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1 | 3305-3308 | 3 6
Time required to decompose SO2Cl2 to half of its initial amount is 60
minutes If the decomposition is a first order reaction, calculate the rate
constant of the reaction Rationalised 2023-24
79
Chemical Kinetics
k = A e -Ea /RT
(3 |
1 | 3306-3309 | 6
Time required to decompose SO2Cl2 to half of its initial amount is 60
minutes If the decomposition is a first order reaction, calculate the rate
constant of the reaction Rationalised 2023-24
79
Chemical Kinetics
k = A e -Ea /RT
(3 18)
where A is the Arrhenius factor or the frequency factor |
1 | 3307-3310 | If the decomposition is a first order reaction, calculate the rate
constant of the reaction Rationalised 2023-24
79
Chemical Kinetics
k = A e -Ea /RT
(3 18)
where A is the Arrhenius factor or the frequency factor It is also called
pre-exponential factor |
1 | 3308-3311 | Rationalised 2023-24
79
Chemical Kinetics
k = A e -Ea /RT
(3 18)
where A is the Arrhenius factor or the frequency factor It is also called
pre-exponential factor It is a constant specific to a particular reaction |
1 | 3309-3312 | 18)
where A is the Arrhenius factor or the frequency factor It is also called
pre-exponential factor It is a constant specific to a particular reaction R is gas constant and Ea is activation energy measured in joules/mole
(J mol –1) |
1 | 3310-3313 | It is also called
pre-exponential factor It is a constant specific to a particular reaction R is gas constant and Ea is activation energy measured in joules/mole
(J mol –1) It can be understood clearly using the following simple reaction
2
2
H
g
I
g
2HI g
According to Arrhenius, this reaction can take place
only when a molecule of hydrogen and a molecule of iodine
collide to form an unstable intermediate (Fig |
1 | 3311-3314 | It is a constant specific to a particular reaction R is gas constant and Ea is activation energy measured in joules/mole
(J mol –1) It can be understood clearly using the following simple reaction
2
2
H
g
I
g
2HI g
According to Arrhenius, this reaction can take place
only when a molecule of hydrogen and a molecule of iodine
collide to form an unstable intermediate (Fig 3 |
1 | 3312-3315 | R is gas constant and Ea is activation energy measured in joules/mole
(J mol –1) It can be understood clearly using the following simple reaction
2
2
H
g
I
g
2HI g
According to Arrhenius, this reaction can take place
only when a molecule of hydrogen and a molecule of iodine
collide to form an unstable intermediate (Fig 3 6) |
1 | 3313-3316 | It can be understood clearly using the following simple reaction
2
2
H
g
I
g
2HI g
According to Arrhenius, this reaction can take place
only when a molecule of hydrogen and a molecule of iodine
collide to form an unstable intermediate (Fig 3 6) It exists
for a very short time and then breaks up to form two
molecules of hydrogen iodide |
1 | 3314-3317 | 3 6) It exists
for a very short time and then breaks up to form two
molecules of hydrogen iodide Fig |
1 | 3315-3318 | 6) It exists
for a very short time and then breaks up to form two
molecules of hydrogen iodide Fig 3 |
1 | 3316-3319 | It exists
for a very short time and then breaks up to form two
molecules of hydrogen iodide Fig 3 6: Formation of HI through
the intermediate
Intermediate
Fig |
1 | 3317-3320 | Fig 3 6: Formation of HI through
the intermediate
Intermediate
Fig 3 |
1 | 3318-3321 | 3 6: Formation of HI through
the intermediate
Intermediate
Fig 3 7: Diagram showing plot of potential
energy vs reaction coordinate
Fig |
1 | 3319-3322 | 6: Formation of HI through
the intermediate
Intermediate
Fig 3 7: Diagram showing plot of potential
energy vs reaction coordinate
Fig 3 |
1 | 3320-3323 | 3 7: Diagram showing plot of potential
energy vs reaction coordinate
Fig 3 8: Distribution curve showing energies
among gaseous molecules
The energy required to form this
intermediate, called activated complex
(C), is known as activation energy (Ea) |
1 | 3321-3324 | 7: Diagram showing plot of potential
energy vs reaction coordinate
Fig 3 8: Distribution curve showing energies
among gaseous molecules
The energy required to form this
intermediate, called activated complex
(C), is known as activation energy (Ea) Fig |
1 | 3322-3325 | 3 8: Distribution curve showing energies
among gaseous molecules
The energy required to form this
intermediate, called activated complex
(C), is known as activation energy (Ea) Fig 3 |
1 | 3323-3326 | 8: Distribution curve showing energies
among gaseous molecules
The energy required to form this
intermediate, called activated complex
(C), is known as activation energy (Ea) Fig 3 7 is obtained by plotting potential
energy vs reaction coordinate |
1 | 3324-3327 | Fig 3 7 is obtained by plotting potential
energy vs reaction coordinate Reaction
coordinate represents the profile of energy
change when reactants change into
products |
1 | 3325-3328 | 3 7 is obtained by plotting potential
energy vs reaction coordinate Reaction
coordinate represents the profile of energy
change when reactants change into
products Some energy is released when the
