“Tug of war is a sport that directly pits two teams against each other in a test of strength. This i — Physical Chemistry — Thermodynamics Chemistry Question
Tug of war
“Tug of war is a sport that directly pits two teams against each other in a test of strength. This is also a traditional game of Vietnamese people”
The following table gives the standard molar Gibbs energy at different temperatures for the reaction (1) below:
SO3(g) SO2 (g) + ½ O2 (g) (1)
Use the Van Hoff’s equation to estimate the lnKp1 at each temperature.
Model Answer
∆rnx G = ∆rnxG o + RT ln (pSO2 pO2^1/2) / pSO3
At equilibrium: ∆rnx G = ∆rnxG o + RT ln (pSO2 pO2^1/2) / pSO3 = 0
∆rG o = – RT ln (pSO2 pO2^1/2) / pSO3 = – RT lnKp1
T / K = T / oC + 273;
T/ K 800 825 900 953 100
ln Kp1 –3.263 –3.007 –1.899 –1.173 – 0.591
Plot ln Kp1 against 1/T to determine the value of ∆rH o in kJ mol–1 assuming that ∆rnxH o does not vary significantly over the given temperature range.
Model Answer
Plot lnKp against 1/T:
y = -10851x + 10.216
R2 = 0.9967
Assuming that ∆rH o is temperature independent, the slope of this plot is -∆rH o / R, so that ∆rnxH o = 90.2 kJ/mol.
Using the best-fit line to plot a lnKp1 versus 1/T, determine the Kp2 for the following reaction (2) at 651.33 oC:
2 SO3(g) 2 SO2 (g) + O2 (g) (2)
Model Answer
2 SO3(g) 2 SO2 (g) + O2 (g) (2)
A best-fit equation is lnKp1 = – 10851(1/T) + 10.216 with R-squared value of 0.9967.
We can use this equation to estimate the Kp1 at (651.33 + 273) = 924.33 K because ∆rnxH o is temperature independent.
ln Kp1 = –10851(1/924.33) + 10.216 → ln Kp1 = –1.523313881 → Kp1 = 0.218
For reaction (2), the equilibrium constant is expressed as:
Kp2 = (pSO2^2 pO2) / pSO3^2 = (Kp1)^2 = (0.218)^2 = 0.047524
An amount of 15.19 g of iron (II) sulfate was heated in an evacuated 1.00 dm3 container to 651.33 oC, in which the following reactions take place:
FeSO4 (s) Fe2O3 (s) + SO3 (g) + SO2(g) (3)
2 SO3(g) 2 SO2 (g) + O2 (g) (4)
When the system has reached equilibrium, the partial pressure of oxygen is of 21.28 mmHg. Calculate the equilibrium pressure of the gases and the value of Kp3 for the reaction (3) at equilibrium.
Model Answer
Reaction (3): 2 FeSO4 (s) Fe2O3 (s) + SO3 (g) + SO2 (g)
Decomposition: - - -
Equilibrium: p – a p + a
Reaction (4): 2 SO3(g) 2 SO2 (g) + O2 (g)
Initial p: p p 0
Change -a +a +a/2
Equilibrium p – a p + a a/2
At equilibrium: partial pressure of oxygen = 21.28 / 760 = 0.028 atm
a/2 = 0.028 atm → a = 0.056 atm
Equilibrium constant for (4):
Kp4 = (pSO2^2 pO2) / pSO3^2 = (Kp1)^2 = (0.218)^2 = 0.047524
Kp4 = (p + a)^2 (a / 2) / (p - a)^2 = (p + 0.056)^2 (0.028) / (p - 0.056)^2 = 0.047524
(p + 0.056)^2 / (p - 0.056)^2 = 1.6973
(p + 0.056) / (p - 0.056) = 1.303
p + 0.056 = 1.303 p - 0.073
0.303 p = 0.12896 → p = 0.425 atm
Equilibrium constant for (3) 2 FeSO4 (s) Fe2O3 (s) + SO3 (g) + SO2 (g)
Kp3 = pSO3 pSO2 = (p – a)(p + a) = (0.425 - 0.056)(0.425 + 0.056) = 0.177
Calculate the percentage of FeSO4 decomposed?
Model Answer
Amount of substance of SO3 = SO2 comes from the decomposition of FeSO4:
p V = n R T, n = pV / RT = (0.425)1 / (0.082 × 924.33) = 5.6 ⋅ 10-3 mol
Amount of substance of FeSO4 decomposed = 2 nSO3 = 0.0112 mol
Mass of FeSO4 decomposed = 0.0112 ×151.91 = 1.70 g
Percentage of FeSO4 decomposed = (1.70 / 15.19) 100 = 11.19 %.