I. In applied thermodynamics, Gibbs free energy plays an important role and can be calculated accord — Physical Chemistry — Thermodynamics Chemistry Question
Applied thermodynamics
I. In applied thermodynamics, Gibbs free energy plays an important role and can be calculated according to the following expression:
ΔGo298 = ΔHo298 – TΔSo298
ΔGo298 - standard free energy change
ΔHo298 - standard enthalpy change
ΔSo298 - standard entropy change
The burning of graphite is represented by two reactions:
C(graphite) + ½ O2 (g) → CO (g) (1)
C(graphite) + O2 (g) → CO2 (g) (2)
The dependence of ΔHo, ΔSo on temperature is as follows:
Reaction (1): ΔHoT (1) (J·mol-1) = –112298.8 + 5.94T;
ΔSoT (1) (J·K-1·mol-1) = 54.0 + 6.21 lnT
Reaction (2): ΔHoT (2) (J·mol-1) = – 393740.1 + 0.77 T;
ΔSoT (2) (J·K-1·mol-1) = 1.54 – 0.77 lnT
Based on the above data:
Derive the expression for the Gibbs free energy as a function of temperature, ΔGoT = f(T) for each reaction.
Model Answer
Based on the above data:
ΔGoT = ΔHoT – TΔSoT
Reaction (1): ΔGoT (1) = (– 112298.8 + 5.94 T) – T(54.0 + 6.21 ln T)
ΔGoT (1) = – 112298.8 – 48.06 T – 6.21 T lnT
Reaction (2): ΔGoT (2) = (– 393740.1 + 0.77 T) – T(1.54 – 0.77 ln T)
ΔGoT (2) = – 393740.1 – 0.77 T + 0.77 T lnT
Predict the changes of ΔGoT with an increase in temperature.
Model Answer
ΔGoT (1) decreases with an increase in temperature.
ΔGoT (2) increases with an increase in temperature.
II. Assume that at 1400 oC, during the course of reactions (1) and (2), CO gas might continue to react with O2 to form the final product CO2.
Write down the reaction (3) for the formation of CO2 from CO gas.
Model Answer
C(graphite) + ½ O2(g) → CO(g) (1)
C(graphite) + O2(g) → CO2(g) (2)
(2) – (1) → CO(g) + ½ O2 → CO2(g) (3)
We have, ΔGoT (3) = ΔGoT (2) – ΔGoT (1)
Calculate ΔGoT (3).
Model Answer
Substitute the values in:
ΔGoT (3) = (– 393740.1 – 0.77 T + 0.77 T lnT) – (– 112298.8 – 48.06 T – 6.21T lnT)
ΔGoT (3) = – 281441.3 + 47.29 T – 6.98 T lnT
At 1673 K: ΔGoT (3) = –115650 J/mol
Determine the equilibrium constant Kp for reaction (3) at the given temperature.
Model Answer
Since ΔGo = –RT lnKp, the equilibrium constant Kp for reaction (3) can be calculated as follows:
-ΔGo1673 (3) / (8.314×1673) = 115650 / (8.314×1673) = 8.313457 = ln Kp,1673 (3)
→ Kp,1673 (3) = 4083
III. In an experiment, NiO powder and CO gas were placed in a closed container which was then heated up to 1400 oC. When the system reached equilibrium, there were four species present: NiO(s), Ni(s), CO(g) and CO2(g). The mole percentages of CO and CO2 are 1 % and 99 %, respectively, and the pressure of the system is 1.0 bar (10^5 Pa).
Write down the reactions in the above experiment.
Model Answer
CO(g) + ½ O2(g) → CO2(g) (3)
NiO(s) + CO(g) → Ni(s) + CO2(g) (4)
(4) – (3) NiO(s) → Ni(s) + ½ O2(g) (5)
Based on the experimental results and the above thermodynamic data, calculate the pressure of O2 in the equilibrium with NiO and Ni at 1400 oC.
Model Answer
At 1673 K, we have:
For reaction (4): Kp(4) = pCO2 / pCO = 99 / 1 = 99
For reaction (3): Kp(3) = pCO2 / (pCO * pO2^1/2) = 4083
or Kp(4) / pO2^1/2 = 99 / pO2^1/2 = 4083
For reaction (5): Kp(5) = pO2^1/2 = Kp(4) / Kp(3) = 99 / 4083 = 0.024247 = 2.4247 * 10^-2
Hence, pO2 = (Kp(5))^2 = (2.4247 * 10^-2)^2 = 5.88 * 10^-4 bar = 58.8 Pa