Calculate $\Delta$S° for the combustion of propane. C₃H₈(g)+ 5O₂(g) $\to$3CO₂(g)+ 4H₂O(g)

Question

Quizplus · Accepted Answer

The enthalpy change for the combustion of propane can be calculated using the given equation:
CSU13B1(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
We can use standard enthalpies of formation to calculate the enthalpy change. The given standard enthalpies of formation are:
ΔHf°(CSU13B1) = -104.7 kJ/mol
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -241.8 kJ/mol
ΔHf°(O2) = 0 kJ/mol

We can calculate the enthalpy change using the following equation:
ΔH° = ΣnΔHf°(products) - ΣmΔHf°(reactants)
where n and m are the stoichiometric coefficients of the products and reactants, respectively.

For the given reaction, n = 1 and m = 5, so:
ΔH° = (3ΔHf°(CO2) + 4ΔHf°(H2O)) - (ΔHf°(CSU13B1) + 5ΔHf°(O2))
ΔH° = (3(-393.5) + 4(-241.8)) - (-104.7 + 5(0))
ΔH° = -2220.1 kJ/mol

We can convert kJ/mol to J/K using the molar mass of propane and Avogadro's number:
1 mol of C3H8 has a mass of 44.09 g
1 J/K = 1 J/mol*K

So,
ΔH° = -2220.1 kJ/mol = -2220.1 * 1000 J/mol
= -2220100 J/mol

We can use the equation:
ΔS° = ΣnS°(products) - ΣmS°(reactants)
to calculate the standard entropy change.

For the given reaction, n = 1 and m = 5, so:
ΔS° = (3S°(CO2) + 4S°(H2O)) - (S°(CSU13B1) + 5S°(O2))
ΔS° = (3(213.6) + 4(69.9)) - (186.3 + 5(205.0))
ΔS° = -205.0 J/K

Finally, we can calculate the standard Gibbs free energy change using the equation:
ΔG° = ΔH° - TΔS°

We can assume that the reaction occurs at standard conditions, so T = 298 K:
ΔG° = -2220100 - (298*(-205.0))
ΔG° = -2220100 + 61010
ΔG° = -2159090 J/mol

We can convert J/mol to J/K using Avogadro's number:
1 J/mol = 1/6.022*10^23 J/K

So,
ΔG° = -2159090 J/mol = -3.588*10^-20 J/K

Now, we can use the equation:
ΔG° = -RTln(K)
to calculate the equilibrium constant, K.

We can assume that the reaction occurs at standard conditions, so T = 298 K and R = 8.314 J/(mol*K):
-3.588*10^-20 = -8.314*298ln(K)
ln(K) = 1.392*10^-20
K = e^ln(K)
K = 1.393*10^-9

Finally, we can calculate ΔG° for the reaction at non-standard conditions using the equation:
ΔG = ΔG° + RTln(Q)

where Q is the reaction quotient. For the given reaction, Q = 1 because we assume that the reactants and products are in their standard states.

So,
ΔG = -2159090 + (8.314*298*ln(1))
ΔG = -2159090 J/mol

We can convert J/mol to J/K using Avogadro's number:
1 J/mol = 1/6.022*10^23 J/K

So,
ΔG = -2159090 J/mol = -3.588*10^-20 J/K

Now, we can use the equation:
ΔG = ΔH - TΔS
to calculate the non-standard enthalpy change:

ΔH = ΔG + TΔS
ΔH = (-3.588*10^-20) + (298*(-205.0))
ΔH = -60654 J/mol

We can convert J/mol to J/K using Avogadro's number:
1 J/mol = 1/6.022*10^23 J/K

So,
ΔH = -60654 J/mol = -1.008*10^-20 J/K

The answer is D) 100.9 J/K.

Calculate Δ\DeltaΔ S° for the Combustion of Propane →\to→ 3CO2(g)+ 4H2O(g) A)-100

Calculate $\Delta$ S° for the Combustion of Propane $\to$ 3CO₂(g)+ 4H₂O(g)

A)-100