Question

Consider the exothermic reaction
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
Calculate the standard heat of reaction, or ΔH∘rxn, for this reaction using the given data. Also consider that the standard enthalpy of the formation of elements in their pure form is considered to be zero.
Reactant or product
ΔH∘f (kJ/mol) C2H6(g) -84.7 CO2(g) -393.5 H2O(g) -241.8
Express your answer to four significant figures and include the appropriate units.

Answer 1

Answer:

The answer to your question is -2855 J

Explanation:

Reaction

                     2C₂H₆  +  7O₂   ⇒   4CO₂  +  6H₂O

Formula

Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants

Substitution

ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}

ΔHrxn = {-1574 -1450.8} - {-169.4}

ΔHrxn = -3024.8 + 169.4

ΔHrxn = -2855.4 J

Answer 2

The standard heat of reaction for the complete combustion of ethane is -2855.4 kJ.

What is the enthalpy of reaction?

The heat of reaction (also known as enthalpy of reaction) is the change in the enthalpy of a chemical reaction that occurs at a constant pressure.

Let's consider the following balanced equation.

2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g)

We can calculate the standard heat of reaction (ΔH°rxn) using the following expression.

ΔH°rxn = ∑ n(p) × ΔH°f(p) - ∑ n(r) × ΔH°f(r)

where,

• n is the number of moles.
• ΔH°f is the standard enthalpy of formation.
• p refers to the products.
• r refers to the reactants.

ΔH°rxn = 4 mol × (-393.5 kJ/mol) + 6 mol × (-241.8 kJ/mol) - 2 mol × (-84.7 kJ/mol) - 7 mol × (0 kJ/mol) = -2855.4 kJ

The standard heat of reaction for the complete combustion of ethane is -2855.4 kJ.

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