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To calculate the reaction enthalpy ([tex]\(\Delta H\)[/tex]) for the reaction:
[tex]\[2 \text{CH}_3\text{OH (l)} + 3 \text{O}_2\text{(g)} \rightarrow 2 \text{CO}_2\text{(g)} + 4 \text{H}_2\text{O (l)}\][/tex]
we will utilize the standard enthalpies of formation for the compounds involved. The standard enthalpy of formation ([tex]\(\Delta H_f^\circ\)[/tex]) is the heat change associated with the formation of one mole of a compound from its elements in their standard states.
The standard enthalpies of formation ([tex]\(\Delta H_f^\circ\)[/tex]) values given are:
- [tex]\(\Delta H_f^\circ\)[/tex] for CH[tex]\(_3\)[/tex]OH (l) = -238.7 kJ/mol
- [tex]\(\Delta H_f^\circ\)[/tex] for O[tex]\(_2\)[/tex] (g) = 0 kJ/mol (since it is in its standard state)
- [tex]\(\Delta H_f^\circ\)[/tex] for CO[tex]\(_2\)[/tex] (g) = -393.5 kJ/mol
- [tex]\(\Delta H_f^\circ\)[/tex] for H[tex]\(_2\)[/tex]O (l) = -285.8 kJ/mol
First, we need to calculate the total enthalpy of the products and the reactants separately.
### Step 1: Calculate the total enthalpy of the products
The products are 2 moles of CO[tex]\(_2\)[/tex] (g) and 4 moles of H[tex]\(_2\)[/tex]O (l). Their enthalpies are calculated as follows:
[tex]\[ \text{Enthalpy of } \text{CO}_2 = 2 \text{ moles} \times (-393.5 \text{ kJ/mol}) = -787.0 \text{ kJ} \][/tex]
[tex]\[ \text{Enthalpy of } \text{H}_2\text{O} = 4 \text{ moles} \times (-285.8 \text{ kJ/mol}) = -1143.2 \text{ kJ} \][/tex]
Total enthalpy of the products:
[tex]\[ \text{Enthalpy}_{\text{products}} = -787.0 \text{ kJ} + -1143.2 \text{ kJ} = -1930.2 \text{ kJ} \][/tex]
### Step 2: Calculate the total enthalpy of the reactants
The reactants are 2 moles of CH[tex]\(_3\)[/tex]OH (l) and 3 moles of O[tex]\(_2\)[/tex] (g). Their enthalpies are calculated as follows (since the enthalpy of formation of O[tex]\(_2\)[/tex] in its standard state is zero):
[tex]\[ \text{Enthalpy of } \text{CH}_3\text{OH} = 2 \text{ moles} \times (-238.7 \text{ kJ/mol}) = -477.4 \text{ kJ} \][/tex]
[tex]\[ \text{Enthalpy of } \text{O}_2 = 3 \text{ moles} \times 0 \text{ kJ/mol} = 0 \text{ kJ} \][/tex]
Total enthalpy of the reactants:
[tex]\[ \text{Enthalpy}_{\text{reactants}} = -477.4 \text{ kJ} + 0 \text{ kJ} = -477.4 \text{ kJ} \][/tex]
### Step 3: Calculate the reaction enthalpy ([tex]\(\Delta H\)[/tex])
The reaction enthalpy is the difference between the total enthalpy of the products and the total enthalpy of the reactants:
[tex]\[ \Delta H = \text{Enthalpy}_{\text{products}} - \text{Enthalpy}_{\text{reactants}} \][/tex]
[tex]\[ \Delta H = -1930.2 \text{ kJ} - (-477.4 \text{ kJ}) = -1930.2 \text{ kJ} + 477.4 \text{ kJ} = -1452.8 \text{ kJ} \][/tex]
### Step 4: Round to the nearest kJ
The final reaction enthalpy, rounded to the nearest kJ, is:
[tex]\[ \Delta H \approx -1453 \text{ kJ} \][/tex]
Thus, the reaction enthalpy for the given reaction under standard conditions is [tex]\(-1453\)[/tex] kJ.
