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Concept explainers
Interpretation: The vapour pressure/composition curve for a mixture of benzene (B) and ethanoic acid (E) at 50 °C for the given vapour pressure data has to be plotted. The activity and activity coefficients of both components on the basis of Raoult’s law have to be deduced. Taking B as a solute, its activity and activity coefficients have to be deduced on the basis of Henry’s law. The excess Gibbs energy of the mixture has to be evaluated over the given composition range.
Concept introduction: The Raoult’s law states that the ratio of the partial pressure of each component to the partial pressure in the vapour pressure when it is present as a pure liquid is equal to its composition in the liquid phase. Henry’s law states that the amount the partial vapour pressure is directly proportional to the amount of substance dissolved. It involves an empirical constant KH called Henry’s constant.
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Answer to Problem 5.2IA
The vapour pressure/composition curve for a mixture of benzene and ethanoic acid for the given vapour pressure data is plotted as below,
The activity and activity coefficient of ethanoic acid on the basis of Raoult’s law is shown in the table below.
xE | aE | γE |
0.016 | 0.016 | 1 |
0.0439 | 0.0439 | 1 |
0.0835 | 0.0835 | 1 |
0.1138 | 0.1138 | 1 |
0.1714 | 0.1714 | 1 |
0.2973 | 0.2973 | 1 |
0.3696 | 0.3696 | 1 |
0.5834 | 0.5834 | 1 |
0.6604 | 0.6604 | 1 |
0.8437 | 0.8437 | 1 |
0.9931 | 0.9931 | 1 |
The activity and activity coefficients of benzene on the basis of Raoult’s law is shown in the table below,
xB | aB | γB |
0.984 | 0.984 | 1 |
0.9561 | 0.9561 | 1 |
0.9165 | 0.9165 | 1 |
0.8862 | 0.8862 | 1 |
0.8286 | 0.8286 | 1 |
0.7027 | 0.7027 | 1 |
0.6304 | 0.6304 | 1 |
0.4166 | 0.4166 | 1 |
0.3396 | 0.3396 | 1 |
0.1563 | 0.1563 | 1 |
0.0069 | 0.0069 | 1 |
The activity and activity coefficients of benzene on the basis of Henry’s law are shown in the table below,
xB | aB | γB |
0.984 | 0.730838 | 0.742721 |
0.9561 | 0.710116 | 0.742721 |
0.9165 | 0.680704 | 0.742721 |
0.8862 | 0.6582 | 0.742721 |
0.8286 | 0.615419 | 0.742721 |
0.7027 | 0.52191 | 0.742721 |
0.6304 | 0.468212 | 0.742721 |
0.4166 | 0.309418 | 0.742721 |
0.3396 | 0.252228 | 0.742721 |
0.1563 | 0.116087 | 0.742721 |
0.0069 | 0.005125 | 0.742721 |
The excess Gibbs energy is −766.3 J mol-1_.
Explanation of Solution
The vapour pressure data for a mixture of benzene (B) and ethanoic acid (E) at 50 °C is given as,
xE | pE/kPa | pB/kPa |
0.016 | 0.484 | 35.05 |
0.0439 | 0.967 | 34.29 |
0.0835 | 1.535 | 33.28 |
0.1138 | 1.89 | 32.64 |
0.1714 | 2.45 | 30.9 |
0.2973 | 3.31 | 28.16 |
0.3696 | 3.83 | 26.08 |
0.5834 | 4.84 | 20.42 |
0.6604 | 5.36 | 18.01 |
0.8437 | 6.76 | 10 |
0.9931 | 7.29 | 0.47 |
The total mole fraction of the components is 1.
xB+xE=1
The mole fraction of benzene (xB) in the solution is calculated in the table as,
xE | xB=1−xE |
0.016 | 0.984 |
0.0439 | 0.9561 |
0.0835 | 0.9165 |
0.1138 | 0.8862 |
0.1714 | 0.8286 |
0.2973 | 0.7027 |
0.3696 | 0.6304 |
0.5834 | 0.4166 |
0.6604 | 0.3396 |
0.8437 | 0.1563 |
0.9931 | 0.0069 |
The total pressure (p) of the solution is,
p=pE+pB (1)
Where,
- pE is the vapour pressure of ethanoic acid.
- pB is the vapour pressure of benzene.
According to Dalton’s law of partial pressures, the partial pressure of benzene (pB') is given by the formula,
pB'=xBp (2)
Where
- xB is the mole fraction of benzene.
Substitute the value of p from equation (1) in equation (2).
pB'=xB(pE+pB) (3)
The mole fraction of benzene (xB) is 0.984.
The vapour pressure of ethanoic acid (pE) is 0.484 kPa.
