PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
2nd Edition
ISBN: 9781285074788
Author: Ball
Publisher: CENGAGE L
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Chapter 7, Problem 7.30E
Interpretation Introduction

Interpretation:

The values of ΔmixG and ΔmixS for an ideal solution are to be predicted.

Concept introduction:

According to Raoult’s law, the vapor pressure of the solution is equal to the product of vapor pressure of the solvent and its mole fraction in the solution. The ideal solution follows the Raoult’s law.

Expert Solution & Answer
Check Mark

Answer to Problem 7.30E

The values of ΔmixG and ΔmixS for an ideal solution are 4213.12J and 13.58JK1 respectively.

Explanation of Solution

The 25.0g of pentane (C5H12), 45.0g of hexane (C6H14), and 55.0g of cyclohexane (C6H12) are mixed to form an ideal solution.

The conversion of temperature in Celsius to Kelvin is given by the formula,

T(K)=T(°C)+273.2

Substitute the value of T=37.0°C in the above formula.

T(K)=37+273.2=310.2K

Thus, the temperature of the ideal solution is 310.2K.

The molar mass of pentane is calculated as,

MolarMassofC5H12=(5×Massofcarbonatom)+(12×Massofhydrogenatom)=(5×12)+(12×1)=60+12=72g

Thus, the molar mass of pentane is 72g.

The molar mass of hexane is calculated as,

MolarMassofC6H14=(6×Massofcarbonatom)+(14×Massofhydrogenatom)=(6×12)+(14×1)=72+14=86g

Thus, the molar mass of hexane is 86g.

The molar mass of cyclohexane is calculated as,

MolarMassofC6H12=(6×Massofcarbonatom)+(12×Massofhydrogenatom)=(6×12)+(12×1)=72+12=84g

Thus, the molar mass of cyclohexane is 84g.

The number of moles is calculated by the formula,

No.ofmoles=GivenMassMolarMass

Substitute the given mass and molar mass of pentane in above formula.

No.ofmoles=2572=0.35mol

Thus, the moles of pentane is 0.35mol.

Substitute the given mass and molar mass of hexane in above formula.

No.ofmoles=4586=0.52mol

Thus, the moles of hexane is 0.52mol.

Substitute the given mass and molar mass of cyclohexane in above formula.

No.ofmoles=5584=0.66mol

Thus, the moles of cyclohexane is 0.66mol.

Hence, the total number of moles is 1.53mol respectively.

The mole fraction is calculated by the formula,

xA=nAnA+nB+nC

Where,

nA is the moles of A.

nB is the moles of B.

nC is the moles of C.

(nA+nB+nC) represents the total moles of the solution.

Substitute the moles of pentane in above formula.

xpentane=npentanenpentane+nhexane+ncyclohexane=0.350.35+0.52+0.66=0.351.53=0.23

Thus, the mole fraction of pentane is 0.23.

Substitute the moles of hexane in above formula.

xhexane=nhexanenpentane+nhexane+ncyclohexane=0.520.35+0.52+0.66=0.521.53=0.34

Thus, the mole fraction of hexane is 0.34.

Substitute the moles of cyclohexane in above formula.

xcyclohexane=ncyclohexanenpentane+nhexane+ncyclohexane=0.660.35+0.52+0.66=0.661.53=0.43

Thus, the mole fraction of cyclohexane is 0.43.

The formula used to calculate ΔmixG is given by,

ΔmixG=RTixilnxi

Where,

R is the gas constant.

T is the temperature of the solution.

xi is the mole fraction of i-th component of a solution.

Substitute the value of xpentane,xhexane and xcyclohexane in the above formula.

ΔmixG=RT(xpentanelnxpentane+xhexanelnxhexane+xcyclohexanelnxcyclohexane)=(8.314Jmol.K)(310.2K)[(0.34ln0.34+0.23ln0.23+0.43ln0.43)]=2753.7Jmol1×1.53mol=4213.12J

Therefore, the ΔmixG of the ideal solution is 4213.12J.

The formula used to calculate ΔmixS is given by,

ΔmixS=Rixilnxi

Where,

R is the gas constant.

xi is the mole fraction of i-th component of a solution.

Substitute the value of xtoluene and xbenzene in the above formula.

ΔmixS=R(xpentanelnxpentane+xhexanelnxhexane+xcyclohexanelnxcyclohexane)=(8.314Jmol.K)[(0.34ln0.34+0.23ln0.23+0.43ln0.43)]=8.878Jmol1K1×1.53mol=13.58JK1

Therefore, the ΔmixS of the ideal solution is 13.58JK1.

Conclusion

The values of ΔmixG and ΔmixS for an ideal solution are 4213.12J and 13.58JK1 respectively.

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PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.

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