The synthesis of hexane from its elements in their most stable form at 1 atm pressureoccurs via the following reaction:6 C (s) + 7 H 2 (g) ® C6 H 14 (l)Calculate the standard enthalpy of formation of hexane using the enthalpies ofcombustion given below:C6 H 14 (l) + 19 ⁄2 O 2 (g) ® 6 CO 2 (g) + 7 H 2 O (l) ΔH = −4163.0 kJ mol-1C (s) + O 2 (g) ® CO 2 (g) ΔH = −393.5 kJ mol-1H 2 (g) + 1 ⁄2 O 2 (g) ® H 2 O (l) ΔH = −285.5 kJ mol-1
Question
The synthesis of hexane from its elements in their most stable form at 1 atm pressureoccurs via the following reaction:6 C (s) + 7 H 2 (g) ® C6 H 14 (l)Calculate the standard enthalpy of formation of hexane using the enthalpies ofcombustion given below:C6 H 14 (l) + 19 ⁄2 O 2 (g) ® 6 CO 2 (g) + 7 H 2 O (l) ΔH = −4163.0 kJ mol-1C (s) + O 2 (g) ® CO 2 (g) ΔH = −393.5 kJ mol-1H 2 (g) + 1 ⁄2 O 2 (g) ® H 2 O (l) ΔH = −285.5 kJ mol-1
Solution
To calculate the standard enthalpy of formation of hexane, we need to use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction and is independent of the path of the reaction.
The enthalpy of formation of hexane is the enthalpy change for the reaction forming hexane from its elements in their standard states.
The enthalpy of combustion of hexane, carbon, and hydrogen are given. We can use these to calculate the enthalpy of formation of hexane.
First, we need to write the combustion reactions for carbon and hydrogen:
C (s) + O2 (g) -> CO2 (g) ΔH = -393.5 kJ/mol H2 (g) + 1/2 O2 (g) -> H2O (l) ΔH = -285.5 kJ/mol
The combustion reaction for hexane is:
C6H14 (l) + 19/2 O2 (g) -> 6 CO2 (g) + 7 H2O (l) ΔH = -4163.0 kJ/mol
We can rearrange these equations to give the formation reaction for hexane:
6 C (s) + 7 H2 (g) -> C6H14 (l)
The enthalpy change for this reaction is the sum of the enthalpy changes for the combustion reactions of the reactants minus the enthalpy change for the combustion reaction of the product:
ΔHf [C6H14 (l)] = 6 ΔHc [C (s)] + 7 ΔHc [H2 (g)] - ΔHc [C6H14 (l)]
Substituting the given values gives:
ΔHf [C6H14 (l)] = 6(-393.5 kJ/mol) + 7(-285.5 kJ/mol) - (-4163.0 kJ/mol)
Solving this gives:
ΔHf [C6H14 (l)] = -2361 kJ/mol + -1998.5 kJ/mol + 4163.0 kJ/mol = -196.5 kJ/mol
So, the standard enthalpy of formation of hexane is -196.5 kJ/mol.
Similar Questions
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
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