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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?
    Most Upvoted Answer
    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).
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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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