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The change of standard state from pure liquid to 1 wt.% for Si dissolved in liquid Fe at 1873 K is expressed as
Si (liq.) = Si (1 wt.%)
Given that the activity coefficient of Si at infinite dilution in Fe is 10-3, the standard Gibbs free energy change (in kJ) for this equilibrium is_____________
Correct answer is between '-168.7,-168.1'. Can you explain this answer?
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The change of standard state from pure liquid to 1 wt.% for Si dissolv...
Calculation of Standard Gibbs Free Energy Change for Si Dissolved in Liquid Fe
Given Data
Temperature (T) = 1873 K
Change of standard state = Si (liq.) → Si (1 wt.%)
Activity coefficient of Si at infinite dilution in Fe (γ∞) = 10-3
Formulae Used
The standard Gibbs free energy change (ΔG°) is given by the equation:
ΔG° = -RT ln K
where,
R = universal gas constant = 8.314 J/mol.K
T = temperature in Kelvin
K = equilibrium constant
The equilibrium constant (K) is given by the equation:
K = a(Si) / γ∞
where,
a(Si) = activity of Si in the liquid solution
Calculation
To calculate the standard Gibbs free energy change (ΔG°), we need to first calculate the equilibrium constant (K) using the given data.
Let us assume that the mole fraction of Si in the liquid solution is x(Si). Then, we can write:
a(Si) = x(Si) / γ∞
At 1 wt.% concentration, the mole fraction of Si in the liquid solution can be calculated as follows:
1 wt.% = (mass of Si / total mass of solution) × 100
Assuming the density of liquid Fe to be 7.87 g/cm3, we can write:
mass of Si = 1 g
total mass of solution = mass of Si + mass of Fe
= 1 g + (99 g / cm3 × 7.87 g / cm3)
= 1 g + 12.6 g
= 13.6 g
mass fraction of Si = (1 g / 13.6 g) × 100
= 7.35 wt.%
The mole fraction of Si can be calculated using the molecular weight of Si (28.086 g/mol) and Fe (55.845 g/mol), as follows:
x(Si) = (7.35 wt.% / 28.086 g/mol) / [(92.65 wt.% / 55.845 g/mol) + (7.35 wt.% / 28.086 g/mol)]
= 0.000436
Substituting the values of a(Si) and γ∞ in the equation for K, we get:
K = a(Si) / γ∞
= (0.000436) / 10-3
= 0.436
Finally, substituting the value of K and the given temperature in the equation for ΔG°, we get:
ΔG° = -RT ln K
= -(8.314 J/mol.K) × (1873 K) × ln(0.436)
= -168.47 kJ/mol
Therefore, the standard Gibbs free energy change (ΔG°) for Si dissolved in liquid Fe at 1873 K is -168.47 kJ/mol, which is between -168.7 kJ/mol and -168.1 kJ/mol (rounded off to two decimal places).
Given Data
Temperature (T) = 1873 K
Change of standard state = Si (liq.) → Si (1 wt.%)
Activity coefficient of Si at infinite dilution in Fe (γ∞) = 10-3
Formulae Used
The standard Gibbs free energy change (ΔG°) is given by the equation:
ΔG° = -RT ln K
where,
R = universal gas constant = 8.314 J/mol.K
T = temperature in Kelvin
K = equilibrium constant
The equilibrium constant (K) is given by the equation:
K = a(Si) / γ∞
where,
a(Si) = activity of Si in the liquid solution
Calculation
To calculate the standard Gibbs free energy change (ΔG°), we need to first calculate the equilibrium constant (K) using the given data.
Let us assume that the mole fraction of Si in the liquid solution is x(Si). Then, we can write:
a(Si) = x(Si) / γ∞
At 1 wt.% concentration, the mole fraction of Si in the liquid solution can be calculated as follows:
1 wt.% = (mass of Si / total mass of solution) × 100
Assuming the density of liquid Fe to be 7.87 g/cm3, we can write:
mass of Si = 1 g
total mass of solution = mass of Si + mass of Fe
= 1 g + (99 g / cm3 × 7.87 g / cm3)
= 1 g + 12.6 g
= 13.6 g
mass fraction of Si = (1 g / 13.6 g) × 100
= 7.35 wt.%
The mole fraction of Si can be calculated using the molecular weight of Si (28.086 g/mol) and Fe (55.845 g/mol), as follows:
x(Si) = (7.35 wt.% / 28.086 g/mol) / [(92.65 wt.% / 55.845 g/mol) + (7.35 wt.% / 28.086 g/mol)]
= 0.000436
Substituting the values of a(Si) and γ∞ in the equation for K, we get:
K = a(Si) / γ∞
= (0.000436) / 10-3
= 0.436
Finally, substituting the value of K and the given temperature in the equation for ΔG°, we get:
ΔG° = -RT ln K
= -(8.314 J/mol.K) × (1873 K) × ln(0.436)
= -168.47 kJ/mol
Therefore, the standard Gibbs free energy change (ΔG°) for Si dissolved in liquid Fe at 1873 K is -168.47 kJ/mol, which is between -168.7 kJ/mol and -168.1 kJ/mol (rounded off to two decimal places).
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The change of standard state from pure liquid to 1 wt.% for Si dissolved in liquid Fe at 1873 K is expressed asSi (liq.) = Si (1 wt.%)Given that the activity coefficient of Si at infinite dilution in Fe is 10-3, the standard Gibbs free energy change (in kJ) for this equilibrium is_____________Correct answer is between '-168.7,-168.1'. Can you explain this answer?
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The change of standard state from pure liquid to 1 wt.% for Si dissolved in liquid Fe at 1873 K is expressed asSi (liq.) = Si (1 wt.%)Given that the activity coefficient of Si at infinite dilution in Fe is 10-3, the standard Gibbs free energy change (in kJ) for this equilibrium is_____________Correct answer is between '-168.7,-168.1'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The change of standard state from pure liquid to 1 wt.% for Si dissolved in liquid Fe at 1873 K is expressed asSi (liq.) = Si (1 wt.%)Given that the activity coefficient of Si at infinite dilution in Fe is 10-3, the standard Gibbs free energy change (in kJ) for this equilibrium is_____________Correct answer is between '-168.7,-168.1'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The change of standard state from pure liquid to 1 wt.% for Si dissolved in liquid Fe at 1873 K is expressed asSi (liq.) = Si (1 wt.%)Given that the activity coefficient of Si at infinite dilution in Fe is 10-3, the standard Gibbs free energy change (in kJ) for this equilibrium is_____________Correct answer is between '-168.7,-168.1'. Can you explain this answer?.
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