Chapter 11, Problem 11.33P

(a)

Interpretation Introduction

Interpretation:

For a binary system, the given equations are to be shown.

Concept Introduction:

For a binary system, equation $10.15$ to be used is:

   ${\overline{M}}_1=M+x_2\frac{dM}{d{x}_1}$ .....(1)

For a binary system, equation $10.16$ to be used is:

   ${\overline{M}}_2=M-x_1\frac{dM}{d{x}_1}$ .....(2)

Explanation of Solution

Given information:

Given equations to be shown are:

   $\begin{array}{c}{\overline{M}}_1^E=x_2^2\left(X+x_1\frac{dX}{d{x}_1}\right)\\ {\overline{M}}_2^E=x_1^2\left(X-x_2\frac{dX}{d{x}_1}\right)\end{array}$

Here,

   $X\equiv \frac{M^E}{x_1{x}_2}$

Rearrange and write the given equation as:

   $\begin{array}{c}{M}^E=x_1{x}_{2}X\mathrm{......}\text{ (3)}\\ M^E=x_1\left(1-x_1\right)X\\ M^E=\left(x_1-x_1^2\right)X\end{array}$

Now, take derivative of this equation with respect to $x_1$ as:

   $\begin{array}{c}\frac{d{M}^E}{d{x}_1}=\frac{d\left(\left(x_1-x_1^2\right)X\right)}{d{x}_1}\\ \frac{d{M}^E}{d{x}_1}=\left(1-2x_1\right)X+\left(x_1-x_1^2\right)\frac{dX}{d{x}_1}\end{array}$

In terms of both $x_1\text{ and }{x}_2$, rewrite the above equation as:

   $\frac{d{M}^E}{d{x}_1}=\left(x_2-x_1\right)X+\left(x_1{x}_2\right)\frac{dX}{d{x}_1}$ .....(4)

For excess property relation, equation (1) can be written as:

   ${\overline{M}}_1^E=M^E+x_2\frac{d{M}^E}{d{x}_1}$

Substitute the values from equations (3) and (4) in the above equation and simplify as:

   $\begin{array}{c}{\overline{M}}_1^E=x_1{x}_{2}X+x_2\left(\left(x_2-x_1\right)X+\left(x_1{x}_2\right)\frac{dX}{d{x}_1}\right)\\ {\overline{M}}_1^E=\cancel{x_1{x}_{2}X}+x_2^{2}X-\cancel{x_1{x}_{2}X}+x_1{x}_2^2\frac{dX}{d{x}_1}\\ {\overline{M}}_1^E=x_2^2\left(X+x_1\frac{dX}{d{x}_1}\right)\end{array}$

Similarly, for excess property relation, equation (2) can be written as:

   ${\overline{M}}_2^E=M^E-x_1\frac{d{M}^E}{d{x}_1}$

Substitute the values from equations (3) and (4) in the above equation and simplify as:

   $\begin{array}{c}{\overline{M}}_2^E=x_1{x}_{2}X-x_1\left(\left(x_2-x_1\right)X+\left(x_1{x}_2\right)\frac{dX}{d{x}_1}\right)\\ {\overline{M}}_2^E=\cancel{x_1{x}_{2}X}-\cancel{x_1{x}_{2}X}+x_1^{2}X-x_2{x}_1^2\frac{dX}{d{x}_1}\\ {\overline{M}}_2^E=x_1^2\left(X-x_2\frac{dX}{d{x}_1}\right)\end{array}$

Hence, the given equations are proved.

(b)

Interpretation Introduction

Interpretation:

The plot of $H^E\text{/}\left(x_1{x}_2\right),{\overline{H}}_1^E,\text{ and }{\overline{H}}_2^E$ from the given heat of mixing data for ${\text{H}}_2{\text{SO}}_4\left(1\right){\text{/H}}_2\text{O}\left(2\right)$ system at ${25}^{\circ }\text{C}$ are to be drawn on a single graph. Also, the reason for diluting the acid by adding acid to water in terms of the plots drawn is to be explained.

