3 files changed, 83 insertions, 0 deletions
diff --git a/249/CH3/EX3.1/3_01.sce b/249/CH3/EX3.1/3_01.sce
new file mode 100755
index 000000000..1bbf163bd
--- /dev/null
+++ b/249/CH3/EX3.1/3_01.sce
@@ -0,0 +1,39 @@
+clear
+clc
+//Given
+t=[0 20 40 60 120 180 300];
+C_A=[10 8 6 5 3 2 1];
+CAo=10;
+//Guessing 1st order kinetics
+//This means log(CAo/C_A) vs t should give a straight line
+for i=1:7
+    k(i)=log(CAo/C_A(i));
+    CA_inv(i)=1/C_A(i);
+end
+//plot(t,k)
+//This doesn't give straight line.
+//Guessing 2nd Order Kinetics so 
+//1/C_A vs t should give a straight line
+//plot(t,CA_inv)
+//Again this doesn't give a straight line
+//Guessing nth order kinetics and using fractional life method with F=80%
+//log Tf=log(0.8^(1-n)-1/(k(n-1)))+(1-n)logCAo
+//plot(t,C_A)
+
+//Picking different values of CAo
+//Time needed for 3 runs,,from graph
+T=[18.5;23;35];
+CAo=[10;5;2];
+for i=1:3
+    CA(i)=0.8*CAo(i);
+    log_Tf(i)=log10(T(i));
+    log_CAo(i)=log10(CAo(i));
+end
+plot(log_CAo,log_Tf)
+xlabel('log CAo');ylabel('log t');
+coeff1=regress(log_CAo,log_Tf);
+n=1-coeff1(2);
+printf("From graph we get slope and intercept for calculating rate eqn")
+k1=((0.8^(1-n))-1)*(10^(1-n))/(18.5*(n-1));
+printf("\n The rate equation is given by %f",k1)
+printf("CA^1.4 mol/litre.sec")
diff --git a/249/CH3/EX3.2/3_02.sce b/249/CH3/EX3.2/3_02.sce
new file mode 100755
index 000000000..a2a1ef1b6
--- /dev/null
+++ b/249/CH3/EX3.2/3_02.sce
@@ -0,0 +1,26 @@
+clear
+clc
+CA=[10;8;6;5;3;2;1];//mol/litre
+T=[0;20;40;60;120;180;300];//sec
+//plot(T,CA)
+//xlabel('Time(sec)');ylabel('CA(mol/litre)');
+//From graph y=-dCA/dt at different points are
+y=[-0.1333;-0.1031;-0.0658;-0.0410;-0.0238;-0.0108;-0.0065];
+//Guessing nth rate order
+//rA=kCA^n
+//log(-dCA/dt)=logk+nlogCA
+for i=1:7
+log_y(i)=log10(y(i));
+log_CA(i)=log10(CA(i));
+end
+plot(log_CA,log_y)
+xlabel('logCA');ylabel('log(-dCA/dt)')
+coeff1=regress(log_CA,log_y);
+n=coeff1(2);
+k=-10^(coeff1(1));
+printf("\n After doing linear regression,the slope and intercept of the graph is %f , %f",coeff(2),coeff(1))
+printf("\n The rate equation is therefore given by %f",k)
+printf("CA^1.375 mol/litre.sec")
+disp('The answer slightly differs from those given in book as regress fn is used for calculating slope and intercept')
+
+
diff --git a/249/CH3/EX3.4/3_04.sce b/249/CH3/EX3.4/3_04.sce
new file mode 100755
index 000000000..e8aacb29d
--- /dev/null
+++ b/249/CH3/EX3.4/3_04.sce
@@ -0,0 +1,18 @@
+clear
+clc
+// At 400k, -rA=2.3*pA^2
+//At 500 k, -rA=2.3*pA^2
+k1=2.3;k2=2.3;T1=400;T2=500;
+//R=82.06*10^-6 m3.atm/mol.k
+R=82.06*10^-6;
+R1=8.314;//m3.pa/mol.k
+E=(log(k2/k1)*R)/(1/T1-1/T2)
+printf("\nRESULT\n")
+printf("E(J/mol)using pressure units is %f",E)
+//pA=CA*RT
+//-rA=2.3(RT)^2*CA^2
+k1=2.3*(R*T1)^2
+k2=2.3*(R*T2)^2
+E=(log(k2/k1)*R1)/(1/T1-1/T2)
+printf("\nE(J/mol)using concentration units is %f",E)
+