Chapter 10, Problem 10.33P

(a)

Interpretation Introduction

Interpretation:

Estimation of parameters a, b and c from the given equation and data of composition versus excess enthalpy.

Concept introduction:

Excess molar properties of mixtures are the non ideal nature of real mixtures. They are generally the difference between the partial molar property of a component in a real mixture and in an ideal mixture.

Parameters a, b and c can be determined by fitting the data given in the question.

Answer to Problem 10.33P

Parameters of the given equations are:

  $\begin{array}{l}a=-530\\ b=-1129\\ c=1050\end{array}$

Explanation of Solution

We know that the mole fraction of the second component in a mixture can be written as

  $x_2=1-x_1$

Which reduces the given equation of excess volume as:

  $\begin{array}{l}{H}^E=x_1{x}_2\left[a+b{x}_1+c{x}_1^2\right]\\ H^E=x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]\end{array}$

For calculating parameters using the best fit metod of data to the equation, we must assume initial parameters as:

a = -500 ; b = -100 and c = 0.1

Now we will estimate values of H^E using given equation and assumed parameters and find the error using formula given below:

  $\text{normalized}\text{error}={\displaystyle \sum {\left[\frac{H_{Given}^E-H_{estimated}^E}{H_{Given}^E}\right]}^2}$

After finding error, we have to minimize the error to best fit the given and estimated data by finding parameter in excel solver as described below:

x1	HE(given)	HE(estimated)	Normalized error
0.0426	-23.3	-23.2149	0.000255
0.0817	-45.7	-45.633	0.00011
0.1177	-66.5	-66.6008	6.04E-05
0.151	-86.6	-85.833	4.78E-05
0.2107	-118.2	-118.781	1.08E-06
0.2624	-144.6	-144.559	5.81E-05
0.3472	-176.6	-178.669	9.94E-07
0.4158	-195.7	-196.933	2.64E-05
0.5163	-204.2	-206.209	1.4E-08
0.6156	-191.7	-193.963	4.17E-05
0.681	-174.1	-174.738	1.05E-05
0.7621	-141	-140.126	3.45E-06
0.8181	-116.8	-110.579	0.001453
0.865	-85.6	-83.2654	1.4E-05
0.9276	-43.5	-44.7103	0.004488
0.9624	-22.6	-23.0104	0.004457
		Sum of error	0.011027
		Estimated	Assumed
	a	-530	-500
	b	-1129	-100
	c	1050	0.1

By the linear fitting of the given data in the given equation, we will get parameters as:

  $\begin{array}{l}a=-530\\ b=-1129\\ c=1050\end{array}$

(b)

Interpretation Introduction

Interpretation:

Estimation the minimum value of excess enthalpy and composition at which it becomes minimum.

Concept introduction:

Excess molar properties of mixtures are the non ideal nature of real mixtures. They are generally the difference between the partial molar property of a component in a real mixture and in an ideal mixture.

Answer to Problem 10.33P

Minimum value of excess enthalpy = -204.4 J/mol at composition 0.512.

Explanation of Solution

We know that the mole fraction of the second component in a mixture can be written as

  $x_2=1-x_1$

Which reduces the given equation of excess enthalpy as:

  $\begin{array}{l}{H}^E=x_1{x}_2\left[a+b{x}_1+c{x}_1^2\right]\\ H^E=x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]\end{array}$...(1)

Put the determined parameter values from part (a) to the equation (1), we get:

  $\begin{array}{l}{H}^E=x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]\\ H^E=x_1\left(1-x_1\right)\left[-530-1129x_1+1050x_1^2\right]\end{array}$

Now differentiate above equation with respect to x₁ and equate it to zero gives:

  $\begin{array}{l}\frac{\partial H^E}{\partial x_1}=0\\ \frac{\partial H^E}{\partial x_1}=\frac{\partial }{\partial x_1}\left[x_1\left(1-x_1\right)\left[-530-1129x_1+1050x_1^2\right]\right]=0\\ \frac{\partial }{\partial x_1}\left[-530x_1-1129x_{{}_1}^2+1050x_1^3+530x_{{}_1}^2+1129x_1^3-1050x_1^4\right]=0\\ -530-1198x_1+6537x_{{}_1}^2-4200x_1^3=0\end{array}$

Now we will find the value of x₁ which satisfies the above equation:

Let x₁ = 0.5

  $\begin{array}{l}-530-1198x_1+6537x_{{}_1}^2-4200x_1^3\\ -530-\left(1198\times 0.5\right)+\left(6537\times {0.5}^2\right)-\left(4200\times {0.5}^3\right)=0\\ -19\approx 0\end{array}$

Near to zero, so composition will be approximately 0.5 or 0.513.

So, at x₁ = 0.513 excess enthalpy will be a minumum and the minimum value of excess enthalpy can be calculated as given below:

  $\begin{array}{l}{H}^E=x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]\\ H_{{}^E}^{\min }=0.513\left(1-0.513\right)\left[-530-\left(1129\times 0.513\right)+\left(1050\times \left({0.513}^2\right)\right)\right]\\ H_{{}^E}^{\min }=-204.4\end{array}$

(c)

Interpretation Introduction

Interpretation:

Derive expression for $\overline{H_{{}^1}^E}$, $\overline{H_2^E}$ and plot with respect to x₁and a plot should be prepared.

