# Chapter 8 Chemical Kinetics and Catalysis¶

## Example 8_3 pgno:552¶

In [20]:
k=1.37*10**-4;#rate constant in per sec#
thalf=5057;#half time of the reaction in sec#
print'If out of each mole of N2O5,x mole of it decomposes at any instant\nThe total pressure in the system is equal to that due to (1-x)moles of undecomposed N2O5,\nx moles of N2O4 and x/2moles of O2 i.e due to 1+(x/2)moles.'
print'\nThe increase in pressure is thus due to x/2moles\nSo the amount of N2O5 that has decomposed at any instant i.e x is proportional to twice the observed increase in pressure'
print'\nThese can be fitted into the kinetic equation for a first order reaction\nk=(2.303/t)*log10(a/(a-x))\nand the value of k can be obtained.The average value of k is 1.37*10**-4per sec'
print'\nk=0.693/t0.5 that will result in t0.5=5057seconds\nThis can also be obtained as the value corresponding to (a-x)=154.1mm,from a graph of t vs (a-x).'

If out of each mole of N2O5,x mole of it decomposes at any instant
The total pressure in the system is equal to that due to (1-x)moles of undecomposed N2O5,
x moles of N2O4 and x/2moles of O2 i.e due to 1+(x/2)moles.

The increase in pressure is thus due to x/2moles
So the amount of N2O5 that has decomposed at any instant i.e x is proportional to twice the observed increase in pressure

These can be fitted into the kinetic equation for a first order reaction
k=(2.303/t)*log10(a/(a-x))
and the value of k can be obtained.The average value of k is 1.37*10**-4per sec

k=0.693/t0.5 that will result in t0.5=5057seconds
This can also be obtained as the value corresponding to (a-x)=154.1mm,from a graph of t vs (a-x).


## Example 8_4 pgno:552¶

In [21]:
Vinfinite=58.3;#volume of nitrogen evolved at infinite time#
V0=19.3;#volume of nitrogen evolved at initial time#
print'Let V0,Vt,Vinfinite be the volumes of N2 evolved at the beginning,at time t and at infinite time(no more collection of N2 is observed)respectively,'
print'\n(Vinfinite-V0)is a measure of the total amount of material that can decompose,\ni.e the initial concentration a.\n(Vinfinite-Vt) is a measure of the amount of material that remains unreacted at time t,i.e (a-x),\nbecause this volume corresponds to the amount of material that can still decompose between time t and infinite time.'
print'\nk=(2.303/t)*(log10((Vinfinte-V0)/(Vinfinite-Vt))).'
print'\nThe average value of k can be found to be 6.54*10**-2per min or 1.09*10**-3per sec'

Let V0,Vt,Vinfinite be the volumes of N2 evolved at the beginning,at time t and at infinite time(no more collection of N2 is observed)respectively,

(Vinfinite-V0)is a measure of the total amount of material that can decompose,
i.e the initial concentration a.
(Vinfinite-Vt) is a measure of the amount of material that remains unreacted at time t,i.e (a-x),
because this volume corresponds to the amount of material that can still decompose between time t and infinite time.

k=(2.303/t)*(log10((Vinfinte-V0)/(Vinfinite-Vt))).

The average value of k can be found to be 6.54*10**-2per min or 1.09*10**-3per sec


## Example 8_5 pgno:553¶

In [22]:
t=12.8;#half life of the particle in hours#
k=0.693/(t*60);#value of k for the experiment in per min#
print'Value of k for the experiment=k=per min',k
print'\n-dN/dt=rate=100*k*N=(0.693/12.8*60)*N\nN,the number of copper atoms required to produce 100Beta particles per minute\n we get N=(100*12.8*60)/0.693=1.108*10**5'
w=63.5;#atomic weight of Cu in grams#
AN=6.023*10**23;
N=1.108*10**5;
W=(N*w)/AN;#weight of Cu in grams#
print'\nWeight of Cu=W==1.17*10**-17grams',W
print'\nSince the maximum activity is 100Beta particles per minute,N=(a-x) at the end of six hours,i.e t=6 and N=1.108*19**5atoms'
print'\nAt zero time N0=a\n a-x=a*exp(-k*t)\nUpon solving the above equation we get N0=a=1.533*10**5atoms\nWeight of Cu to start with=1.66*10**-17grams.'
print'\nInitial activity=138.30000 disintegrations per minute',

Value of k for the experiment=k=per min 0.00090234375

-dN/dt=rate=100*k*N=(0.693/12.8*60)*N
N,the number of copper atoms required to produce 100Beta particles per minute
we get N=(100*12.8*60)/0.693=1.108*10**5

Weight of Cu=W==1.17*10**-17grams 1.16815540428e-17

Since the maximum activity is 100Beta particles per minute,N=(a-x) at the end of six hours,i.e t=6 and N=1.108*19**5atoms

At zero time N0=a
a-x=a*exp(-k*t)
Upon solving the above equation we get N0=a=1.533*10**5atoms

Initial activity=138.30000 disintegrations per minute


## Example 8_7 pgno:556¶

In [23]:
print'The kinetic equation for a second order reaction involving unequal concentrations of the reactants is\nk2=(2.303/(t*(a-b)))*log10((b*(a-x))/(a*(b-x)))'
k=2.312*10**-4;
p=4.94*10**-3;#value of (a-b)#
k2=(k*2.303)/p;#second order reaction rate#
print'\nSecond order reaction rate=k2=litre per mol per second',round(k2,5)

The kinetic equation for a second order reaction involving unequal concentrations of the reactants is
k2=(2.303/(t*(a-b)))*log10((b*(a-x))/(a*(b-x)))

Second order reaction rate=k2=litre per mol per second 0.10778


## Example 8_8 pgno:557¶

In [24]:
t1half=37.00;#half time for the first order reaction#
t2half=19.2;#half time for the second order reaction#
t3half=9.45;#half time for the third order reaction#
print'to know the order of the equation we can use 2^(n-1)=t1half/t2half'
print'\nby solving for first and second order n=1.95\nby solving for second and third order n=2.02'
print'\nby solving for first and third order n=1.98\nSo the order of the reaction=n=2'

to know the order of the equation we can use 2^(n-1)=t1half/t2half

by solving for first and second order n=1.95
by solving for second and third order n=2.02

by solving for first and third order n=1.98
So the order of the reaction=n=2


## Example 8_10 pgno:561¶

In [25]:
from math import log10
k=8.676*10**-3;#average value of k in per min#
print'Average value of k=8.676*10**-3per min'
r0=22.4;
rt=0;
rinfinite=-11.1;
t=(2.303/k)*log10((r0-rinfinite)/(rt-rinfinite));
print'\nThe time at which the mixture is optically inactive=t=min',round(t,1)#here in textbook the answer is given wrong,but by solving we get the same result as executed#

Average value of k=8.676*10**-3per min

The time at which the mixture is optically inactive=t=min 127.3


## Example 8_13 pgno:565¶

In [26]:
from math import log10
k=1.;
a=10.;
thalf=10**-1.88;#half time of the reaction#
n=1-(log10(thalf/k)/log10(a));#order of the reaction#
print'order of the reaction after solving is n=',n
print'\nHence the order of the reaction',3

order of the reaction after solving is n= 2.88

Hence the order of the reaction 3


## Example 8_14 pgno:567¶

In [27]:
thalf1=49;#half life for 0.02M compound#
thalf2=100;#half life for 0.04M compound#
M1=0.02;#initial concentration of the compound#
M2=0.01;#Final concentration of the compound#
print'Since thalf is directly proportional to concentration(doubling the concentration increases thalf by a factor 2),\nthe reaction is Zero order.\nThe half time corresponding to a concentration of 0.01M will be 24.5mins'

Since thalf is directly proportional to concentration(doubling the concentration increases thalf by a factor 2),
the reaction is Zero order.
The half time corresponding to a concentration of 0.01M will be 24.5mins


## Example 8_18 pgno:581¶

In [28]:
from math import log10
R=1.987;#universal gas constant#
T1=293.;#initial temperature in kelvin#
T2=303.;#Final temperature in kelvin#
K1=6.68*10**-3;#rate constant corresponding to T1 in per min#
K2=1.31*10**-2;#rate constant corresponding to T2 in per min#
E=(2.303*R*T1*T2*log10(K2/K1))/(T2-T1);#energy of activation in Kcal per mol#
print'Energy of activation=E=cal per mol=11.88Kcal per mol',round(E)

Energy of activation=E=cal per mol=11.88Kcal per mol 11883.0


## Example 8_19 pgno:582¶

In [29]:
R=1.987;#universal gas constant#
T1=350.;#initial temperature in kelvin#
T2=360.;#Final temperature in kelvin#
E=40000.;#energy of activation in cal per mol#
K=10**(E*((T2-T1)/(T1*T2))/(2.303*R));#ratio of v2/v1#
print'The ratio of V2 and V1 is K=V2/V1=',round(K,3)

