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-c	348
c=c	614
c=o	799 (in co₂)
o-o	146
o=o	495
h-o	463
h-h	436

options:
○ +6092 kj
○ +808 kj
○ -808 kj
○ -1798 kj

Explanation:

Step1: Identify Bonds Broken/Formed

Reaction: $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}$
Bonds broken: 4 C-H, 2 O=O.
Bonds formed: 2 C=O (in $\ce{CO2}$), 4 O-H (in 2 $\ce{H2O}$).

Step2: Calculate Energy for Bonds Broken

Energy to break bonds:
$4\times\ce{C-H} + 2\times\ce{O=O} = 4\times413 + 2\times495$
$= 1652 + 990 = 2642\ \text{kJ}$

Step3: Calculate Energy for Bonds Formed

Energy released from forming bonds:
$2\times\ce{C=O (in CO2)} + 4\times\ce{O-H} = 2\times799 + 4\times463$
$= 1598 + 1852 = 3450\ \text{kJ}$

Step4: Calculate $\Delta H_{\text{rxn}}$

$\Delta H_{\text{rxn}} = \text{Bonds broken} - \text{Bonds formed} = 2642 - 3450 = -808\ \text{kJ}$

Answer:

-808 kJ (Option: -808 kJ)