Question

given the tabulated bond enthalpies, estimate the enthalpy of reaction (δhᵣₓₙ) for the combustion of methane: ch₄(g) + 2 o₂(g) → co₂(g) + 2 h₂o(g) δhᵣₓₙ = ? bond enthalpy (kj mol⁻¹) c-h 413 c-c 348 c=c 614 c=o (in co₂) 799 o-o 146 o=o 495 h-o 463 h-h 436 options: +808 kj, -1798 kj, +6092 kj, -808 kj

Explanation:

Step1: Identify bonds broken/formed

Reaction: \( \text{CH}_4(g) + 2\text{O}_2(g)
ightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \)
Bonds broken: 4 C-H (\( 413 \, \text{kJ/mol} \)), 2 O=O (\( 495 \, \text{kJ/mol} \)).
Bonds formed: 2 C=O (in \( \text{CO}_2 \), \( 799 \, \text{kJ/mol} \)), 4 H-O (\( 463 \, \text{kJ/mol} \)).

Step2: Calculate energy for bonds broken

Energy broken: \( (4 \times 413) + (2 \times 495) \)
\( = 1652 + 990 = 2642 \, \text{kJ} \).

Step3: Calculate energy for bonds formed

Energy formed: \( (2 \times 799) + (4 \times 463) \)
\( = 1598 + 1852 = 3450 \, \text{kJ} \).

Step4: Find \( \Delta H_{\text{rxn}} \)

\( \Delta H_{\text{rxn}} = \text{Energy broken} - \text{Energy formed} \)
\( = 2642 - 3450 = -808 \, \text{kJ} \).

Answer:

-808 kJ