Given below are the standard enthalpies of formation of liquid hexene and oxygen gas included in this chemical equation,C6H12 (l) + 9O2 (g) → 6CO2 (g) + 6H2O (l).Compute for the enthalpy changes of reaction of the given chemical reaction and identify what kind of thermochemical reaction it is. Substance InvolvedEnthalpy of FormationC6H12 (l)-151.9 kJ/molO2 (g)0 kJ/molCO2 (g)-393.52 kJ/molH2O (l)-285.8 kJ/mol
Question
Given below are the standard enthalpies of formation of liquid hexene and oxygen gas included in this chemical equation,C6H12 (l) + 9O2 (g) → 6CO2 (g) + 6H2O (l).Compute for the enthalpy changes of reaction of the given chemical reaction and identify what kind of thermochemical reaction it is. Substance InvolvedEnthalpy of FormationC6H12 (l)-151.9 kJ/molO2 (g)0 kJ/molCO2 (g)-393.52 kJ/molH2O (l)-285.8 kJ/mol
Solution
To calculate the enthalpy change of the reaction, we need to use the formula:
ΔH = Σ ΔHf (products) - Σ ΔHf (reactants)
The enthalpy of formation (ΔHf) of the products are:
- For CO2 (g): -393.52 kJ/mol
- For H2O (l): -285.8 kJ/mol
The enthalpy of formation (ΔHf) of the reactants are:
- For C6H12 (l): -151.9 kJ/mol
- For O2 (g): 0 kJ/mol
Substituting these values into the formula, we get:
ΔH = [6(-393.52 kJ/mol) + 6(-285.8 kJ/mol)] - [1(-151.9 kJ/mol) + 9(0 kJ/mol)]
ΔH = [-2361.12 kJ + -1714.8 kJ] - [-151.9 kJ]
ΔH = -4075.92 kJ + 151.9 kJ
ΔH = -3924.02 kJ
The negative sign indicates that the reaction is exothermic, meaning it releases heat. Therefore, the enthalpy change of the reaction is -3924.02 kJ and it is an exothermic reaction.
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