text{H}_2\text{O} \)[/tex]):
[tex]\[ \Delta H_{\text{products}} = 4 \times \Delta H_f \left( \text{CO}_2 \right) + 2 \times \Delta H_f \left( \text{H}_2\text{O} \right) \][/tex]
From the table, [tex]\( \Delta H_f \left( \text{CO}_2 \right) = -393.5 \, \text{kJ/mol} \)[/tex] and [tex]\( \Delta H_f \left( \text{H}_2\text{O} \right) = -241.8 \, \text{kJ/mol} \)[/tex]:
[tex]\[ \Delta H_{\text{products}} = 4 \times (-393.5) + 2 \times (-241.8) = -1574.0 + (-483.6) = -2057.6 \, \text{kJ} \][/tex]
### Step 3: Calculate the enthalpy change for the reaction
The enthalpy change ([tex]\( \Delta H_{\text{combustion}} \)[/tex]) for the combustion of 2 moles of acetylene is the difference between the enthalpy of the products and the reactants:
[tex]\[ \Delta H_{\text{combustion}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} = -2057.6 - 453.6 = -2511.2 \, \text{kJ} \][/tex]
### Step 4: Calculate the enthalpy of combustion per mole of acetylene
Since this enthalpy change is for 2 moles of acetylene, we need to divide by 2 to find the enthalpy change for 1 mole of acetylene:
[tex]\[ \Delta H_{\text{combustion per mole}} = \frac{\Delta H_{\text{combustion}}}{2} = \frac{-2511.2}{2} = -1255.6 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy of combustion of 1 mole of acetylene is [tex]\( \boxed{-1255.6 \, \text{kJ/mol}} \)[/tex].
[tex]\[ 2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]
First, we need to determine the enthalpy change for the combustion of 2 moles of acetylene by finding the enthalpy change of the reactants and products.
### Step 1: Calculate the enthalpy of the reactants
The reactants are 2 moles of acetylene ([tex]\( \text{C}_2\text{H}_2 \)[/tex]) and 5 moles of oxygen ([tex]\( \text{O}_2 \)[/tex]). Since oxygen is in its standard state, its enthalpy of formation is zero:
[tex]\[ \Delta H_{\text{reactants}} = 2 \times \Delta H_f \left( \text{C}_2\text{H}_2 \right) + 5 \times \Delta H_f \left( \text{O}_2 \right) \][/tex]
From the table, [tex]\( \Delta H_f \left( \text{C}_2\text{H}_2 \right) = 226.8 \, \text{kJ/mol} \)[/tex] and [tex]\( \Delta H_f \left( \text{O}_2 \right) = 0 \, \text{kJ/mol} \)[/tex]:
[tex]\[ \Delta H_{\text{reactants}} = 2 \times 226.8 + 5 \times 0 = 453.6 \, \text{kJ} \][/tex]
### Step 2: Calculate the enthalpy of the products
The products are 4 moles of carbon dioxide ([tex]\( \text{CO}_2 \)[/tex]) and 2 moles of water ([tex]\( \text{H}_2\text{O} \)[/tex]):
[tex]\[ \Delta H_{\text{products}} = 4 \times \Delta H_f \left( \text{CO}_2 \right) + 2 \times \Delta H_f \left( \text{H}_2\text{O} \right) \][/tex]
From the table, [tex]\( \Delta H_f \left( \text{CO}_2 \right) = -393.5 \, \text{kJ/mol} \)[/tex] and [tex]\( \Delta H_f \left( \text{H}_2\text{O} \right) = -241.8 \, \text{kJ/mol} \)[/tex]:
[tex]\[ \Delta H_{\text{products}} = 4 \times (-393.5) + 2 \times (-241.8) = -1574.0 + (-483.6) = -2057.6 \, \text{kJ} \][/tex]
### Step 3: Calculate the enthalpy change for the reaction
The enthalpy change ([tex]\( \Delta H_{\text{combustion}} \)[/tex]) for the combustion of 2 moles of acetylene is the difference between the enthalpy of the products and the reactants:
[tex]\[ \Delta H_{\text{combustion}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} = -2057.6 - 453.6 = -2511.2 \, \text{kJ} \][/tex]
### Step 4: Calculate the enthalpy of combustion per mole of acetylene
Since this enthalpy change is for 2 moles of acetylene, we need to divide by 2 to find the enthalpy change for 1 mole of acetylene:
[tex]\[ \Delta H_{\text{combustion per mole}} = \frac{\Delta H_{\text{combustion}}}{2} = \frac{-2511.2}{2} = -1255.6 \, \text{kJ/mol} \][/tex]
Therefore, the enthalpy of combustion of 1 mole of acetylene is [tex]\( \boxed{-1255.6 \, \text{kJ/mol}} \)[/tex].
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