The ratio of V2 and V1 is K=V2/V1= 4.94


## Example 8_20 pgno:582¶

In [30]:
from math import log10
R=1.987;#universal gas constant#
T1=313.;#initial temperature in kelvin#
T2=333.;#Final temperature in kelvin#
t1=15.;#time for 20% reaction at 313K in mins#
t2=3.;#time for 20%reaction at 333K in mins#
K=t1/t2;#ratio of K2/K1#
E=(2.303*R*T1*T2*log10(K))/(1000*(T2-T1));#energy of activation in Kcal per mol#
print'Energy of activation=E=Kcal per mol',round(E,2)

Energy of activation=E=Kcal per mol 16.67


## Example 8_21 pgno:583¶

In [31]:
R=1.987;#universal gas constant#
print'From the graph slope=(-0.92/0.30)=(-E/(2.303*R))\nGraphical evaluation of A requires the determination of the intercept on the y axis corresponding to 1/T=0\nOne can also calculate A from k=A*exp(-E/(R*T))'
E=(0.92*R*2.303)/(0.30*10**3);
print'\nEnergy of activation=E==14.04Kcal per mol',E
k=2.31*10**-2;
T=273;#temperature in kelvin#
print'\nwe can find the value of A using log10(k)=log10(A)-(E/(2.303*R*T))\nUpon solving we get A=4.015*10**9litre per mol per second'

From the graph slope=(-0.92/0.30)=(-E/(2.303*R))
Graphical evaluation of A requires the determination of the intercept on the y axis corresponding to 1/T=0
One can also calculate A from k=A*exp(-E/(R*T))

Energy of activation=E==14.04Kcal per mol 0.0140332537333

we can find the value of A using log10(k)=log10(A)-(E/(2.303*R*T))
Upon solving we get A=4.015*10**9litre per mol per second


## Example 8_22 pgno:594¶

In [32]:
from math import exp
R=1.987;#universal gas constant#
T=556.;#temperature in kelvin#
E=44.;#Energy in Kcal#
dS=-10;#entropy change in cal per deg#
k=(exp(2)*exp(dS/R)*exp(-E/(R*T))*10**13.07);#rate constant for the reaction#
print'Rate constant for the reaction=litre per mol per sec',k

Rate constant for the reaction=litre per mol per sec 5.44017531908e+11


## Example 8_23 pgno:595¶

In [33]:
from math import log10,exp
R=1.987;#universal gas constant#
T=473;#temperature in kelvin#
A=2.75*10**15;#frequency factor in per sec#
K=1.38*10**-16;#boltzmans constant#
h=6.625*10**-27;#planks constant#
dn=0;
dS=4.57*(log10(A)-log10(exp(1))-log10(9.85*10**12));#entropy change in cal per deg#
print'The entropy of activation=dS==9.19eu',dS
print'\nSince A is independent of concentration units dS does not sepend on the concentration units used\nand hence the standard state.\nHowever if the time were expressed in different units A will assume a different value\nand consequently the value of dS will be different\nIf time were expressed in minutes A=2.75*10**15*60 per min\ndS=9.19+4.57*log10(60)=17.32eu\nfor bimolecular reaction e**2=7.4*10**10 \nso dS will result in dS=-10.1eu or mol per litre.'
print'\nIf the concentration were expressed in mol per millilitre A would be 7.4*10**13 \nso dS will result in dS=-10.1+13.6=3.5eu or mol per millilitre\nIf the concentration were expressed in molecules per millilitre the value of A will be multiplied by 6.023*10**23 \nso dS would result in as dS=-10.1-94.9=-105eu or -105molecules per millilitre'

The entropy of activation=dS==9.19eu 9.19302105513

Since A is independent of concentration units dS does not sepend on the concentration units used
and hence the standard state.
However if the time were expressed in different units A will assume a different value
and consequently the value of dS will be different
If time were expressed in minutes A=2.75*10**15*60 per min
dS=9.19+4.57*log10(60)=17.32eu
for bimolecular reaction e**2=7.4*10**10
so dS will result in dS=-10.1eu or mol per litre.

If the concentration were expressed in mol per millilitre A would be 7.4*10**13
so dS will result in dS=-10.1+13.6=3.5eu or mol per millilitre
If the concentration were expressed in molecules per millilitre the value of A will be multiplied by 6.023*10**23
so dS would result in as dS=-10.1-94.9=-105eu or -105molecules per millilitre