Question

Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)BOND ΔHo (kJ/mol)O-O 142H-O 459C-H 411C=O 799O=O 498C-O 358ΔHo = ? kJ/molIs the reaction endothermic or exothermic?

Answer 1

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,$ΔH° = ?

We are given with:

$\Delta H_{O-O}=142 kJ/mol$

$\Delta H_{O=O}=498 kJ/mol$

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{C-H}=411 kJ/mol$

$\Delta H_{C-O}=358 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH° =  

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})$

$\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)$

$\Delta H^o=-794kJ$

$\Delta H^o>0$ endothermic reaction

$\Delta H^o$ exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,