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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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treedbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3718/CH5
parentb1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff)
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-rw-r--r--3718/CH5/EX5.1/Ex5_1.sce11
-rw-r--r--3718/CH5/EX5.10/Ex5_10.sce17
-rw-r--r--3718/CH5/EX5.11/Ex5_11.sce11
-rw-r--r--3718/CH5/EX5.12/Ex5_12.sce15
-rw-r--r--3718/CH5/EX5.2/Ex5_2.sce11
-rw-r--r--3718/CH5/EX5.3/Ex5_3.sce24
-rw-r--r--3718/CH5/EX5.4/Ex5_4.sce12
-rw-r--r--3718/CH5/EX5.5/Ex5_5.sce7
-rw-r--r--3718/CH5/EX5.6/Ex5_6.sce16
-rw-r--r--3718/CH5/EX5.7/Ex5_7.sce12
-rw-r--r--3718/CH5/EX5.8/Ex5_8.sce16
11 files changed, 152 insertions, 0 deletions
diff --git a/3718/CH5/EX5.1/Ex5_1.sce b/3718/CH5/EX5.1/Ex5_1.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 1
+clc;
+
+//Declaration of Variables
+K = 3.5 * 10 ** - 2 // Rate constant
+
+// Solution
+mprintf("First order reaction = 0.693 / K\n")
+t = 0.693 / K
+mprintf(" Time taken for half the initial concentration to react:%.1f minutes", t)
diff --git a/3718/CH5/EX5.10/Ex5_10.sce b/3718/CH5/EX5.10/Ex5_10.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 10
+clc;
+
+//Declaration of Constant
+R = 1.987 //in cal per K per mol
+
+//Declaration of Variables
+K2_K1 = 4 // factor increase
+T1 = 27 //in C
+T2 = 47 //in C
+
+// Solution
+T1 = T1 + 273.0
+T2 = T2 + 273.0
+Ea = log10(4) * 2.303 * R * (T1 * T2 / (T2 - T1))
+mprintf("The activation energy for the reaction is %.2e cal /mol",Ea)
diff --git a/3718/CH5/EX5.11/Ex5_11.sce b/3718/CH5/EX5.11/Ex5_11.sce
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index 000000000..4f03204d2
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+++ b/3718/CH5/EX5.11/Ex5_11.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 11
+clc;
+
+//Declaration of Variables
+a = 1 //in mole
+x = 3 / 4.0 // reaction completed
+
+// Solution
+K = (2.303 / 6) * log10(1 / (1 - x))
+mprintf("The rate constant is :%.3f / min",K)
diff --git a/3718/CH5/EX5.12/Ex5_12.sce b/3718/CH5/EX5.12/Ex5_12.sce
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index 000000000..21a720919
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+++ b/3718/CH5/EX5.12/Ex5_12.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 12
+clc;
+
+// Solution
+mprintf("Let the initial concentration be 100, when x = 25,t = 30 minutes\n")
+a = 100
+x = 25.0
+t = 30
+K = 2.303 / t * log10(a / (a - x))
+t05 = 0.683 / K
+t = 2.303 / K * log10(a / x)
+mprintf(" K = %.2e / min\n",K)
+mprintf(" T0.5 = %.2f min\n",t05)
+mprintf(" t = %.1f min",t)
diff --git a/3718/CH5/EX5.2/Ex5_2.sce b/3718/CH5/EX5.2/Ex5_2.sce
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index 000000000..04c757c80
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+++ b/3718/CH5/EX5.2/Ex5_2.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 2
+clc;
+
+//Declaration of Variables
+t = 40 //in minutes
+
+// Solution
+mprintf("Rate constant = 0.693 / t\n")
+K = 0.693 / t
+mprintf(" Rate constant = %.4f / min",K)
diff --git a/3718/CH5/EX5.3/Ex5_3.sce b/3718/CH5/EX5.3/Ex5_3.sce
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index 000000000..b95280d30
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 3
+clc;
+
+//Declaration of Variables
+t0 = 37.0 //in cm cube of KMnO4
+t5 = 29.8 //in cm cube of KMnO4
+t15 = 19.6 //in cm cube of KMnO4
+t25 = 12.3 //in cm cube of KMnO4
+t45 = 5.00 //in cm cube of KMnO4
+
+// Solution
+K5 = 2.303 / 5 * log10(t0 / t5)
+K15 = 2.303 / 15 * log10(t0 / t15)
+K25 = 2.303 / 25 * log10(t0 / t25)
+K45 = 2.303 / 45 * log10(t0 / t45)
+
+mprintf("At t = 5 min, K = %.3e /min\n",K5)
+mprintf(" At t = 15 min, K = %.3e /min\n",K15)
+mprintf(" At t = 25 min, K = %.3e /min\n",K25)
+mprintf(" At t = 45 min, K = %.3e /min\n",K45)
+mprintf(" As the different values of K are nearly same, the reaction is of first-order\n")
+K = (K45 + K25 + K5 + K15) / 4
+mprintf(" The average value of K = %.3e /min",K)
diff --git a/3718/CH5/EX5.4/Ex5_4.sce b/3718/CH5/EX5.4/Ex5_4.sce
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index 000000000..019c82985
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+++ b/3718/CH5/EX5.4/Ex5_4.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 4
+clc;
+
+//Declaration of Variables
+t = 60 //in min
+x = "0.5a"
+K = 5.2 * 10 ** - 3 //in per mol L per min
+
+// Solution
+a = 1 / (t * K)
+mprintf("Initial concentration = %.3f mol / L",a)
diff --git a/3718/CH5/EX5.5/Ex5_5.sce b/3718/CH5/EX5.5/Ex5_5.sce
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index 000000000..a7a1d537e
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 5
+clc;
+
+// Solution
+t = ((2.303 * log10(100 / (100 - 99.9))) / (2.303 * log10(100 / (100 - 50))))
+mprintf("99.9 percent / 50 percent =%.1f",t)
diff --git a/3718/CH5/EX5.6/Ex5_6.sce b/3718/CH5/EX5.6/Ex5_6.sce
new file mode 100644
index 000000000..4bd645e2f
--- /dev/null
+++ b/3718/CH5/EX5.6/Ex5_6.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 6
+clc;
+
+//Declaration of Constants
+R = 1.987 //in cal per K per mol
+
+//Declaration of Variables
+T1 = 273.0 //in K
+T2 = 303.0 //in K
+K1 = 2.45 * 10 ** -5
+K2 = 162 * 10 ** -5
+
+// Solution
+Ea = log10(K2 / K1) * R * 2.303 / ((T2 - T1) / (T1 * T2))
+mprintf("The activation energy of the reaction is %d cal / mol",Ea)
diff --git a/3718/CH5/EX5.7/Ex5_7.sce b/3718/CH5/EX5.7/Ex5_7.sce
new file mode 100644
index 000000000..8c910d166
--- /dev/null
+++ b/3718/CH5/EX5.7/Ex5_7.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 7
+clc;
+
+//Declaration of Variables
+t05 = 30 //in minutes
+a = 0.1 //in M
+
+// Solution
+mprintf("For second order reaction,\n t0.5 = 1 / Ka\n")
+K = 1 / (a * t05)
+mprintf(" The rate constant is %.3f mol per lit per min",K)
diff --git a/3718/CH5/EX5.8/Ex5_8.sce b/3718/CH5/EX5.8/Ex5_8.sce
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index 000000000..bc995af68
--- /dev/null
+++ b/3718/CH5/EX5.8/Ex5_8.sce
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+//Chapter 5: Chemical Kinetics and Catalysis
+//Problem: 8
+clc;
+
+//Declaration of Variables
+T = 500 //in C
+Pi = 350 //in torr
+r1 = 1.07 //in torr / s
+r2 = 0.76 //in torr / s
+
+// Solution
+mprintf("1.07 = k(0.95a)^n\n")
+mprintf(" 0.76 = k(0.80a)^n\n")
+n = log(r1 / r2) / log(0.95 / 0.80)
+n=ceil(n)
+mprintf(" Hence, order of reaction is %d",n)