Question

Combine the following equations to determine the enthalpy change for the combustion of 1 mole of propane. Assume that solid carbon is graphite. 3C(s,graphite)+4H2​(g)--->C3​H8​(g)ΔHo = -103.8 kJ/mol C(s,graphite)+O2​(g)--->CO2​(g)ΔHo= -393.5 kJ/mol H2​(g)+1/2​ O2​(g)--->H2​O(g)ΔHo= -285.8 kJ/mol____ kJ/mol

Answer 1

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

                              +

C₃H₈ (g)   ⇒  3 C (s) + 4 H₂ (g)          ΔHº = - (-103.8 Kj ) =    + 103.8 kJ

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

                                     +

C₃H₈ (g)  + 3 O₂ (g)    ⇒  3 CO₂ (g)  + 4 H₂ (g)                ΔHº  =  -1076.5 kJ

C₃H₈ (g)   +    5 O₂ (g)     ⇒  3 CO₂ (g)      + 4 H₂O (g)       ΔHº = -2219.7 kJ