# The derivation of Gibbs energy of mixing  involves the mole fractions of two components A and B in an ideal solution, xAand xB, respectively and the total amount, n. Following the method used in Derivation 6.3. The initial Gibbs energy of the unmixed components is Gi = nAµA* + nBµB*; after mixing,  the chemical potentials in eqn 6.11:  μA = μAϴ + RTlnxA are used. Use this equation to calculate the Gibbs energy of mixing of 0.5 mole A and 0.9 mole B at 300 K. ΔG = ____________ kJ.

The derivation of Gibbs energy of mixing  involves the mole fractions of two components A and B in an ideal solution, xAand xB, respectively and the total amount, n. Following the method used in Derivation 6.3. The initial Gibbs energy of the unmixed components is Gi = nAµA* + nBµB*; after mixing,  the chemical potentials in eqn 6.11:  μA = μAϴ + RTlnxA are used.

Use this equation to calculate the Gibbs energy of mixing of 0.5 mole A and 0.9 mole B at 300 K.

ΔG = ____________ kJ. (550 words)

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