Answer:
A. -166.6 kJ/mol
B. -127.7 kJ/mol
C. -133.9 kJ/mol
Explanation:
Let's consider the oxidation of sulfur dioxide.
2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ
The Gibbs free energy (ΔG) can be calculated using the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature (25 + 273.15 = 298.15 K)
Q is the reaction quotient
The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:
![[]=\frac{P}{R.T}](https://tex.z-dn.net/?f=%5B%5D%3D%5Cfrac%7BP%7D%7BR.T%7D)
<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>
<em>Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>
![[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M](https://tex.z-dn.net/?f=%5BSO_%7B2%7D%5D%3D%5BO_%7B2%7D%5D%3D%5Cfrac%7B130atm%7D%7B%280.08206atm.L%2Fmol.K%29.298K%7D%20%3D5.32M)
![[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M](https://tex.z-dn.net/?f=%5BSO_%7B3%7D%5D%3D%5Cfrac%7B2.0atm%7D%7B%280.08206atm.L%2Fmol.K%29.298K%7D%20%3D0.0818M)
![Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BSO_3%5D%5E%7B2%7D%20%7D%7B%5BSO_%7B2%7D%5D%5E%7B2%7D.%5BO_%7B2%7D%5D%20%7D%20%3D%5Cfrac%7B0.0818%5E%7B2%7D%20%7D%7B5.32%5E%7B3%7D%20%7D%20%3D4.44%20%5Ctimes%2010%5E%7B-5%7D)
ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol
<em>Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.</em>
<em />
![[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M](https://tex.z-dn.net/?f=%5BSO_%7B2%7D%5D%3D%5Cfrac%7B5.0atm%7D%7B%280.08206atm.L%2Fmol.K%29.298K%7D%3D0.204M)
![[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M](https://tex.z-dn.net/?f=%5BO_%7B2%7D%5D%3D%5Cfrac%7B3.0atm%7D%7B%280.08206atm.L%2Fmol.K%29.298K%7D%3D0.123M)
![[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M](https://tex.z-dn.net/?f=%5BSO_%7B3%7D%5D%3D%5Cfrac%7B30atm%7D%7B%280.08206atm.L%2Fmol.K%29.298K%7D%3D1.23M)
![Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BSO_3%5D%5E%7B2%7D%20%7D%7B%5BSO_%7B2%7D%5D%5E%7B2%7D.%5BO_%7B2%7D%5D%20%7D%20%3D%5Cfrac%7B1.23%5E%7B2%7D%20%7D%7B0.204%5E%7B2%7D.0.123%20%7D%20%3D296)
ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol
<em>Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.</em>
<em />
![[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M](https://tex.z-dn.net/?f=%5BSO_%7B2%7D%5D%3D%5BO_%7B2%7D%5D%3D%5BSO_%7B3%7D%5D%3D%5Cfrac%7B1.0atm%7D%7B%280.08206atm.L%2Fmol.K%29.298K%7D%3D0.0409M)
![Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BSO_3%5D%5E%7B2%7D%20%7D%7B%5BSO_%7B2%7D%5D%5E%7B2%7D.%5BO_%7B2%7D%5D%20%7D%20%3D%5Cfrac%7B0.0409%5E%7B2%7D%20%7D%7B0.0409%5E%7B3%7D%7D%20%3D24.4)
ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol