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Calculate the enthalpy change of the reaction C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l) using the given standard enthalpies of formation.

Chemistry

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Answer

B. -2222 kJ/mol\text{B. -2222 kJ/mol}

Step-by-step solution

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To calculate the enthalpy change of the reaction (ΔHrxn\Delta H_{rxn}), we use the formula:

ΔHrxn=∑nΔHf∘(products)−∑mΔHf∘(reactants)\Delta H_{rxn} = \sum n \Delta H_f^\circ(\text{products}) - \sum m \Delta H_f^\circ(\text{reactants})

where nn and mm are the stoichiometric coefficients for the products and reactants, respectively.

The given reaction is:

C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})

Given standard enthalpies of formation (ΔHf∘\Delta H_f^\circ):

  • ΔHf∘(C3H8(g))=−104 kJ/mol\Delta H_f^\circ(\text{C}_3\text{H}_8(\text{g})) = -104 \text{ kJ/mol}
  • ΔHf∘(H2O(l))=−286 kJ/mol\Delta H_f^\circ(\text{H}_2\text{O}(\text{l})) = -286 \text{ kJ/mol}
  • ΔHf∘(CO2(g))=−394 kJ/mol\Delta H_f^\circ(\text{CO}_2(\text{g})) = -394 \text{ kJ/mol}
  • ΔHf∘(O2(g))=0 kJ/mol\Delta H_f^\circ(\text{O}_2(\text{g})) = 0 \text{ kJ/mol} (since it is an element in its standard state)

  1. Calculate the sum of the enthalpies of formation for the products.

    ∑nΔHf∘(products)=[3×ΔHf∘(CO2(g))]+[4×ΔHf∘(H2O(l))]\sum n \Delta H_f^\circ(\text{products}) = [3 \times \Delta H_f^\circ(\text{CO}_2(\text{g}))] + [4 \times \Delta H_f^\circ(\text{H}_2\text{O}(\text{l}))] =[3×(−394 kJ/mol)]+[4×(−286 kJ/mol)]= [3 \times (-394 \text{ kJ/mol})] + [4 \times (-286 \text{ kJ/mol})] =−1182 kJ−1144 kJ= -1182 \text{ kJ} - 1144 \text{ kJ} =−2326 kJ= -2326 \text{ kJ}

  2. Calculate the sum of the enthalpies of formation for the reactants.

    ∑mΔHf∘(reactants)=[1×ΔHf∘(C3H8(g))]+[5×ΔHf∘(O2(g))]\sum m \Delta H_f^\circ(\text{reactants}) = [1 \times \Delta H_f^\circ(\text{C}_3\text{H}_8(\text{g}))] + [5 \times \Delta H_f^\circ(\text{O}_2(\text{g}))] =[1×(−104 kJ/mol)]+[5×(0 kJ/mol)]= [1 \times (-104 \text{ kJ/mol})] + [5 \times (0 \text{ kJ/mol})] =−104 kJ+0 kJ= -104 \text{ kJ} + 0 \text{ kJ} =−104 kJ= -104 \text{ kJ}

  3. Calculate ΔHrxn\Delta H_{rxn}.

    ΔHrxn=∑nΔHf∘(products)−∑mΔHf∘(reactants)\Delta H_{rxn} = \sum n \Delta H_f^\circ(\text{products}) - \sum m \Delta H_f^\circ(\text{reactants}) =(−2326 kJ)−(−104 kJ)= (-2326 \text{ kJ}) - (-104 \text{ kJ}) =−2326 kJ+104 kJ= -2326 \text{ kJ} + 104 \text{ kJ} =−2222 kJ/mol= -2222 \text{ kJ/mol}

    The calculated value matches option B.

    The final answer is B. -2222 kJ/mol\text{B. -2222 kJ/mol}.

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