One mole of a gas, with a total pressure of 0.7 atm, containing 40% X2 gas...

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One mole of a gas, with a total pressure of 0.7 atm, containing 40% X2 gas and 60% Y2 gas (by volume) is bubbled through 100 grams of liquid metal M in a crucible at 1100 K at 0.7 atm. The metal only gives reaction with X2 gas (Y2 is inert) according to the given equilibrium reaction below. Equilibrium is attained between the gas mixture and the liquid in the exit stream. M (solid, liquid, gas) + X2 (g) = MX2 (s) Available thermodynamic data: Normal melting temperature = 510°C, AHm = 23430 J/mol Normal boiling temperature 1285°C, AH, = 155400 J/mol For simplicity use the following form for the vapor pressure equation: InP = -+B (atm) The temperature dependences of the Gibbs free energy for the reaction above for three different phases of M (solid, liquid or gas) are given below: (the phases are unknown, you have to find which equation belongs to which phase) AG°T = - 314000 – 0.3TINT + 151T J AG°r =- 552000 – 0.1TINT + 435T J AG°T=- 235000 – 0.7TINT + 107T J What is the remaining liquid metal M in the crucible if the total pressure does not change throughout the process? One mole of a gas, with a total pressure of 0.7 atm, containing 40% X2 gas and 60% Y2 gas (by volume) is bubbled through 100 grams of liquid metal M in a crucible at 1100 K at 0.7 atm. The metal only gives reaction with X2 gas (Y2 is inert) according to the given equilibrium reaction below. Equilibrium is attained between the gas mixture and the liquid in the exit stream. M (solid, liquid, gas) + X2 (g) = MX2 (s) Available thermodynamic data: Normal melting temperature = 510°C, AHm = 23430 J/mol Normal boiling temperature 1285°C, AH, = 155400 J/mol For simplicity use the following form for the vapor pressure equation: InP = -+B (atm) The temperature dependences of the Gibbs free energy for the reaction above for three different phases of M (solid, liquid or gas) are given below: (the phases are unknown, you have to find which equation belongs to which phase) AG°T = - 314000 – 0.3TINT + 151T J AG°r =- 552000 – 0.1TINT + 435T J AG°T=- 235000 – 0.7TINT + 107T J What is the remaining liquid metal M in the crucible if the total pressure does not change throughout the process?

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Solution Cne mole J a Jas 40 X2 and 60 gas by Volyme through liquid meta... View the full answer