Question

given the following tabulated data, calculate the enthalpy of reaction (δhᵣₓₙ) for the combustion of propane:
c₃h₈(g) + 5 o₂(g) → 3 co₂(g) + 4 h₂o(l)
δhᵣₓₙ = ?

compound	δhᵢ° (kj mol⁻¹)
c₂h₄(g)	-85.0
c₃h₈(g)	-103.9
co(g)	-110.5
co₂(g)	-393.5
h₂o(l)	-285.8

options:

• 6754 kj
• 2560 kj
• 2324 kj
• 2220 kj

Explanation:

Step1: Recall Hess's Law

The enthalpy of reaction ($\Delta H_{\text{rxn}}$) can be calculated using the formula:
$\Delta H_{\text{rxn}} = \sum n\Delta H_f^{\circ}(\text{products}) - \sum m\Delta H_f^{\circ}(\text{reactants})$,
where $n$ and $m$ are the stoichiometric coefficients.

Step2: Identify Reactants and Products

Reaction: $\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}$
- Reactants: $\ce{C3H8(g)}$ (coefficient = 1), $\ce{O2(g)}$ (coefficient = 5; $\Delta H_f^{\circ}(\ce{O2(g)}) = 0$ kJ/mol, as it is an element in its standard state).
- Products: $\ce{CO2(g)}$ (coefficient = 3), $\ce{H2O(l)}$ (coefficient = 4).

Step3: Extract $\Delta H_f^{\circ}$ Values

From the table:
- $\Delta H_f^{\circ}(\ce{C3H8(g)}) = -103.9$ kJ/mol,
- $\Delta H_f^{\circ}(\ce{CO2(g)}) = -393.5$ kJ/mol,
- $\Delta H_f^{\circ}(\ce{H2O(l)}) = -285.8$ kJ/mol.

Step4: Calculate $\sum n\Delta H_f^{\circ}(\text{products})$

For $\ce{CO2(g)}$: $3 \times (-393.5)$ kJ/mol = $-1180.5$ kJ/mol.
For $\ce{H2O(l)}$: $4 \times (-285.8)$ kJ/mol = $-1143.2$ kJ/mol.
Sum of products: $-1180.5 + (-1143.2) = -2323.7$ kJ/mol (≈ -2324 kJ/mol).

Step5: Calculate $\sum m\Delta H_f^{\circ}(\text{reactants})$

For $\ce{C3H8(g)}$: $1 \times (-103.9)$ kJ/mol = $-103.9$ kJ/mol.
For $\ce{O2(g)}$: $5 \times 0 = 0$ kJ/mol.
Sum of reactants: $-103.9 + 0 = -103.9$ kJ/mol.

Step6: Compute $\Delta H_{\text{rxn}}$

$\Delta H_{\text{rxn}} = (-2323.7) - (-103.9) = -2323.7 + 103.9 = -2219.8$ kJ/mol (≈ -2220 kJ/mol). Wait, no—wait, correction: Wait, the formula is products - reactants. Wait, let's recheck:

Wait, $\Delta H_{\text{rxn}} = [3\Delta H_f(\ce{CO2}) + 4\Delta H_f(\ce{H2O})] - [\Delta H_f(\ce{C3H8}) + 5\Delta H_f(\ce{O2})]$.

Substitute values:
$[3(-393.5) + 4(-285.8)] - [(-103.9) + 5(0)]$
$= [ -1180.5 - 1143.2 ] - [ -103.9 ]$
$= (-2323.7) - (-103.9)$
$= -2323.7 + 103.9 = -2219.8$ kJ/mol ≈ -2220 kJ/mol. Wait, but the options include -2324 kJ. Wait, did I mix up reactants and products? Wait no—wait, no: Wait, the reactant is $\ce{C3H8}$, so the formula is (sum of products) - (sum of reactants). Wait, sum of reactants: $\Delta H_f(\ce{C3H8}) + 5\Delta H_f(\ce{O2}) = -103.9 + 0 = -103.9$. Sum of products: $3(-393.5) + 4(-285.8) = -1180.5 - 1143.2 = -2323.7$. Then $\Delta H_{\text{rxn}} = (-2323.7) - (-103.9) = -2219.8 ≈ -2220$? But wait, the options have -2324. Wait, maybe I made a mistake. Wait, no—wait, the reaction is combustion of propane, which is exothermic. Wait, let's re-express the formula: $\Delta H_{\text{rxn}} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$. So products are CO2 and H2O, reactants are C3H8 and O2. So:

Products: 3*(-393.5) + 4*(-285.8) = -1180.5 - 1143.2 = -2323.7
Reactants: 1*(-103.9) + 5*0 = -103.9
Thus, $\Delta H_{\text{rxn}} = (-2323.7) - (-103.9) = -2219.8 ≈ -2220$ kJ/mol. Wait, but the options include -2324. Wait, maybe the question’s table or my reading is wrong? Wait, no—wait, maybe I flipped the formula. Wait, no: Hess’s law is $\Delta H_{\text{rxn}} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$. Let me check again.

Wait, another way: The combustion of propane: $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$. The standard enthalpy of combustion of propane is known to be around -2220 kJ/mol (or -2219.9 kJ/mol). So the closest option is -2220 kJ? Wait, but the options are -6754, -2560, -2324, -2220. Wait, my calculation gives ~-2220, which matches the last option. Wait, but earlier I thought -2324, but that was a miscalculation. Wait, no—wait, 3*(-393.5) is -1180.5, 4*(-285.8) is -1143.2. Sum: -1180.5 -1143.2 = -2323.7. Then subtract reactants: -2323.7 - (-103.9) = -2323.7 + 103.9 = -2219.8 ≈ -2220. Yes, so the correct answer is -2220 kJ. Wait, but the option is -2220 kJ. Wait, but let me check the table again. Wait, the table for $\ce{C3H8}$ is -103.9, $\ce{CO2}$ -393.5, $\ce{H2O(l)}$ -285.8. So the calculation is correct. So the answer is -2220 kJ.

Answer:

-2220 kJ (Option: -2220 kJ)