SAT High School
Answers
Answer 1
Considering the normal of standard enthalpy of formation, the molar enthalpy for the reaction is -2043.98 kJ.
The balanced reaction is:
C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O (g)
The normal or standard enthalpy of formation (also sometimes called the normal heat of formation), is represented by ΔHf and is the enthalpy change when a mole of compound is formed from its elements in the normal state (that is, in the state of aggregation and more stable allotropic form to which said elements are found under standard conditions).
The enthalpy of formation may be positive, an endothermic reaction, or negative, an exothermic reaction.
In summary, the enthalpy of formation of a compound is defined as the enthalpy of reaction of the reaction in which a mole of said compound is formed from its elements in their most stable states of aggregation.
The use of enthalpies of formation allows to calculate enthalpies of reaction as follows: The standard enthalpy of a reaction is equal to the sum of the standard molar enthalpies of formation of the products minus the sum of the standard molar enthalpies of the formation of the reactants, multiplied respectively by the stoichiometric coefficients that appear in the fitted equation.
In this case, you have ΔHf values in kJ/mol for:
C₃H₈(g) = -103.8O₂(g) = 0CO₂(g) = -393.5H₂O(g) =- 241.82
So, being: ΔH° = ΣnHfº (products) - ΣmHfº (reactants); n and m are total moles
and replacing the values:
ΔH° = [4 moles× (-241.82 [tex]\frac{kJ}{mol}[/tex]) + 3 moles× (-393.5 [tex]\frac{kJ}{mol}[/tex])] - [5 moles× (0 [tex]\frac{kJ}{mol}[/tex]) + 1 moles× (-103.8 [tex]\frac{kJ}{mol}[/tex])]
Solving:
ΔH° = [-967.28 kJ - 1180.5 kJ] - [0 kJ - 103.8 kJ]
ΔH° = -2147.78 kJ - [-103.8 kJ]
ΔH° = -2147.78 kJ + 103.8 kJ
ΔH° = -2043.98 kJ
In summary, the molar enthalpy for the reaction is -2043.98 kJ.
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Solution Points on a graph
The points which lie on the graph can be tested for as follows;
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