complex decomposes to form products |
1 | 3326-3329 | 7 is obtained by plotting potential
energy vs reaction coordinate Reaction
coordinate represents the profile of energy
change when reactants change into
products Some energy is released when the
complex decomposes to form products So, the final enthalpy of the reaction
depends upon the nature of reactants
and products |
1 | 3327-3330 | Reaction
coordinate represents the profile of energy
change when reactants change into
products Some energy is released when the
complex decomposes to form products So, the final enthalpy of the reaction
depends upon the nature of reactants
and products All the molecules in the reacting
species do not have the same kinetic
energy |
1 | 3328-3331 | Some energy is released when the
complex decomposes to form products So, the final enthalpy of the reaction
depends upon the nature of reactants
and products All the molecules in the reacting
species do not have the same kinetic
energy Since it is difficult to predict the
behaviour of any one molecule with
precision, Ludwig Boltzmann and James
Clark Maxwell used statistics to predict
the behaviour of large number of
molecules |
1 | 3329-3332 | So, the final enthalpy of the reaction
depends upon the nature of reactants
and products All the molecules in the reacting
species do not have the same kinetic
energy Since it is difficult to predict the
behaviour of any one molecule with
precision, Ludwig Boltzmann and James
Clark Maxwell used statistics to predict
the behaviour of large number of
molecules According to them, the
distribution of kinetic energy may be
described by plotting the fraction of
molecules (NE/NT) with a given kinetic
energy (E) vs kinetic energy (Fig |
1 | 3330-3333 | All the molecules in the reacting
species do not have the same kinetic
energy Since it is difficult to predict the
behaviour of any one molecule with
precision, Ludwig Boltzmann and James
Clark Maxwell used statistics to predict
the behaviour of large number of
molecules According to them, the
distribution of kinetic energy may be
described by plotting the fraction of
molecules (NE/NT) with a given kinetic
energy (E) vs kinetic energy (Fig 3 |
1 | 3331-3334 | Since it is difficult to predict the
behaviour of any one molecule with
precision, Ludwig Boltzmann and James
Clark Maxwell used statistics to predict
the behaviour of large number of
molecules According to them, the
distribution of kinetic energy may be
described by plotting the fraction of
molecules (NE/NT) with a given kinetic
energy (E) vs kinetic energy (Fig 3 8) |
1 | 3332-3335 | According to them, the
distribution of kinetic energy may be
described by plotting the fraction of
molecules (NE/NT) with a given kinetic
energy (E) vs kinetic energy (Fig 3 8) Here, NE is the number of molecules with
energy E and NT is total number
of molecules |
1 | 3333-3336 | 3 8) Here, NE is the number of molecules with
energy E and NT is total number
of molecules The peak of the curve corresponds to
the most probable kinetic energy, i |
1 | 3334-3337 | 8) Here, NE is the number of molecules with
energy E and NT is total number
of molecules The peak of the curve corresponds to
the most probable kinetic energy, i e |
1 | 3335-3338 | Here, NE is the number of molecules with
energy E and NT is total number
of molecules The peak of the curve corresponds to
the most probable kinetic energy, i e ,
kinetic energy of maximum fraction of
molecules |
1 | 3336-3339 | The peak of the curve corresponds to
the most probable kinetic energy, i e ,
kinetic energy of maximum fraction of
molecules There are decreasing number
of molecules with energies higher or
lower than this value |
1 | 3337-3340 | e ,
kinetic energy of maximum fraction of
molecules There are decreasing number
of molecules with energies higher or
lower than this value When the
Rationalised 2023-24
80
Chemistry
Fig |
1 | 3338-3341 | ,
kinetic energy of maximum fraction of
molecules There are decreasing number
of molecules with energies higher or
lower than this value When the
Rationalised 2023-24
80
Chemistry
Fig 3 |
1 | 3339-3342 | There are decreasing number
of molecules with energies higher or
lower than this value When the
Rationalised 2023-24
80
Chemistry
Fig 3 10: A plot between ln k and 1/T
In Fig |
1 | 3340-3343 | When the
Rationalised 2023-24
80
Chemistry
Fig 3 10: A plot between ln k and 1/T
In Fig 3 |
1 | 3341-3344 | 3 10: A plot between ln k and 1/T
In Fig 3 10, slope = –
Ea
R
and intercept = ln
A |
1 | 3342-3345 | 10: A plot between ln k and 1/T
In Fig 3 10, slope = –
Ea
R
and intercept = ln
A So we can calculate Ea and A using these values |
1 | 3343-3346 | 3 10, slope = –
Ea
R
and intercept = ln
A So we can calculate Ea and A using these values At temperature T1, equation (3 |
1 | 3344-3347 | 10, slope = –
Ea
R
and intercept = ln
A So we can calculate Ea and A using these values At temperature T1, equation (3 19) is
ln k1 = –
a
1
E
RT
+ ln A
(3 |
1 | 3345-3348 | So we can calculate Ea and A using these values At temperature T1, equation (3 19) is
ln k1 = –
a
1
E
RT
+ ln A
(3 20)
At temperature T2, equation (3 |