The vapour pressure of benzene (pB) is 35.05 kPa.
Substitute the value of xB, pE and pB in equation(3).
pB'=0.984(0.484 kPa+35.05 kPa)=34.965 kPa
Similarly, the values of pB' and p are calculated for all values of xB in the table as shown,
xB=1−xE | p/kPa | pB'/kPa |
0.984 | 35.534 | 34.96546 |
0.9561 | 35.257 | 33.70922 |
0.9165 | 34.815 | 31.90795 |
0.8862 | 34.53 | 30.60049 |
0.8286 | 33.35 | 27.63381 |
0.7027 | 31.47 | 22.11397 |
0.6304 | 29.91 | 18.85526 |
0.4166 | 25.26 | 10.52332 |
0.3396 | 23.37 | 7.936452 |
0.1563 | 16.76 | 2.619588 |
0.0069 | 7.76 | 0.053544 |
The graph between mole fraction of benzene (xB) and the partial pressure of benzene (pB') is plotted as,
According to Raoult’s law the relation between the partial pressure of benzene is given by the expression,
pB'=xBp*B
Where,
- p*B is the vapour pressure of the pure substance.
As the graph between the partial pressure of benzene and mole fraction of benzene is a straight line, hence, it follows Raoult’s law.
According to Raoult’s law the relation between the partial pressure of ethanoic acid is given by the expression,
pE'=xEp*E (4)
Where,
- p*E is the vapour pressure of the pure substance.
- xE is the mole fraction of ethanoic acid.
- pE' is the partial pressure of ethanoic acid.
The vapour pressure of pure ethanoic acid (p*E) at 50 °C is 7.615 kPa.
The mole fraction of ethanoic acid (xE) is 0.0160.
Substitute the value of p*E and xE in equation (4).
pE'=0.016×7.615 kPa=0.1218 kPa
The mole fraction of ethanoic acid in vapour (yE) is given by the formula,
yE=pE'p (5)
The partial pressure of ethanoic acid (pE') is 0.1218 kPa.
The total pressure of the mixture (p) when mole fraction of ethanoic acid is 0.016 is 35.534 kPa
Substitute the value of pE' and p in equation (5).
yE=0.1218 kPa35.534 kPA=0.00342
The activity of ethanoic acid (aE) is given as,
aE=yEpp*E (6)
Substitute the values of yE, p, and p*E in equation (6).
aE=0.00342×35.534 kPa7.615 kPa=0.0159
The activity coefficient (γE) of ethanoic acid is given as,
γE=aExE (7)
Substitute the value of aE and xE in equation (7).
γE=0.01590.016=0.99
Similarly, the values of aE and γE are calculated for all values of xE in the table as shown,
xE | pE'/kPa | yE | aE | γE |
0.016 | 0.12184 | 0.003429 | 0.016 | 1 |
0.0439 | 0.334299 | 0.009482 | 0.0439 | 1 |
0.0835 | 0.635853 | 0.018264 | 0.0835 | 1 |
0.1138 | 0.866587 | 0.025097 | 0.1138 | 1 |
0.1714 | 1.305211 | 0.039137 | 0.1714 | 1 |
0.2973 | 2.26394 | 0.07194 | 0.2973 | 1 |
0.3696 | 2.814504 | 0.094099 | 0.3696 | 1 |
0.5834 | 4.442591 | 0.175875 | 0.5834 | 1 |
0.6604 | 5.028946 | 0.215188 | 0.6604 | 1 |
0.8437 | 6.424776 | 0.38334 | 0.8437 | 1 |
0.9931 | 7.562457 | 0.974543 | 0.9931 | 1 |
According to Raoult’s law the relation between the partial pressure of benzene is given by the expression,
pB'=xBp*B (8)
Where,
- p*B is the vapour pressure of the pure substance.
- xB is the mole fraction of benzene.
- pB' is the partial pressure of benzene.
The vapour pressure of pure benzene (p*B) at 50 °C is 34.66 kPa.
The mole fraction of benzene (xB) is 0.984.
Substitute the value of p*B and xB in equation (8).
pB'=0.984×33.66 kPa=33.121kPa
The mole fraction of benzene in vapour (yB) is given by the formula,
yB=pB'p (9)
The partial pressure of benzene (pB') is 33.121 kPa.
The total pressure of the mixture (p) when mole fraction of benzene is 0.984 is 35.534 kPa.
Substitute the value of pB' and p in equation (9).
yB=33.121 kPa35.534 kPA=0.932
The activity of benzene (aB) is given as,
aB=yBpp*B (10)
Substitute the values of yB, p, and p*B in equation (10).
aB=0.932×35.534 kPa33.66 kPa=0.984
The activity coefficient (γB) of benzene is given as,
γB=aBxB (11)
Substitute the value of aB and xB in equation (11).
