Respuesta :
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:
[tex][]=\frac{P}{R.T}[/tex]
Calculate ΔG at 25°C given the following sets of partial pressures.
Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.
[tex][SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M[/tex]
[tex][SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M[/tex]
[tex]Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}[/tex]
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
Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.
[tex][SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M[/tex]
[tex][O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M[/tex]
[tex][SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M[/tex]
[tex]Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296[/tex]
ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol
Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.
[tex][SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M[/tex]
[tex]Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4[/tex]
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