The enthalpy of a binary liquid system of species 1 and 2 at fixed T and P is represented by the equation: H = 400x1 + 600x2 + x1x2(40x1 + 20x2). Determine expressions for H1 and H... The enthalpy of a binary liquid system of species 1 and 2 at fixed T and P is represented by the equation: H = 400x1 + 600x2 + x1x2(40x1 + 20x2). Determine expressions for H1 and H2 as functions of x1, numerical values for the pure-species enthalpies H1 and H2, and numerical values for the partial enthalpies at infinite dilution H1∞ and H2∞.
Understand the Problem
The question is asking to derive expressions for the partial enthalpies of a binary liquid system and to compute specific numerical values for the enthalpies of pure species and their values at infinite dilution. This involves understanding the relationship between the given enthalpy equation and the properties of the components in the mixture.
Answer
Expressions: - \(H_1 = 400x_1 + 600(1-x_1) + x_1(1-x_1)(20x_1 + 20)\) - \(H_2 = 600 + 400(1-x_1) + x_1(1-x_1)(20x_1 + 20)\) Values: - \(H_1 = 400 \, J \cdot mol^{-1}\), - \(H_2 = 600 \, J \cdot mol^{-1}\) Infinite dilution: - \(H_1^\infty = 600 \, J \cdot mol^{-1}\), - \(H_2^\infty = 400 \, J \cdot mol^{-1}\)
Answer for screen readers
Expressions for the partial enthalpies are:
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(H_1 = 400x_1 + 600(1-x_1) + x_1(1-x_1)(20x_1 + 20))
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(H_2 = 600 + 400(1-x_1) + x_1(1-x_1)(20x_1 + 20))
Numerical values are:
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(H_1 = 400 , J \cdot mol^{-1})
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(H_2 = 600 , J \cdot mol^{-1})
Partial enthalpies at infinite dilution:
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(H_1^\infty = 600 , J \cdot mol^{-1})
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(H_2^\infty = 400 , J \cdot mol^{-1})
Steps to Solve
- Define the Components and Variables
Let (H) be the enthalpy given by:
$$ H = 400x_1 + 600x_2 + x_1x_2(40x_1 + 20x_2) $$
where (x_1) is the mole fraction of species 1 and (x_2) is the mole fraction of species 2. Since (x_2 = 1 - x_1), we can rewrite the equation in terms of (x_1).
- Rewrite H in terms of (x_1)
Substituting (x_2):
$$ H = 400x_1 + 600(1-x_1) + x_1(1-x_1)(40x_1 + 20(1-x_1)) $$
- Simplify the Equation
Expanding the equation gives:
$$ H = 400x_1 + 600 - 600x_1 + x_1(1-x_1)(40x_1 + 20 - 20x_1) $$
This simplifies to:
$$ H = (400 - 600)x_1 + 600 + x_1(1-x_1)(20x_1 + 20) $$
- Calculate the Pure Species Enthalpies (H_1) and (H_2)
For pure species 1 ((x_1 = 1, x_2 = 0)):
$$ H_1 = H|_{x_1=1} = 400(1) + 600(0) + 0 = 400 , J \cdot mol^{-1} $$
For pure species 2 ((x_1 = 0, x_2 = 1)):
$$ H_2 = H|_{x_1=0} = 400(0) + 600(1) + 0 = 600 , J \cdot mol^{-1} $$
- Find Partial Enthalpies (H_1^\infty) and (H_2^\infty) at Infinite Dilution
At infinite dilution, we want to find the partial molar enthalpies.
To find (H_1^\infty), set (x_2 = 1) (or (x_1 \to 0)):
$$ H_1^\infty = H|_{x_1=0} = 400(0) + 600(1) + 0 = 600 , J \cdot mol^{-1} $$
To find (H_2^\infty), set (x_1 = 0):
$$ H_2^\infty = H|_{x_1=1} = 400(1) + 600(0) + 0 = 400 , J \cdot mol^{-1} $$
Expressions for the partial enthalpies are:
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(H_1 = 400x_1 + 600(1-x_1) + x_1(1-x_1)(20x_1 + 20))
-
(H_2 = 600 + 400(1-x_1) + x_1(1-x_1)(20x_1 + 20))
Numerical values are:
-
(H_1 = 400 , J \cdot mol^{-1})
-
(H_2 = 600 , J \cdot mol^{-1})
Partial enthalpies at infinite dilution:
-
(H_1^\infty = 600 , J \cdot mol^{-1})
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(H_2^\infty = 400 , J \cdot mol^{-1})
More Information
This calculation uses the concept of partial molar enthalpy, which is important in understanding thermodynamic properties of mixtures. The numerical enthalpy values reflect the energy content of each pure component at standard conditions, while the infinite dilution values indicate how the components behave when mixed at extremely low concentrations.
Tips
- Not properly substituting (x_2 = 1 - x_1) in the enthalpy equation.
- Forgetting to simplify the terms after substitution, leading to errors in the final expressions.
- Confusing partial enthalpies with pure species enthalpies.
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