At 25°C,the standard enthalpy of combustion of gaseous propane (C₃H₈) is -2219.0 kJ per mole of propane,and the standard enthalpy of combustion of gaseous propylene (C₃H₆) is -2058.3 kJ per mole of propylene.What is the standard enthalpy change for the following reaction at 25°C?
C₃H₆(g) + H₂(g) → C₃H₈(g)
Substance
ΔH°_f (kJ/mol)
CO₂(g)
-393.5
H₂O(l)
-285.8

Question

Quizplus · Accepted Answer

We can use Hess's law to find the enthalpy change for the given reaction: C₃H₆(g) + H₂(g) + 5/2O2(g) → C₃H₈(g) + H2O(l) + CO2(g) First, we need to balance the equation with the given enthalpies: C₃H₆(g) + 9/2O2(g) → 3CO2(g) + 2H2O(g) (ΔH° = -2058.3 kJ/mol) H₂(g) + 1/2O2(g) → HSO4(g) (ΔH° = -249.2 kJ/mol) HSO4(g) + H2O(l) → HSO4-(aq) + H3O+(aq) (ΔH° = -56.2 kJ/mol) 3CO2(g) + 4H3O+(aq) + 4HSO4-(aq) → C₃H₈(g) + 12H2O(l) + 4SO4-(aq) (ΔH° = -2219.0 kJ/mol) Adding up these reactions and canceling out the intermediates gives: H₂O(l) + 9/2O2(g) → CO2(g) + H2O(g) + HSO4-(aq) (ΔH° = -311.5 kJ/mol) C₃H₆(g) + H₂(g) + 5/2O2(g) → C₃H₈(g) + H2O(l) + CO2(g) (ΔH° = -1965.5 kJ/mol) Lastly, we need to convert the product H2O(g) to H2O(l) using the enthalpy of vaporization: H2O(g) → H2O(l) (ΔH° = 40.7 kJ/mol) Adding this to the previous equation gives: C₃H₆(g) + H₂(g) → C₃H₈(g) (ΔH° = -125.1 kJ/mol) Therefore, the answer is C, which is the enthalpy change for the given reaction.

At 25°C,the Standard Enthalpy of Combustion of Gaseous Propane (C3H8)is

At 25°C,the Standard Enthalpy of Combustion of Gaseous Propane (C₃H₈)is