Question

please provide the matlab code

Required information A reversible chemical reaction 2A + B C can be characterized by the equilibrium relationship K = C where the nomenclature cirepresents the concentration of constituent i. Suppose that we define a variable x as representing the number of moles of C that are produced. Conservation of mass can be used to reformulate the equilibrium relationship as K- (ccott) (6,0-2x) (00x) where the subscript O designates the initial concentration of each constituent. Take K= 0.016, Ca,0 = 42, Cb,0 = 28, and Cc,0 = 4. Determine the value of x graphically. (Please upload your response/solution using the controls below.)

Answer 1

From the given equation and values we have:

  

KE (cco +) -2.0(Co-)

(4+ c) =0.016 = (42 - 212/28 - x)

0.016 ~ (42 - 2.c)2 x (28 - 1) = (4+1)

(28.224 +0.064.r? – 2.688.r) x (28 - 1) = (4+1)

(790.272+1.792.22 – 75.264.- 28.224.- 0.064.3 +2.688.72) = (4+1)

  

790.272+1.792.12 – 75.264.2 – 28.224.2 -0.064.3 +2.688.-4-=0

  

一—104.488x +4.48.2 -0.064.3 + 786.272 = 0

  

0.064.13 – 4.48.2 + 104.488.– 786.272 = 0

Let,    (1)

f(0) = 0.064.23 – 4.48.2+ 104.488.– 786.272 = 0

Clearly, is a polynomial of degree '3'. The root of this cubic equation or polynomial can be determined either by analytical approach or by using graphical method. When the graph of the polynomial is plot against the value of then the coordinates of points the graph of the polynomial cuts the x-axis are the roots of the given polynomial.

Now plotting, , for different values of x,

(2)! = 1

Method I: Using MS-Excel,

The First column is made x and given arbitrary values as shown in the table below:

The Function is made according to the equation no. (1) as: y=0.064*A2^3-4.48*A2^2+104.488*A2-786.272, where A2 is the location of x in second row, first column.

Table:1

x	y
5	-367.832
5.5	-336.46
6	-306.8
6.5	-278.804
7	-252.424
7.5	-227.612
8	-204.32
8.5	-182.5
9	-162.104
9.5	-143.084
10	-125.392
10.5	-108.98
11	-93.8
11.5	-79.804
12	-66.944
12.5	-55.172
13	-44.44
13.5	-34.7
14	-25.904
14.5	-18.004
15	-10.952
15.5	-4.7
16	0.8
16.5	5.596
17	9.736
17.5	13.268
18	16.24
18.5	18.7
19	20.696
19.5	22.276
20	23.488

The Graph is then plotted between x and y and we obtained the graph as:

Cubic Equation By Inspection the roots of the equation are,x=15.9 11 12 13 14 15 16 17 18 19 20 Y =f(x) The point at which graph of polynomial cuts the x-axis gives the solution or roots at that value of x coordinate.

Another full view graph is obtained as:

Cubic Polynomial By Inspection the roots of the equation are,x=15.9 10 20 3035 (x)} = A The point at which graph of polynomial cuts the x-axis gives the solution or roots at that value of x coordinate. -400

Steps:

1. Go to insert graph, choose scatter and then click on the graph.

2. right click on the blank graph and choose select data,

3. select the data points in x and y axis corresponding data.

4. plot the data, and resize the graph according to convenience.

Hence the graph cut the x-axis at x=15.9 (approx.) hence the value of x is 15.9.

Method 2: Using MATLAB

Now , give the command as follows:

p=[0.064 -4.48 104.488 -786.272];

r=roots(p)

Ans:r =
27.0385 + 6.3621i
27.0385 - 6.3621i
15.9230 + 0.0000i

Now the roots are confirmed, neglecting the imaginary roots, 15.923 is the correct root.

Also looking graphically as follows:

x=5:0.5:25;
y=0.064*x.^3-4.48*x.^2+104.488*x-786.272;
plot(x,y);

File Tools BOD Q by + -300 -350 -400 4 6 8 10 12 14 16 18 20 22 24 26

Clearly, x=15.9 is the correct answer.