Question

Calculate the standard enthalpy of reaction for the combustion of propane. NOTE: This equation is not balanced. Round to the nearest whole number.
C₃H₈(g) + O₂ --> CO₂(g) + H₂O(l)
kJ/mol

Compound	Hf (kJ/mole)
C3H8(g)	-105
CO2(g)	-394
H2O(l)	-284

Answer 1

Given a balanced equation , C₃H_8(g) + 5 O _2(g)   3 CO _2(g) + 4 H₂O (l)

The standard enthalpy change of reaction is given as , r H ⁰ = H ⁰_f ( products ) -  H ⁰_f ( reactants )

First calculate H ⁰_f ( products ).

H ⁰_f ( products ) = 3 H ⁰_f CO _2(g) + 4 H ⁰_f H₂O (l)

H ⁰_f ( products ) = [ 3 ( - 394 ) + 4 ( - 284 ) ] k J / mol

H ⁰_f ( products ) = - 2318 k J / mol

Now calculate H ⁰_f ( reactants )

H ⁰_f ( reactants ) = H ⁰_f  C₃H_8(g) + 5 H ⁰_f  O _2(g)

H ⁰_f ( reactants ) =[ ( - 105 ) + 5 ( 0) ] k J / mol

H ⁰_f ( reactants ) = - 105 k J / mol

We have , r H ⁰ = H ⁰_f ( products ) -  H ⁰_f ( reactants )

$\therefore$ r H ⁰ = ( - 2318 k J / mol ) - ( - 105 k J / mol )

$\therefore$ r H ⁰ = - 2213 k J / mol

ANSWER : C₃H_8(g) + 5 O _2(g)   3 CO _2(g) + 4 H₂O (l)   r H ⁰ = - 2213 k J / mol