The excess partial molar free energy of zinc in liquid Cu-Zn alloys at 1027°C (1300 K) can be represented as 2 G^ (cal/mole) = -5150 (1 -xZn) . Calculate the activity of copper at 1027 C (1300 K) in an equiatomic solution.
Question
The excess partial molar free energy of zinc in liquid Cu-Zn alloys at 1027°C (1300 K) can be represented as 2 G^ (cal/mole) = -5150 (1 -xZn) . Calculate the activity of copper at 1027 C (1300 K) in an equiatomic solution.
Solution
The activity of a component in a solution can be calculated using the Gibbs-Duhem equation. The excess partial molar free energy of a component is related to its activity by the equation:
a_i = exp((G_i^E)/(RT))
where: a_i is the activity of the component, G_i^E is the excess partial molar free energy of the component, R is the gas constant, and T is the temperature in Kelvin.
Given that the excess partial molar free energy of zinc (G_Zn^E) in the Cu-Zn alloy is -5150 (1 - x_Zn) cal/mole, we can calculate the activity of copper (a_Cu) as follows:
Step 1: Convert the excess partial molar free energy of zinc to joules/mole. 1 cal = 4.184 J, so G_Zn^E = -5150 * 4.184 (1 - x_Zn) J/mole.
Step 2: Since the solution is equiatomic, the mole fraction of zinc (x_Zn) is 0.5. Substitute this into the equation to get G_Zn^E = -5150 * 4.184 * (1 - 0.5) J/mole = -10788 J/mole.
Step 3: Substitute G_Zn^E, R, and T into the equation for a_i. Use R = 8.314 J/(molK) and T = 1300 K. This gives a_Zn = exp((-10788)/(8.3141300)).
Step 4: The activities of the components in a binary solution sum to 1, so a_Cu = 1 - a_Zn.
Step 5: Substitute the calculated value of a_Zn into the equation to get a_Cu.
This will give you the activity of copper in the Cu-Zn alloy at 1027°C.
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