[tex]\[2 \text{CH}_3\text{OH (l)} + 3 \text{O}_2\text{(g)} \rightarrow 2 \text{CO}_2\text{(g)} + 4 \text{H}_2\text{O (l)}\][/tex]
we will utilize the standard enthalpies of formation for the compounds involved. The standard enthalpy of formation ([tex]\(\Delta H_f^\circ\)[/tex]) is the heat change associated with the formation of one mole of a compound from its elements in their standard states.
The standard enthalpies of formation ([tex]\(\Delta H_f^\circ\)[/tex]) values given are:
- [tex]\(\Delta H_f^\circ\)[/tex] for CH[tex]\(_3\)[/tex]OH (l) = -238.7 kJ/mol
- [tex]\(\Delta H_f^\circ\)[/tex] for O[tex]\(_2\)[/tex] (g) = 0 kJ/mol (since it is in its standard state)
- [tex]\(\Delta H_f^\circ\)[/tex] for CO[tex]\(_2\)[/tex] (g) = -393.5 kJ/mol
- [tex]\(\Delta H_f^\circ\)[/tex] for H[tex]\(_2\)[/tex]O (l) = -285.8 kJ/mol
First, we need to calculate the total enthalpy of the products and the reactants separately.
### Step 1: Calculate the total enthalpy of the products
The products are 2 moles of CO[tex]\(_2\)[/tex] (g) and 4 moles of H[tex]\(_2\)[/tex]O (l). Their enthalpies are calculated as follows:
[tex]\[ \text{Enthalpy of } \text{CO}_2 = 2 \text{ moles} \times (-393.5 \text{ kJ/mol}) = -787.0 \text{ kJ} \][/tex]
[tex]\[ \text{Enthalpy of } \text{H}_2\text{O} = 4 \text{ moles} \times (-285.8 \text{ kJ/mol}) = -1143.2 \text{ kJ} \][/tex]
Total enthalpy of the products:
[tex]\[ \text{Enthalpy}_{\text{products}} = -787.0 \text{ kJ} + -1143.2 \text{ kJ} = -1930.2 \text{ kJ} \][/tex]
### Step 2: Calculate the total enthalpy of the reactants
The reactants are 2 moles of CH[tex]\(_3\)[/tex]OH (l) and 3 moles of O[tex]\(_2\)[/tex] (g). Their enthalpies are calculated as follows (since the enthalpy of formation of O[tex]\(_2\)[/tex] in its standard state is zero):
[tex]\[ \text{Enthalpy of } \text{CH}_3\text{OH} = 2 \text{ moles} \times (-238.7 \text{ kJ/mol}) = -477.4 \text{ kJ} \][/tex]
[tex]\[ \text{Enthalpy of } \text{O}_2 = 3 \text{ moles} \times 0 \text{ kJ/mol} = 0 \text{ kJ} \][/tex]
Total enthalpy of the reactants:
[tex]\[ \text{Enthalpy}_{\text{reactants}} = -477.4 \text{ kJ} + 0 \text{ kJ} = -477.4 \text{ kJ} \][/tex]
### Step 3: Calculate the reaction enthalpy ([tex]\(\Delta H\)[/tex])
The reaction enthalpy is the difference between the total enthalpy of the products and the total enthalpy of the reactants:
[tex]\[ \Delta H = \text{Enthalpy}_{\text{products}} - \text{Enthalpy}_{\text{reactants}} \][/tex]
[tex]\[ \Delta H = -1930.2 \text{ kJ} - (-477.4 \text{ kJ}) = -1930.2 \text{ kJ} + 477.4 \text{ kJ} = -1452.8 \text{ kJ} \][/tex]
### Step 4: Round to the nearest kJ
The final reaction enthalpy, rounded to the nearest kJ, is:
[tex]\[ \Delta H \approx -1453 \text{ kJ} \][/tex]
Thus, the reaction enthalpy for the given reaction under standard conditions is [tex]\(-1453\)[/tex] kJ.
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