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Excess Gibbs Free Energy using NRTL Equation Formula

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Excess Gibbs Free Energy is the Gibbs energy of a solution in excess of what it would be if it were ideal. Check FAQs

G E = (x 1 \cdot x 2 \cdot [R] \cdot T NRTL) \cdot ((\frac{(\exp (- \frac{α \cdot b 21}{[R]} \cdot T NRTL)) \cdot (\frac{b 21}{[R] \cdot T NRTL})}{x 1 + x 2 \cdot \exp (- \frac{α \cdot b 21}{[R]} \cdot T NRTL)}) + (\frac{(\exp (- \frac{α \cdot b 12}{[R]} \cdot T NRTL)) \cdot (\frac{b 12}{[R] \cdot T NRTL})}{x 2 + x 1 \cdot \exp (- \frac{α \cdot b 12}{[R]} \cdot T NRTL)}))

G^E - Excess Gibbs Free Energy?x₁ - Mole Fraction of Component 1 in Liquid Phase?x₂ - Mole Fraction of Component 2 in Liquid Phase?T_NRTL - Temperature for NRTL model?α - NRTL Equation Coefficient (α)?b₂₁ - NRTL Equation Coefficient (b21)?b₁₂ - NRTL Equation Coefficient (b12)?[R] - Universal gas constant?[R] - Universal gas constant?[R] - Universal gas constant?[R] - Universal gas constant?[R] - Universal gas constant?[R] - Universal gas constant?[R] - Universal gas constant?

Excess Gibbs Free Energy using NRTL Equation Example

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Here is how the Excess Gibbs Free Energy using NRTL Equation equation looks like with Values.

Here is how the Excess Gibbs Free Energy using NRTL Equation equation looks like with Units.

Here is how the Excess Gibbs Free Energy using NRTL Equation equation looks like.

0.0255 = (0.4 \cdot 0.6 \cdot 8.3145 \cdot 550) \cdot ((\frac{(\exp (- \frac{0.15 \cdot 0.12}{8.3145} \cdot 550)) \cdot (\frac{0.12}{8.3145 \cdot 550})}{0.4 + 0.6 \cdot \exp (- \frac{0.15 \cdot 0.12}{8.3145} \cdot 550)}) + (\frac{(\exp (- \frac{0.15 \cdot 0.19}{8.3145} \cdot 550)) \cdot (\frac{0.19}{8.3145 \cdot 550})}{0.6 + 0.4 \cdot \exp (- \frac{0.15 \cdot 0.19}{8.3145} \cdot 550)}))

0.0255J = (0.4 \cdot 0.6 \cdot 8.3145 \cdot 550K) \cdot ((\frac{(\exp (- \frac{0.15 \cdot 0.12J/mol}{8.3145} \cdot 550K)) \cdot (\frac{0.12J/mol}{8.3145 \cdot 550K})}{0.4 + 0.6 \cdot \exp (- \frac{0.15 \cdot 0.12J/mol}{8.3145} \cdot 550K)}) + (\frac{(\exp (- \frac{0.15 \cdot 0.19J/mol}{8.3145} \cdot 550K)) \cdot (\frac{0.19J/mol}{8.3145 \cdot 550K})}{0.6 + 0.4 \cdot \exp (- \frac{0.15 \cdot 0.19J/mol}{8.3145} \cdot 550K)}))

0.0255 = (0.4 \cdot 0.6 \cdot 8.3145 \cdot 550) \cdot ((\frac{(\exp (- \frac{0.15 \cdot 0.12}{8.3145} \cdot 550)) \cdot (\frac{0.12}{8.3145 \cdot 550})}{0.4 + 0.6 \cdot \exp (- \frac{0.15 \cdot 0.12}{8.3145} \cdot 550)}) + (\frac{(\exp (- \frac{0.15 \cdot 0.19}{8.3145} \cdot 550)) \cdot (\frac{0.19}{8.3145 \cdot 550})}{0.6 + 0.4 \cdot \exp (- \frac{0.15 \cdot 0.19}{8.3145} \cdot 550)}))

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Excess Gibbs Free Energy using NRTL Equation Solution

Follow our step by step solution on how to calculate Excess Gibbs Free Energy using NRTL Equation?

