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+//Example 3.2

+clear;

+clc;

+

+//Given 

+delH1 = -393.5//H1 is the heat of reaction in the formation of CARBON DIOXIDE in kJ (i)

+delH2 = -110.5//H2 is the heat of reaction in the formation of CARBON MONOXIDE in kJ (ii)

+delH3 = -890.35//H3 is the heat of reaction in the combustion of METHANE in kJ (iii)

+delH4 = -85.3//H4 is the heat of reaction in the formation of SILVER CHLORIDE in kJ (iv)

+R = 0.008314;//R is gas constant in kJ K^-1 mol^-1

+T = 298;//T is temperature in K

+ 

+//To determine the heat of formation

+delv1= 1-(1);//change in moles in reaction (i)

+delE1 = delH1 - (delv1*R*T);//E1 is the internal energy (i) in kJ (1st law of thermodynamics)

+mprintf('(i) change in internal energy = %f kJ',delE1);

+delv2=1-(0.5);//change in moles in reaction (ii)

+delE2 = delH2 - (delv2*R*T);//E2 is the internal energy (ii) in kJ (1st law of thermodynamics)

+mprintf('\n (ii) change in internal energy = %f kJ',delE2);

+delv3=1-(1+2);//change in moles in reaction (iii)

+delE3 = delH3 - (delv3*R*T);//E3 is the internal energy (iii) in kJ (1st law of thermodynamics)

+mprintf('\n (iii) change in internal energy = %f kJ',delE3);

+delv4=0-(1);//change in moles in reaction (iv)

+delE4 = delH4 - (delv4*R*T);//E4 is the internal energy (iv) in kJ (1st law of thermodynamics)

+mprintf('\n (iv) change in internal energy = %f kJ',delE4);

+//end
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