To calculate the quantity of heat produced when 13.0 g of propane (C₃H₈) is combusted in air under standard conditions, we need to follow these steps:
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Write the balanced combustion reaction for propane:
The complete combustion of propane is represented by the following reaction:
\[ C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g) \]
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Find the standard enthalpy of formation (\( \Delta H_f^\circ \)) values:
- For \(C_3H_8(g)\): \(\Delta H_f^\circ = -103.85 , \text{kJ/mol} \)
- For \(CO_2(g)\): \(\Delta H_f^\circ = -393.5 , \text{kJ/mol} \)
- For \(H_2O(g)\): \(\Delta H_f^\circ = -241.8 , \text{kJ/mol} \)
- For \(O_2(g)\): \(\Delta H_f^\circ = 0 , \text{kJ/mol} \) (since it's a standard state element)
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Calculate the standard enthalpy change (\( \Delta H_{rxn} \)) for the reaction:
Using the equation:
\[ \Delta H_{rxn} = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]
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Products: 3 moles of \(CO_2\) and 4 moles of \(H_2O\): \[ \Delta H_{products} = [3(-393.5) + 4(-241.8)] \text{ kJ/mol} \]
\[ \Delta H_{products} = -1180.5 - 967.2 = -2147.7 \text{ kJ/mol} \]
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Reactants: 1 mole of \(C_3H_8\) and 5 moles of \(O_2\): \[ \Delta H_{reactants} = [(-103.85) + 5(0)] \text{ kJ/mol} = -103.85 \text{ kJ/mol} \]
Now we can find \( \Delta H_{rxn} \):
\[ \Delta H_{rxn} = -2147.7 - (-103.85) = -2043.85 \text{ kJ/mol} \]
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Determine the number of moles of propane in 13.0 g:
The molar mass of propane (C₃H₈) is:
\[ \text{C: } 3 \times 12.01 \text{ g/mol} + \text{H: } 8 \times 1.008 \text{ g/mol} = 36.046 \text{ g/mol} \]
Now, calculate the number of moles of propane in 13.0 g:
\[ \text{moles of } C_3H_8 = \frac{13.0 \text{ g}}{36.046 \text{ g/mol}} \approx 0.360 \text{ moles} \]
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Calculate heat produced from the combustion of 13.0 g of propane:
Since we have \( \Delta H_{rxn} \) for 1 mole, we multiply by the number of moles of propane combusted:
\[ \text{Heat produced} = \Delta H_{rxn} \times \text{moles of } C_3H_8 \]
\[ \text{Heat produced} = -2043.85 \text{ kJ/mol} \times 0.360 \text{ moles} \approx -737.8 \text{ kJ} \]
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Final Answer:
The quantity of heat produced when 13.0 g of propane is completely combusted in air under standard conditions is approximately 737.8 kJ (with the sign indicating it’s an exothermic reaction).
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