FIRST Step Consider the formula

G E = (x 1 \cdot x 2 \cdot [R] \cdot T NRTL) \cdot ((\frac{(\exp (- \frac{α \cdot b 21}{[R]} \cdot T NRTL)) \cdot (\frac{b 21}{[R] \cdot T NRTL})}{x 1 + x 2 \cdot \exp (- \frac{α \cdot b 21}{[R]} \cdot T NRTL)}) + (\frac{(\exp (- \frac{α \cdot b 12}{[R]} \cdot T NRTL)) \cdot (\frac{b 12}{[R] \cdot T NRTL})}{x 2 + x 1 \cdot \exp (- \frac{α \cdot b 12}{[R]} \cdot T NRTL)}))

Next Step Substitute values of Variables

∴

G E = (0.4 \cdot 0.6 \cdot [R] \cdot 550K) \cdot ((\frac{(\exp (- \frac{0.15 \cdot 0.12J/mol}{[R]} \cdot 550K)) \cdot (\frac{0.12J/mol}{[R] \cdot 550K})}{0.4 + 0.6 \cdot \exp (- \frac{0.15 \cdot 0.12J/mol}{[R]} \cdot 550K)}) + (\frac{(\exp (- \frac{0.15 \cdot 0.19J/mol}{[R]} \cdot 550K)) \cdot (\frac{0.19J/mol}{[R] \cdot 550K})}{0.6 + 0.4 \cdot \exp (- \frac{0.15 \cdot 0.19J/mol}{[R]} \cdot 550K)}))

Next Step Substitute values of Constants

∴

G E = (0.4 \cdot 0.6 \cdot 8.3145 \cdot 550K) \cdot ((\frac{(\exp (- \frac{0.15 \cdot 0.12J/mol}{8.3145} \cdot 550K)) \cdot (\frac{0.12J/mol}{8.3145 \cdot 550K})}{0.4 + 0.6 \cdot \exp (- \frac{0.15 \cdot 0.12J/mol}{8.3145} \cdot 550K)}) + (\frac{(\exp (- \frac{0.15 \cdot 0.19J/mol}{8.3145} \cdot 550K)) \cdot (\frac{0.19J/mol}{8.3145 \cdot 550K})}{0.6 + 0.4 \cdot \exp (- \frac{0.15 \cdot 0.19J/mol}{8.3145} \cdot 550K)}))

Next Step Prepare to Evaluate

∴

G E = (0.4 \cdot 0.6 \cdot 8.3145 \cdot 550) \cdot ((\frac{(\exp (- \frac{0.15 \cdot 0.12}{8.3145} \cdot 550)) \cdot (\frac{0.12}{8.3145 \cdot 550})}{0.4 + 0.6 \cdot \exp (- \frac{0.15 \cdot 0.12}{8.3145} \cdot 550)}) + (\frac{(\exp (- \frac{0.15 \cdot 0.19}{8.3145} \cdot 550)) \cdot (\frac{0.19}{8.3145 \cdot 550})}{0.6 + 0.4 \cdot \exp (- \frac{0.15 \cdot 0.19}{8.3145} \cdot 550)}))

Next Step Evaluate

∴

G E = 0.0255091211453841J

LAST Step Rounding Answer

∴

G E = 0.0255J

Excess Gibbs Free Energy using NRTL Equation Formula Elements

Variables

Constants

Functions

Excess Gibbs Free Energy

Excess Gibbs Free Energy is the Gibbs energy of a solution in excess of what it would be if it were ideal.

Symbol: G^E

Measurement: EnergyUnit: J

Note: Value can be positive or negative.

Formulas to find Excess Gibbs Free Energy

Mole Fraction of Component 1 in Liquid Phase

The mole fraction of component 1 in liquid phase can be defined as the ratio of the number of moles a component 1 to the total number of moles of components present in the liquid phase.

Symbol: x₁

Measurement: NAUnit: Unitless

Note: Value should be between 0 to 1.

Formulas to find Mole Fraction of Component 1 in Liquid Phase

Mole Fraction of Component 2 in Liquid Phase

The mole fraction of component 2 in liquid phase can be defined as the ratio of the number of moles a component 2 to the total number of moles of components present in the liquid phase.

Symbol: x₂

Measurement: NAUnit: Unitless

Note: Value should be between 0 to 1.

Formulas to find Mole Fraction of Component 2 in Liquid Phase

Temperature for NRTL model

Temperature for NRTL model is the degree or intensity of heat present in a substance or object.

Symbol: T_NRTL

Measurement: TemperatureUnit: K

Note: Value can be positive or negative.

Formulas to find Temperature for NRTL model

NRTL Equation Coefficient (α)

NRTL Equation Coefficient (α) is the coefficient used in the NRTL equation which is parameter specific to a particular pair of species.

Symbol: α

Measurement: NAUnit: Unitless

Note: Value can be positive or negative.

Formulas to find NRTL Equation Coefficient (α)

NRTL Equation Coefficient (b21)

The NRTL Equation Coefficient (b21) is the coefficient used in the NRTL equation for component 2 in the binary system. It's independent of concentration and temperature.

Symbol: b₂₁

Measurement: Energy Per MoleUnit: J/mol

Note: Value can be positive or negative.

Formulas to find NRTL Equation Coefficient (b21)

NRTL Equation Coefficient (b12)

The NRTL Equation Coefficient (b12) is the coefficient used in the NRTL equation for component 1 in the binary system. It's independent of concentration and temperature.

Symbol: b₁₂

Measurement: Energy Per MoleUnit: J/mol

Note: Value can be positive or negative.

Formulas to find NRTL Equation Coefficient (b12)

Universal gas constant