Question

The equilibrium constant for the reaction 2 C3H6(g) ↔ C2H4(g) + C4H8(g) Is found to fit the expression: lnK = A + B/T + C/T2, between 300 and 600 K, where A = -1.04, B = -1088 K, C = 1.51x105 K2. Calculate the standard reaction enthalpy and standard reaction entropy at 400 K.

Answer 1

The standard reaction enthalpy and the standard reaction enthalpy at 400 K for this reaction are equal to 0.35 kJ and 0.000125 kJ respectively.

How to calculate Gibbs's free energy?

Mathematically, the Gibbs's free energy for this chemical reaction can be calculated by using this formula:

ΔG° = -RTlnK = ΔH° - TΔS°    ......equation 1.

First of all, we would determine the value of lnK from the given expression at 390 K and 410 K respectively:

lnK = A + B/T + C/T²

lnK = -1.04 - 1088/T + 1.51 × 10⁵/T²    

lnK = -1.04 - 1088/390 + 1.51 × 10⁵/390²

lnK = -1.04 - 2.79 + 0.99

lnK = -2.84.

At T = 410 K, we have:    

lnK' = -1.04 - 1088/410 + 1.51 × 10⁵/410²

lnK' = -1.04 - 2.65 + 0.90

lnK' = -2.79.

For the standard reaction enthalpy, we have:

lnK' - lnK = ΔH°/R(1/T - 1/T')

-2.79 - (-2.84) = ΔH°/8.314(1/390 - 1/410)

-2.79 + 2.84 = ΔH°/8.314(0.00256 - 0.00244)

0.05 = ΔH°/8.314(0.0012)

0.4157 = 0.0012ΔH°

ΔH° = 0.4157/0.0012

ΔH° = 346.42 ≈ 0.35 kJ.

Next, we would determine the Gibbs's free energy at each temperature:

At T = 390 K, we have:

ΔG° = -RTlnK

Δ₁G° = -8.314 × 390 × (-2.84)

Δ₁G° = 9.21 kJ/mol.

At T = 410 K, we have:

Δ₂G° = -8.314 × 410 × 2.79

Δ₂G° = 9.51 kJ/mol.

For the standard reaction enthalpy at 400 K, we have:

ΔG° = ΔH° - TΔS°

9.51 - 9.21 = 0.35 - 400ΔS°

0.30 = 0.35 - 400ΔS°

400ΔS° = 0.35 - 0.30

400ΔS° = 0.05

ΔS° = 0.05/400

ΔS° = 0.000125 kJ.

Read more on Gibbs's free energy here: brainly.com/question/18752494

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