BALANCING AND STUDYING EFFECTS OF A CHEMICAL COMBUSTION EQUATION IN MATLAB

STOICHIOMETRIC COMBUSTION

 

DESCRIPTION:

 

Stoichiometric combustion defines the ideal combustion that needs to take place for a given chemical equation. This is actually the general form of equation for alkanes.

The general form might differ for each of the Carbon family: Alkanes, Alkenes and Alkynes as shown below,

 

FOR ALKANES, C will have 4 single bonds.

Alkanes: CnH2n+2

FOR ALKENES, C will have 2 double bonds.

Alkenes: CnH2n

FOR ALKYNES, C will have a triple bond.

Alkynes: CnH2n−2 

Based on this concept, we should be balancing the equation.

  

The Stoichiometric co-efficient refers to the co-efficient of moles needed in a chemical equation to require balance through out. So, to find the stoichiometric co-efficient of 'ar' given here, we need to balance the whole equation with respect to the elements present in the reactant and the product sides.

 

 

SOLVING PROCEDURE FOR ALKANES:

 

1. Number of moles of C in LHS and RHS:
n = a

2.  Number of moles of H in LHS and RHS:
⦁ Alkanes: 2n+2 = 2b

3. Number of moles of N in LHS and RHS:
3.76ar = 2c

4. Number of moles of O in LHS and RHS:
2ar = 2a+b

 

Solving for ar value:

2*ar = (2*a) + b

2*ar = (2*n) + (2*n +1)

ar = n + (n+1)/2

ar = ((3*n)+1)/2

 

SOLVING PROCEDURE FOR ALKENES:

 

`CnH2n + ar(O2 + 3.76N2) = a(CO2) + b(H2O) + c(N2)`

 

1. Number of moles of C in LHS and RHS:
n = a

2.  Number of moles of H in LHS and RHS:
⦁ Alkenes: 2n = 2b

3. Number of moles of N in LHS and RHS:
3.76ar = 2c

4. Number of moles of O in LHS and RHS:
2ar = 2a+b

 

Solving for ar value:

2*ar = (2*a) + b

2*ar = (2*n) + (n)

ar = (3*n)/2

 

SOLVING PROCEDURE FOR ALKYNES:

 

`CnH2n-2 + ar(O2 + 3.76N2) = a(CO2) + b(H2O) + c(N2)`

 

1. Number of moles of C in LHS and RHS:
n = a

2.  Number of moles of H in LHS and RHS:
⦁ Alkynes: 2n-2 = 2b

3. Number of moles of N in LHS and RHS:
3.76ar = 2c

4. Number of moles of O in LHS and RHS:
2ar = 2a+b

 

Solving for ar value:

2*ar = (2*a) + b

2*ar = (2*n) + (n-1)

ar = n + ((n-1)/2)

ar = ((3*n)-1)/2

 

 

SOLVED FORMULA:

Alkanes: ar = [(3/2) x n] + 0.5

Alkenes: ar = [3/2 x n]

Alkynes: ar = [(3/2) x n] - 0.5
 

 

CODE:

close all
clear all
clc

% a = n;
% b = n+1;
% c = ar;
n = linspace(1,10,10)
% for the given equation in the challenge

    
%assingning ar value for alkane, alkene and alkyne
ar_alkane = (3*n+1)/2;
ar_alkene = 3*n/2;
ar_alkyne = (3*n-1)/2;

figure(1)
plot(n,ar_alkane,'b','linewidth',1.5)
hold on
plot(n,ar_alkene,'g','linewidth',1.5)
plot(n,ar_alkyne,'r','linewidth',1.5)
legend('Alkane','Alkene','Alkyne','location','northwest')
xlabel('no of moles')
ylabel('stoichiometric co-eff')
title('Stoichiometric Combustion')

 

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