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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+//To calculate the values of Enthalpy and Entropy of Reaction
+
+//Example 9.4
+
+clc;
+
+clear;
+
+T=[872,973,1073,1173];//Temperatures in Kelvin
+
+Kp=[1.8*10^-4,1.8*10^-3,1.08*10^-2,0.0480];//Equilibrium Constant
+
+for i=1:4
+ x(i)=1/T(i);//Defining x-axis of the graph as x=1/T
+end
+
+for i=1:4
+ y(i)=log(Kp(i));//Defining y-axis of the graph as y=log(Kp)
+end
+
+plot(x,y);//Plotting the Graph between 1/T and log(Kp)
+
+xlabel("K/T", "fontsize", 2);//Putting the x-axis as K/T
+
+ylabel("ln(Kp)", "fontsize", 2);//Putting the y-axis as ln(Kp)
+
+m=-(y(2)-y(1))/(x(2)-x(1));//Slope of the Graph
+
+R=8.314;//Universal Gas Constant in J K^-1 mol^-1
+
+delH=R*m/1000;//Change in Enthalpy in kJ mol^-1
+
+c=12.954;//y-Intercept of the Graph
+
+delS=R*c;//Change in Entropy in J K^-1 mol^-1
+
+printf("Change in Enthalpy of reaction = %.2f*10^2 kJ mol^-1",delH*10^-2);
+
+printf("\n Entropy Change for the reaction = %.0f J K^-1 mol^-1",delS)