Answer:
-2219.7 kJ/mol
Explanation:
Our strategy here is to combine these equations to get the desired equation for the combustion of 1 mol C₃H₈ from the given equations:
C₃H₈ (g) + 5 O₂ (g) ⇒ 3 CO₂ (g) + 4 H₂O (g)
The first equation given is:
3 C (s) + 4 H₂ (g) ⇒ C₃H₈ (g) ΔHº = -103.8 Kj
We see right away we need to reverse this equation since we want C₃H₈ (g) as a reactant:
C₃H₈ (g) ⇒ 3 C (s) + 4 H₂ (g) ΔHº = - (-103.8 Kj ) = + 103.8 kJ
Now we need to cancel out 3 C(s) , so take three times the second given equation and addd to the first:
3 [C(s)+O2(g) ⇒ CO2(g)ΔHo= -393.5 kJ ]
3 C (s) + 3 O₂ (g) ⇒ 3 CO₂ (g) ΔHº = 3 (-393.5 kJ ) = -1180.5 kJ
+
<u>C₃H₈ (g) ⇒ 3 C (s) + 4 H₂ (g) ΔHº = - (-103.8 Kj ) = + 103.8 kJ</u>
C₃H₈ (g) + 3 O₂ (g) ⇒ 3 CO₂ (g) + 4 H₂ (g) ΔHº = -1076.5 kJ
Now we need to cancel out the 4 H₂ from the last equation, so we will multiply the third equation and add it to the last one:
4 ( H₂ (g) + 1/2 O₂ (g) ⇒ H₂O (g) ΔHº = -285.8 kJ )
4 H₂ (g) + 2 O₂ (g) ⇒ 4 H₂O (g) ΔHº = 4(-285.8 kJ ) = -1143.2 kJ
+
<u>C₃H₈ (g) + 3 O₂ (g) ⇒ 3 CO₂ (g) + 4 H₂ (g) ΔHº = -1076.5 kJ</u>
C₃H₈ (g) + 5 O₂ (g) ⇒ 3 CO₂ (g) + 4 H₂O (g) ΔHº = -2219.7 kJ
