1) An ideal solution is made from 60.0 g of benzene and 60.0 g of toluene....

Question

Question

1) An ideal solution is made from 60.0 g of benzene and 60.0 g of toluene. Find the values of AG mix, and AS mix at 298 K and

Answer 1

Answer #1

a) For ideal solutions, enthalpy of mixing, $\Delta$ H_mix = 0 so their components are miscible in all proportions

$\Delta$ S_mix = -nR[xlnx + (1-x)ln(1-x)]

Where x is the mole fraction of component 1 while 1-x is the mole fraction of component 2

The mixture contains 60 g of benzene and 60 g of Toluene

Molecular weight of Benzene = 78.11

Molecular weight of Toluene = 92.14

Number of moles of Benzene = 60/78.11 = 0.768

Number of moles of Toluene = 60/92.14 = 0.6511

Total number of moles, n = 0.768 + 0.6511 = 1.41933

Mole fraction of benzene = 0.768/(0.768+0.6511), x= 0.5412

Mole fraction of toluene = 1-x = 1-0.5412 = 0.458

Therefore, $\Delta$ S_mix = - 1.41933 * 8.314 [0.5412ln(0.5412) + 0.458ln(0.458)] = 8.319 J

$\Delta$ G_mix = $\Delta$ H_mix - T $\Delta$ S_mix = 0 - 298 *8.319 = -2425.49 J

$\Delta$ G_mix = -2425.49 J

As can be seen $\Delta$ G_mix is negative, therefore, the mixing is spontaneous.

Answer 2

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