γB=0.9840.984=1
Similarly, the values of aB and γB are calculated for all values of xB in the table as shown,
xB | pB'/kPa | yB | aB | γB |
0.984 | 33.12144 | 0.932106 | 0.984 | 1 |
0.9561 | 32.18233 | 0.912793 | 0.9561 | 1 |
0.9165 | 30.84939 | 0.886095 | 0.9165 | 1 |
0.8862 | 29.82949 | 0.863872 | 0.8862 | 1 |
0.8286 | 27.89068 | 0.836302 | 0.8286 | 1 |
0.7027 | 23.65288 | 0.751601 | 0.7027 | 1 |
0.6304 | 21.21926 | 0.709437 | 0.6304 | 1 |
0.4166 | 14.02276 | 0.555137 | 0.4166 | 1 |
0.3396 | 11.43094 | 0.489129 | 0.3396 | 1 |
0.1563 | 5.261058 | 0.313906 | 0.1563 | 1 |
0.0069 | 0.232254 | 0.02993 | 0.0069 | 1 |
According to Henry’s law the partial pressure of benzene (pB') is given as,
pB'=xBKH (12)
Where
- pB' is the vapour pressure of benzene in the solution.
- KH is Henry’s constant.
- xB is the mole fraction of benzene in the solution.
From the graph plotted between the partial pressure of benzene and mole fraction of benzene,
When mole fraction of benzene (xB) is 0.4, the partial pressure of benzene (pB') is 10 kPa.
Substitute the value of xB and pB' in equation (12).
10 kPa=0.4×KHKH=25 kPa
The mole fraction of benzene in the vapour phase is given as,
yB=KHxBp (13)
The mole fraction of benzene (xB) is 0.984.
The total pressure of the mixture (p) when mole fraction of benzene is 0.984 is 35.534 kPa.
Substitute the value of KH, xB, and p in the equation (8).
yB=25 kPa×0.98435.534 kPa=0.6922
The activity of benzene (aB) is given as,
aB=yBpp*B (10)
Substitute the values of yB, p, and p*B in equation (10).
aB=0.6922×35.534 kPa33.66 kPa=0.73
The activity coefficient (γB) of benzene is given as,
γB=aBxB (11)
Substitute the value of aB and xB in equation (11).
γB=0.7430.984=0.742
Similarly, the values of aB and γB are calculated for all values of xB in the table.
xB | p/kPa | yB | aB | γB |
0.984 | 35.534 | 0.692295 | 0.730838 | 0.742721 |
0.9561 | 35.257 | 0.67795 | 0.710116 | 0.742721 |
0.9165 | 34.815 | 0.658121 | 0.680704 | 0.742721 |
0.8862 | 34.53 | 0.641616 | 0.6582 | 0.742721 |
0.8286 | 33.35 | 0.621139 | 0.615419 | 0.742721 |
0.7027 | 31.47 | 0.55823 | 0.52191 | 0.742721 |
0.6304 | 29.91 | 0.526914 | 0.468212 | 0.742721 |
0.4166 | 25.26 | 0.412312 | 0.309418 | 0.742721 |
0.3396 | 23.37 | 0.363286 | 0.252228 | 0.742721 |
0.1563 | 16.76 | 0.233144 | 0.116087 | 0.742721 |
0.0069 | 7.76 | 0.022229 | 0.005125 | 0.742721 |
The excess Gibbs energy is given as,
ΔGe=xERT lnγE+xBRTlnγB (14)
Where,
- R is the gas constant (8.314 J K-1 mol-1).
- T is the temperature in Kelvin.
The temperature of the mixture is 50 °C
The temperature values in degree Celsius is converted to Kelvin using the equation given below as,
T(K)=T(°C)+273.15
Substitute the temperature value in the above equation as follows.
T(K)=T(°C)+273.15=50+273.15=323.15 K
The value of mole fraction of benzene (xB) is 0.984.
The value of mole fraction of ethanoic acid (xE) is 0.016.
The activity coefficient of benzene is (γB) is 0.7427.
The activity coefficient of ethanoic acid (γE) is 1.
Substitute the value of R, T, xB, xE, γB and γE in equation (14).
ΔGe=xERT lnγE+xBRTlnγB=(0.016×8.314 J K-1 mol-1×323 Kln 1+0.984×8.314 J K-1 mol-1×323 K ln 0.7427)=0+0.984×8.314 J K-1 mol-1×323 K×(-0.29)=−766.3 J mol-1
Thus the excess Gibbs energy is −766.3 J mol-1_.
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Chapter 5 Solutions
Atkins' Physical Chemistry
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