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

The enthalpy of combustion for one mole of propane, given the provided formation data, is -2220.9 kJ/mol. This is calculated by combining and manipulating the given enthalpies of reactants and products to match the combustion reaction of propane.

Step-by-step explanation:

To calculate the enthalpy of combustion for one mole of propane (C3H8), we first need to write the combustion reaction of propane:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Now, we'll combine the given reactions to match the combustion reaction:

3C(s, graphite) + 4H2(g) → C3H8(g) ΔHo = -103.8 kJ/mol (Reverse this equation to represent the decomposition of propane)

C(s, graphite) + O2(g) → CO2(g) ΔHo= -393.5 kJ/mol (Multiply by 3 to match the 3 moles of CO2 produced)

H2(g) + ½O2(g) → H2O(g) ΔHo= -285.8 kJ/mol (Multiply by 4 to match the 4 moles of H2O produced)

The sum of the enthalpies for these steps will be the desired enthalpy change for the combustion of propane. Let's calculate:

1) Reversing the first equation changes its sign: +103.8 kJ/mol

2) Three times the second equation: 3(-393.5 kJ/mol) = -1180.5 kJ/mol

3) Four times the third equation: 4(-285.8 kJ/mol) = -1143.2 kJ/mol

Adding these together, the enthalpy of combustion for propane is:

+103.8 kJ/mol - 1180.5 kJ/mol - 1143.2 kJ/mol = -2220.9 kJ/mol

Answer 2

-2219.7 kJ/mol

Step-by-step 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