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3 years ago

15 g of gold and 25 g of silver are mixed to form a single-phase ideal solid solution.

Chemistry

1 answer:

Tatiana [17]3 years ago

8 0

Answer:

1. How many moles of solution are there. Ans: 0.3079193mol

2. Mole fraction for gold : 0.2473212

Mole fraction for silver: 0.7526787

3. Molar entropy of mixing for gold: 2.87285j/k

Molar entropy of mixing for silver: 1.77804j/k

4. Total entropy of mixing: 4.65089j/k

5. Molar free energy: -2325.445kj

6. Chemical potential for silver: -1750.31129j/mol

Chemical potential for gold: -575.13185j/mol

Explanation:

(1)

molar mass of silver = 107.8682g/mol

Molar mass of gold= 196.96657g/mol

Therefore mole = mass/molar mass

For silver: 25g/107.8682g/mol = 0.2317643mol

For gold: 15g/196.96657g/mol= 0.076155mol

Total number of mole= 0.2317643+0.076155= 0.30791193mol

(2)

Mole fraction for silver= 0.2317643/0.3079193= 0.7526787

Mole fraction for gold=0.076155/0.3079193=0.2473212

(3)

The molar entropy mixing ∆Sm= -RXi×lnXi

R= gas constant= 8.3144598

Xi = mole fraction

For silver:

-8.3144598×0.7526787( ln0.7526787)= 1.77804j/k

For gold:

-8.3144598×0.2473212( ln0.2473212)= 2.87285k/j

(4)

Total entropy= 1.77804+2.87285=4.65089k/j

(5)

Molar free energy change at 500°C

G=H-TS

Where G= Gibbs free energy

H= enthalpy,. T= Temperature, S= entropy

H=0, T=500+ 273=773k, S=4.65089

Therefore

G= 0- 773x4.65089= -3595.138kj

(6)

Chemical potential = Gibbs free energy × mole fraction

For silver:

-3595.138×0.7526787=-2705.9837j/mol

For gold:

-3595.138×0.2473212= -889.15381j/mol

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vesna_86 [32]

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Explanation:

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