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.
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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