Concept introduction:

Excess molar properties of mixtures are the non ideal nature of real mixtures. They are generally the difference between the partial molar property of a component in a real mixture and in an ideal mixture.

Partial molar properties can be derived from equations:

  $\begin{array}{l}\overline{H_1^E}=H^E+\left(1-x_1\right)\frac{d{H}^E}{d{x}_1}\\ \overline{H_2^E}=V^E-\left(x_1\right)\frac{d{H}^E}{d{x}_1}\end{array}$

Answer to Problem 10.33P

Expression of partial molar properties is:

  $\overline{H_1^E}={\left(1-x_1\right)}^2\left[a+2b{x}_1+3c{x}_1^2\right]$

  $\overline{H_2^E}=x_1^2\left[\left(a-b\right)+2\left(b-c\right)x_1+3c{x}_1^2\right]$

The graph plotted between partial molar properties and composition.

Explanation of Solution

We know that the mole fraction of the second component in a mixture can be written as

  $x_2=1-x_1$

Which reduces the given equation of excess enthalpy as:

  $\begin{array}{l}{H}^E=x_1{x}_2\left[a+b{x}_1+c{x}_1^2\right]\\ H^E=x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]\end{array}$...(1)

Put the determined parameter values from part (a) to the equation (1), we get:

  $\begin{array}{l}{H}^E=x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]\\ H^E=x_1\left(1-x_1\right)\left[-530-1129x_1+1050x_1^2\right]\end{array}$

Now differentiate the above equation with respect to x₁ and get:

  $\begin{array}{l}\frac{\partial H^E}{\partial x_1}=\frac{\partial }{\partial x_1}\left[x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]\right]\\ \frac{\partial H^E}{\partial x_1}=\frac{\partial }{\partial x_1}\left[a{x}_1+b{x}_{{}_1}^2+c{x}_1^3-a{x}_{{}_1}^2-b{x}_1^3-c{x}_1^4\right]\\ \frac{\partial H^E}{\partial x_1}=a+2\left(b-a\right)x_1+3\left(c-b\right)x_{{}_1}^2-4c{x}_1^3\end{array}$

The expression for partial molar enthalpy can be written as follows:

  $\begin{array}{l}\overline{H_1^E}=H^E+\left(1-x_1\right)\frac{d{H}^E}{d{x}_1}\\ \overline{H_1^E}=x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]+\left(1-x_1\right)\left[a+2\left(b-a\right)x_1+3\left(c-b\right)x_{{}_1}^2-4c{x}_1^3\right]\\ \boxed{\overline{H_1^E}={\left(1-x_1\right)}^2\left[a+2b{x}_1+3c{x}_1^2\right]}\\ \overline{H_2^E}=H^E-\left(x_1\right)\frac{d{H}^E}{d{x}_1}\\ \overline{H_2^E}=x_1\left(1-x_1\right)\left[a+b{x}_1+c{x}_1^2\right]-\left(x_1\right)\left[a+2\left(b-a\right)x_1+3\left(c-b\right)x_{{}_1}^2-4c{x}_1^3\right]\\ \boxed{\overline{H_2^E}=x_1^2\left[\left(a-b\right)+2\left(b-c\right)x_1+3c{x}_1^2\right]}\end{array}$

Plot of partial molar properties derived above with x₁ can be drawn as follows:

We know that,

  $\begin{array}{l}a=-530\\ b=-1129\\ c=1050\end{array}$

So, the partial molar equations become:

  $\begin{array}{l}\overline{H_1^E}={\left(1-x_1\right)}^2\left[a+2b{x}_1+3c{x}_1^2\right]\\ \overline{H_1^E}={\left(1-x_1\right)}^2\left[-530-2258x_1+3150x_1^2\right]\\ \overline{H_2^E}=x_1^2\left[\left(a-b\right)+2\left(b-c\right)x_1+3c{x}_1^2\right]\\ \overline{H_2^E}=x_1^2\left[\left(-530+1129\right)+2\left(-1129-1050\right)x_1+3150x_1^2\right]\\ \overline{H_2^E}=x_1^2\left[599-4358x_1+3150x_1^2\right]\end{array}$

x1	HE	H1(E )	H2(E )
0.0426	-23.3	-568.736	0.760504
0.0817	-45.7	-584.771	1.762019
0.1177	-66.5	-585.497	1.796789
0.151	-86.6	-576.017	0.291058
0.2107	-118.2	-539.462	-7.96384
0.2624	-144.6	-492.7	-22.5599
0.3472	-176.6	-398.129	-64.4174
0.4158	-195.7	-315.445	-115.569
0.5163	-204.2	-200.304	-216.279
0.6156	-191.7	-107.319	-337.296
0.681	-174.1	-61.7537	-421.072
0.7621	-141	-23.8448	-518.49
0.8181	-116.8	-8.90102	-574.265
0.865	-85.6	-2.30111	-608.877
0.9276	-43.5	0.450114	-630.781
0.9624	-22.6	0.303215	-627.571

Plot drawn according to data calculated in the above table as: