diff options
Diffstat (limited to '2465/CH5')
| -rw-r--r-- | 2465/CH5/EX5.1/Example_1.sce | 23 | ||||
| -rw-r--r-- | 2465/CH5/EX5.10/Example_10.sce | 29 | ||||
| -rw-r--r-- | 2465/CH5/EX5.11/Example_11.sce | 23 | ||||
| -rw-r--r-- | 2465/CH5/EX5.12/Example_12.sce | 23 | ||||
| -rw-r--r-- | 2465/CH5/EX5.15/Example_15.sce | 17 | ||||
| -rw-r--r-- | 2465/CH5/EX5.16/Example_16.sce | 19 | ||||
| -rw-r--r-- | 2465/CH5/EX5.17/Example_17.sce | 17 | ||||
| -rw-r--r-- | 2465/CH5/EX5.2/Example_2.sce | 19 | ||||
| -rw-r--r-- | 2465/CH5/EX5.3/Example_3.sce | 19 | ||||
| -rw-r--r-- | 2465/CH5/EX5.4/Example_4.sce | 15 | ||||
| -rw-r--r-- | 2465/CH5/EX5.5/Example_5.sce | 17 | ||||
| -rw-r--r-- | 2465/CH5/EX5.6/Example_6.sce | 17 | ||||
| -rw-r--r-- | 2465/CH5/EX5.7/Example_7.sce | 27 | ||||
| -rw-r--r-- | 2465/CH5/EX5.9/Example_9.sce | 15 |
14 files changed, 280 insertions, 0 deletions
diff --git a/2465/CH5/EX5.1/Example_1.sce b/2465/CH5/EX5.1/Example_1.sce new file mode 100644 index 000000000..aa4ffc6d9 --- /dev/null +++ b/2465/CH5/EX5.1/Example_1.sce @@ -0,0 +1,23 @@ +//Chapter-5,Example 1,Page 121
+clc();
+close();
+
+//for 1st order reaction
+//k = (1/t)*log(a/(a-x))
+
+a= 46.1 //time value
+
+//time intervals
+
+t=[ 5 10 20 30 50]
+
+x=[ 37.1 29.8 19.6 12.3 5.0]
+
+k = (1 ./t).*log(a./(x))
+
+printf('value of k are ' )
+
+disp(k)
+
+printf('since k values are fairly constant by putting in 1nd order rate equation. \nHence decomposition of H2O2 is of 1st order.')
+
diff --git a/2465/CH5/EX5.10/Example_10.sce b/2465/CH5/EX5.10/Example_10.sce new file mode 100644 index 000000000..d2c27a4d5 --- /dev/null +++ b/2465/CH5/EX5.10/Example_10.sce @@ -0,0 +1,29 @@ +//Chapter-5,Example 10,Page 125
+clc();
+close();
+
+T1=50 //time in sec
+
+T2 = 25 //time in sec
+
+a1=0.5 //initial concentration
+
+a2= 1
+
+// (T1/T2) = (a2/a1)^(n-1)
+//therefore (50/25) =(1/0.5)^(n-1)
+// 2=2^(n-1)
+// n=2
+//hence its 2nd order
+
+t_half= T1
+
+k=1/(a1*t_half)
+
+//assume y= a-x
+
+y=0.2*a1 //remaining concentration
+
+t=(a1-y)/(a1*k*(y))
+
+printf('the time taken is %.f sec ',t)
diff --git a/2465/CH5/EX5.11/Example_11.sce b/2465/CH5/EX5.11/Example_11.sce new file mode 100644 index 000000000..06b492323 --- /dev/null +++ b/2465/CH5/EX5.11/Example_11.sce @@ -0,0 +1,23 @@ +//Chapter-5,Example 11,Page 126
+clc();
+close();
+
+a=0.1 //initial concentration of reactants
+
+x=0.2*a
+
+t=40 //time
+
+k=x/(a*t*(a-x))
+
+t_half=1/(a*k)
+
+x1=0.75*a
+
+t1=x1/(k*a*(a-x1))
+
+printf('the rate constant is k = %.4f l/mol.min',k)
+
+printf('\n the half life period is %.f mins',t_half)
+
+printf('\n the time required to complete 75 percent reaction is %.f mins',t1)
diff --git a/2465/CH5/EX5.12/Example_12.sce b/2465/CH5/EX5.12/Example_12.sce new file mode 100644 index 000000000..608882bf0 --- /dev/null +++ b/2465/CH5/EX5.12/Example_12.sce @@ -0,0 +1,23 @@ +//Chapter-5,Example 12,Page 126
+clc();
+close();
+
+a1=100
+
+x1=1
+
+t1=1
+
+k=2.303*log10(a1/(a1-x1))/t1
+
+t2=60 //time in minutes
+
+a2=100
+
+//assume (a2-x2)= y
+
+y= 1/(10^(k*t2/2.303)/a2)
+
+printf('the undecomposed is %.2f ',y)
+
+//mistake in textbook
diff --git a/2465/CH5/EX5.15/Example_15.sce b/2465/CH5/EX5.15/Example_15.sce new file mode 100644 index 000000000..0b9f3a800 --- /dev/null +++ b/2465/CH5/EX5.15/Example_15.sce @@ -0,0 +1,17 @@ +//Chapter-5,Example 15,Page 128
+clc();
+close();
+
+K1=2.45*10^-5 //rate constant at 273 K
+
+K2=162*10^-5 //rate constant at 303 K
+
+T1=273 //temperature in Kelvin
+
+T2=303 //temperature in Kelvin
+
+R=1.987 //gas constant
+
+Ea= log10(K2/K1)*2.303*R*T1*T2/(T2-T1)
+
+printf('the activation energy is Ea = %.f cal/mole' ,Ea)
diff --git a/2465/CH5/EX5.16/Example_16.sce b/2465/CH5/EX5.16/Example_16.sce new file mode 100644 index 000000000..90d6670a2 --- /dev/null +++ b/2465/CH5/EX5.16/Example_16.sce @@ -0,0 +1,19 @@ +//Chapter-5,Example 16,Page 128
+clc();
+close();
+
+t_half = 600 // half life
+
+K=0.693/t_half
+
+Ea=98600 //activation energy
+
+A= 4*10^13 //Arrhenius factor
+
+R=8.316 //gas constant
+
+T=Ea/(2.303*R*log10(A/K))
+
+printf('temperature is %.f K',T)
+
+//mistake in textbook
diff --git a/2465/CH5/EX5.17/Example_17.sce b/2465/CH5/EX5.17/Example_17.sce new file mode 100644 index 000000000..7156929ab --- /dev/null +++ b/2465/CH5/EX5.17/Example_17.sce @@ -0,0 +1,17 @@ +//Chapter-5,Example 17,Page 129
+clc();
+close();
+
+K1=5*10^-3 //rate constant at 800 degrees
+
+Ea=4.5*10^4 //activation energy
+
+T1=800+273 //temperature in Kelvin
+
+T2=875+273 //temperature in Kelvin
+
+R=8.314 //gas constant
+
+K2=K1*10^(Ea*(T2-T1)/(2.303*R*T1*T2))
+
+printf('the value of K2 = %.4f l/mol.sec',K2)
diff --git a/2465/CH5/EX5.2/Example_2.sce b/2465/CH5/EX5.2/Example_2.sce new file mode 100644 index 000000000..e1d462398 --- /dev/null +++ b/2465/CH5/EX5.2/Example_2.sce @@ -0,0 +1,19 @@ +//Chapter-5,Example 2,Page 122
+clc();
+close();
+
+t=[7.18 18 27.05] //time in minute
+
+r=[ 21.4 17.7 15] //rotation in degrees
+
+r_0=24.09
+
+r_a=-10.74
+
+k=(1 ./t).*log10((r_0-r_a)./(r-r_a))
+
+printf('values of k')
+
+disp(k)
+
+printf('since k values are fairly constant by putting in 1nd order rate equation. \nHence hydrolysis of methyl acetate is of 1st order.')
diff --git a/2465/CH5/EX5.3/Example_3.sce b/2465/CH5/EX5.3/Example_3.sce new file mode 100644 index 000000000..08a5f99ec --- /dev/null +++ b/2465/CH5/EX5.3/Example_3.sce @@ -0,0 +1,19 @@ +//Chapter-5,Example 3,Page 122
+clc();
+close();
+
+t=[75 119 183] //time in minute
+
+V=[24.20 26.60 29.32] //volume of alkali used
+
+V_0=19.24
+
+V_a=42.03
+
+k=(2.303 ./t).*log10((V_a-V_0)./(V_a-V))
+
+printf('values of k')
+
+disp(k)
+
+printf('since k values are fairly constant by putting in 1nd order rate equation. \nHence hydrolysis of methyl acetate is of 1st order.')
diff --git a/2465/CH5/EX5.4/Example_4.sce b/2465/CH5/EX5.4/Example_4.sce new file mode 100644 index 000000000..9f7f895e6 --- /dev/null +++ b/2465/CH5/EX5.4/Example_4.sce @@ -0,0 +1,15 @@ +//Chapter-5,Example 4,Page 123
+clc();
+close();
+
+t= 30 //time in minutes
+
+a=100
+
+x= 25
+
+k=(2.303/t)*log10(a/(a-x))
+
+t_half=0.693/k
+
+printf('the time of 50 percent completion of reaction is %.2f mins',t_half)
diff --git a/2465/CH5/EX5.5/Example_5.sce b/2465/CH5/EX5.5/Example_5.sce new file mode 100644 index 000000000..371a55fec --- /dev/null +++ b/2465/CH5/EX5.5/Example_5.sce @@ -0,0 +1,17 @@ +//Chapter-5,Example 5,Page 123
+clc();
+close();
+
+t_half=17 //half life period
+
+k=0.693/t_half //rate constant
+
+a=100
+
+x= 75
+
+t=(2.303/k)*log10(a/(a-x))
+
+printf('the rate constant is k = %.5f /min',k)
+
+printf('\n the time taken t = %.1f min', t)
diff --git a/2465/CH5/EX5.6/Example_6.sce b/2465/CH5/EX5.6/Example_6.sce new file mode 100644 index 000000000..f607293bc --- /dev/null +++ b/2465/CH5/EX5.6/Example_6.sce @@ -0,0 +1,17 @@ +//Chapter-5,Example 6,Page 123
+clc();
+close();
+
+t_half=1600 //half life period
+
+k=0.693/t_half //rate constant
+
+a=100
+
+x= 80
+
+t=(2.303/k)*log10(a/(a-x))
+
+printf('\n the time taken t = %.2f years', t)
+
+//mistake in textbook
diff --git a/2465/CH5/EX5.7/Example_7.sce b/2465/CH5/EX5.7/Example_7.sce new file mode 100644 index 000000000..728c0cf49 --- /dev/null +++ b/2465/CH5/EX5.7/Example_7.sce @@ -0,0 +1,27 @@ +//Chapter-5,Example 7,Page 124
+clc();
+close();
+
+//for 2st order reaction
+//k = (1/a*t)*(x/(a-x))
+
+a= 16
+
+//time intervals
+
+t=[ 5 15 25 35]
+
+//assume y = a-x
+
+y=[ 10.24 6.13 4.32 3.41] //volume of acid
+
+x=a-y
+
+k = (1 ./(a*t)).*(x./(y))
+
+printf('value of k are ' )
+
+disp(k)
+
+printf('since k values are fairly constant by putting in 2nd order rate equation. \nHence dhydrolysis of methyl acetate is of 2st order.')
+
diff --git a/2465/CH5/EX5.9/Example_9.sce b/2465/CH5/EX5.9/Example_9.sce new file mode 100644 index 000000000..9e528c241 --- /dev/null +++ b/2465/CH5/EX5.9/Example_9.sce @@ -0,0 +1,15 @@ +//Chapter-5,Example 9,Page 125
+clc();
+close();
+
+ //final concentration is half of initial concentration
+//therefore t =t_half
+t= 60 //time in minutes
+
+t_half=t
+
+k=5.2*10^-3 //rste constant
+
+a=1/(k*t_half) //for 2nd order reaction
+
+printf('the initial concentration is %.2f mol/litre',a)
|