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Sagot :
To find the enthalpy change (ΔH) for the combustion of candle wax (paraffin), we need to evaluate the enthalpy of formation for both reactants and products involved in the reaction.
### Step-by-Step Solution:
1. Identify the substances and their respective standard enthalpy of formation (ΔH_f^°):
- For [tex]\( \text{C}_{25}\text{H}_{52} \)[/tex] (s): [tex]\( \Delta H_f^° = -751.5 \ \text{kJ/mol} \)[/tex]
- For [tex]\( \text{O}_2 \)[/tex] (g): [tex]\( \Delta H_f^° = 0 \ \text{kJ/mol} \)[/tex] (because it is a standard element)
- For [tex]\( \text{CO}_2 \)[/tex] (g): [tex]\( \Delta H_f^° = -393.5 \ \text{kJ/mol} \)[/tex]
- For [tex]\( \text{H}_2\text{O} \)[/tex] (l): [tex]\( \Delta H_f^° = -241.8 \ \text{kJ/mol} \)[/tex]
2. Determine the number of moles for each substance in the balanced equation:
- Moles of [tex]\( \text{C}_{25}\text{H}_{52} \)[/tex] (s): [tex]\( 1 \)[/tex]
- Moles of [tex]\( \text{O}_2 \)[/tex] (g): [tex]\( \frac{76}{2} = 38 \)[/tex] (derived from balancing the equation)
- Moles of [tex]\( \text{CO}_2 \)[/tex] (g): [tex]\( 25 \)[/tex]
- Moles of [tex]\( \text{H}_2\text{O} \)[/tex] (l): [tex]\( 26 \)[/tex]
3. Calculate the total enthalpy of formation for the reactants and the products:
For the reactants:
- [tex]\( \text{C}_{25}\text{H}_{52}: \ 1 \times (-751.5 \ \text{kJ/mol}) \)[/tex]
- [tex]\( \text{O}_2: \ 38 \times 0 \ \text{kJ/mol} \)[/tex]
[tex]\[ \text{Total enthalpy of reactants} = -751.5 \ \text{kJ/mol} \][/tex]
For the products:
- [tex]\( \text{CO}_2: \ 25 \times (-393.5 \ \text{kJ/mol}) \)[/tex]
- [tex]\( \text{H}_2\text{O}: \ 26 \times (-241.8 \ \text{kJ/mol}) \)[/tex]
[tex]\[ \text{Total enthalpy of products} = (25 \times -393.5) + (26 \times -241.8) \ = -9837.5 + (-6286.8) = -16124.3 \ \text{kJ/mol} \][/tex]
4. Determine the enthalpy change ΔH for the reaction:
[tex]\[ \Delta H = \text{Total enthalpy of products} - \text{Total enthalpy of reactants} \][/tex]
[tex]\[ \Delta H = -16124.3 \ \text{kJ/mol} - (-751.5 \ \text{kJ/mol}) = -16124.3 \ \text{kJ/mol} + 751.5 \ \text{kJ/mol} = -15372.8 \ \text{kJ/mol} \][/tex]
Therefore, the enthalpy change (ΔH) for the combustion of candle wax (paraffin) is:
[tex]\[ \Delta H = -15372.8 \ \text{kJ/mol} \][/tex]
### Step-by-Step Solution:
1. Identify the substances and their respective standard enthalpy of formation (ΔH_f^°):
- For [tex]\( \text{C}_{25}\text{H}_{52} \)[/tex] (s): [tex]\( \Delta H_f^° = -751.5 \ \text{kJ/mol} \)[/tex]
- For [tex]\( \text{O}_2 \)[/tex] (g): [tex]\( \Delta H_f^° = 0 \ \text{kJ/mol} \)[/tex] (because it is a standard element)
- For [tex]\( \text{CO}_2 \)[/tex] (g): [tex]\( \Delta H_f^° = -393.5 \ \text{kJ/mol} \)[/tex]
- For [tex]\( \text{H}_2\text{O} \)[/tex] (l): [tex]\( \Delta H_f^° = -241.8 \ \text{kJ/mol} \)[/tex]
2. Determine the number of moles for each substance in the balanced equation:
- Moles of [tex]\( \text{C}_{25}\text{H}_{52} \)[/tex] (s): [tex]\( 1 \)[/tex]
- Moles of [tex]\( \text{O}_2 \)[/tex] (g): [tex]\( \frac{76}{2} = 38 \)[/tex] (derived from balancing the equation)
- Moles of [tex]\( \text{CO}_2 \)[/tex] (g): [tex]\( 25 \)[/tex]
- Moles of [tex]\( \text{H}_2\text{O} \)[/tex] (l): [tex]\( 26 \)[/tex]
3. Calculate the total enthalpy of formation for the reactants and the products:
For the reactants:
- [tex]\( \text{C}_{25}\text{H}_{52}: \ 1 \times (-751.5 \ \text{kJ/mol}) \)[/tex]
- [tex]\( \text{O}_2: \ 38 \times 0 \ \text{kJ/mol} \)[/tex]
[tex]\[ \text{Total enthalpy of reactants} = -751.5 \ \text{kJ/mol} \][/tex]
For the products:
- [tex]\( \text{CO}_2: \ 25 \times (-393.5 \ \text{kJ/mol}) \)[/tex]
- [tex]\( \text{H}_2\text{O}: \ 26 \times (-241.8 \ \text{kJ/mol}) \)[/tex]
[tex]\[ \text{Total enthalpy of products} = (25 \times -393.5) + (26 \times -241.8) \ = -9837.5 + (-6286.8) = -16124.3 \ \text{kJ/mol} \][/tex]
4. Determine the enthalpy change ΔH for the reaction:
[tex]\[ \Delta H = \text{Total enthalpy of products} - \text{Total enthalpy of reactants} \][/tex]
[tex]\[ \Delta H = -16124.3 \ \text{kJ/mol} - (-751.5 \ \text{kJ/mol}) = -16124.3 \ \text{kJ/mol} + 751.5 \ \text{kJ/mol} = -15372.8 \ \text{kJ/mol} \][/tex]
Therefore, the enthalpy change (ΔH) for the combustion of candle wax (paraffin) is:
[tex]\[ \Delta H = -15372.8 \ \text{kJ/mol} \][/tex]
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