Concept Introduction:

For mixing process, equation for enthalpy is:

   $\Delta H=H^E$ .....(5)

In terms of a third-degree polynomial equation in $x_1$, $H^E$ can be expressed as:

   $H^E=a+b{x}_1+c{x}_1^2+d{x}_1^3$ .....(6)

Here, $a,b,c,\text{ and }d$ are the constants and can be found by polynomial regression using data for $H^E\text{ and }{x}_1$ .

Answer to Problem 11.33P

The plot of $H^E\text{/}\left(x_1{x}_2\right),{\overline{H}}_1^E,\text{ and }{\overline{H}}_2^E$ from the given heat of mixing data for ${\text{H}}_2{\text{SO}}_4\left(1\right){\text{/H}}_2\text{O}\left(2\right)$ system at ${25}^{\circ }\text{C}$ drawn on a single graph is:

From the plot of ${\overline{H}}_1^E,\text{ and }{\overline{H}}_2^E$, when water is added to acid then overall heat released is enormous due to which water splashes out and hazards may occur. When small amount of acid is added to water, then the heat released by the acid is dissipated into the water whose excess enthalpy is less negative than that of sulfuric acid at low acid mole fraction.

Explanation of Solution

Given information:

The heat of mixing data for the ${\text{H}}_2{\text{SO}}_4\left(1\right){\text{/H}}_2\text{O}\left(2\right)$ system at ${25}^{\circ }\text{C}$ is given as:

$x_1$	$-\Delta H\left(\text{kJ/kg}\right)$
$0.10$	$73.27$
$0.20$	$144.21$
$0.30$	$208.64$
$0.40$	$262.83$
$0.50$	$302.84$
$0.60$	$323.31$
$0.70$	$320.98$
$0.80$	$279.58$
$0.85$	$237.25$
$0.90$	$178.87$
$0.95$	$100.71$

Here, $x_1$ is the mole fraction of ${\text{H}}_2{\text{SO}}_4$ .

From the equation (5), the given heat of mixing for the values of $x_1$ is same as excess enthalpy. Thus,

Now, plot the graph of $H^E\text{ versus }{x}_1$ and use the regression tool of excel and fit the curve into the third-degree polynomial and compare the equation with the form of equation (6) as:

   $H^E=-27.858-412.29(x_1)-1006.9{(x_1)}^2+1423.5{(x_1)}^3$ .....(7)

Now, calculate the derivative of equation (7) as:

   $\frac{d{H}^E}{d{x}_1}=-412.29-2013.8(x_1)+4270.5{(x_1)}^2$ .....(8)

Now, calculate the value of $H^E\text{/}\left(x_1{x}_2\right),{\overline{H}}_1^E,\text{ and }{\overline{H}}_2^E$ using the formulas shown below and use the calculated values as required. Then, tabulate the data in the excel sheet as shown:

   $\begin{array}{c}{H}^E\text{/}\left(x_1{x}_2\right)=\frac{\Delta H}{x_1{x}_2}\\ {\overline{H}}_1^E=H^E+x_2\frac{d{H}^E}{d{x}_1}\\ {\overline{H}}_2^E=H^E-x_1\frac{d{H}^E}{d{x}_1}\end{array}$

Now, plot the graph for $H^E\text{/}\left(x_1{x}_2\right),{\overline{H}}_1^E,\text{ and }{\overline{H}}_2^E$ versus $x_1$ as:

From the plot of ${\overline{H}}_1^E,\text{ and }{\overline{H}}_2^E$, when water is added to acid then overall heat released is enormous due to which water splashes out and hazards may occur. When small amount of acid is added to water, then the heat released by the acid is dissipated into the water whose excess enthalpy is less negative than that of sulfuric acid at low acid mole fraction.