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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:
<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>
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 />
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 />
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
Answer : The correct option is, 0.21 moles
Solution : Given,
Molarity of the solution = 0.85 M = 0.85 mole/L
Volume of solution = 250 ml = 0.25 L
Molarity : It is defined as the number of moles of solute present in one liter of the solution.
Formula used :
Now put all the given values in this formula, we get the moles of solute of the solution.
Therefore, the moles of solute is, 0.21 moles