1 | 3346-3349 | At temperature T1, equation (3 19) is
ln k1 = –
a
1
E
RT
+ ln A
(3 20)
At temperature T2, equation (3 19) is
ln k2 = –
a
2
E
RT
+ ln A
(3 |
1 | 3347-3350 | 19) is
ln k1 = –
a
1
E
RT
+ ln A
(3 20)
At temperature T2, equation (3 19) is
ln k2 = –
a
2
E
RT
+ ln A
(3 21)
(since A is constant for a given reaction)
k1 and k2 are the values of rate constants at
temperatures T1 and T2 respectively |
1 | 3348-3351 | 20)
At temperature T2, equation (3 19) is
ln k2 = –
a
2
E
RT
+ ln A
(3 21)
(since A is constant for a given reaction)
k1 and k2 are the values of rate constants at
temperatures T1 and T2 respectively Fig |
1 | 3349-3352 | 19) is
ln k2 = –
a
2
E
RT
+ ln A
(3 21)
(since A is constant for a given reaction)
k1 and k2 are the values of rate constants at
temperatures T1 and T2 respectively Fig 3 |
1 | 3350-3353 | 21)
(since A is constant for a given reaction)
k1 and k2 are the values of rate constants at
temperatures T1 and T2 respectively Fig 3 9: Distribution curve showing temperature
dependence of rate of a reaction
temperature is raised, the maximum
of the curve moves to the higher
energy value (Fig |
1 | 3351-3354 | Fig 3 9: Distribution curve showing temperature
dependence of rate of a reaction
temperature is raised, the maximum
of the curve moves to the higher
energy value (Fig 3 |
1 | 3352-3355 | 3 9: Distribution curve showing temperature
dependence of rate of a reaction
temperature is raised, the maximum
of the curve moves to the higher
energy value (Fig 3 9) and the curve
broadens out, i |
1 | 3353-3356 | 9: Distribution curve showing temperature
dependence of rate of a reaction
temperature is raised, the maximum
of the curve moves to the higher
energy value (Fig 3 9) and the curve
broadens out, i e |
1 | 3354-3357 | 3 9) and the curve
broadens out, i e , spreads to the right
such that there is a greater proportion
of molecules with much higher
energies |
1 | 3355-3358 | 9) and the curve
broadens out, i e , spreads to the right
such that there is a greater proportion
of molecules with much higher
energies The area under the curve
must
be
constant
since
total
probability must be one at all times |
1 | 3356-3359 | e , spreads to the right
such that there is a greater proportion
of molecules with much higher
energies The area under the curve
must
be
constant
since
total
probability must be one at all times We can mark the position of Ea on
Maxwell Boltzmann distribution curve
(Fig |
1 | 3357-3360 | , spreads to the right
such that there is a greater proportion
of molecules with much higher
energies The area under the curve
must
be
constant
since
total
probability must be one at all times We can mark the position of Ea on
Maxwell Boltzmann distribution curve
(Fig 3 |
1 | 3358-3361 | The area under the curve
must
be
constant
since
total
probability must be one at all times We can mark the position of Ea on
Maxwell Boltzmann distribution curve
(Fig 3 9) |
1 | 3359-3362 | We can mark the position of Ea on
Maxwell Boltzmann distribution curve
(Fig 3 9) Increasing the temperature of the substance increases the fraction
of molecules, which collide with energies greater than Ea |
1 | 3360-3363 | 3 9) Increasing the temperature of the substance increases the fraction
of molecules, which collide with energies greater than Ea It is clear
from the diagram that in the curve at (t + 10), the area showing the
fraction of molecules having energy equal to or greater than activation
energy gets doubled leading to doubling the rate of a reaction |
1 | 3361-3364 | 9) Increasing the temperature of the substance increases the fraction
of molecules, which collide with energies greater than Ea It is clear
from the diagram that in the curve at (t + 10), the area showing the
fraction of molecules having energy equal to or greater than activation
energy gets doubled leading to doubling the rate of a reaction In the Arrhenius equation (3 |
1 | 3362-3365 | Increasing the temperature of the substance increases the fraction
of molecules, which collide with energies greater than Ea It is clear
from the diagram that in the curve at (t + 10), the area showing the
fraction of molecules having energy equal to or greater than activation
energy gets doubled leading to doubling the rate of a reaction In the Arrhenius equation (3 18) the factor e -Ea /RT
corresponds to
the fraction of molecules that have kinetic energy greater than Ea |
1 | 3363-3366 | It is clear
from the diagram that in the curve at (t + 10), the area showing the
fraction of molecules having energy equal to or greater than activation
energy gets doubled leading to doubling the rate of a reaction In the Arrhenius equation (3 18) the factor e -Ea /RT
corresponds to
the fraction of molecules that have kinetic energy greater than Ea Taking natural logarithm of both sides of equation (3 |
1 | 3364-3367 | In the Arrhenius equation (3 18) the factor e -Ea /RT
corresponds to
the fraction of molecules that have kinetic energy greater than Ea Taking natural logarithm of both sides of equation (3 18)
ln k = –
Ea
RT + ln A
(3 |
1 | 3365-3368 | 18) the factor e -Ea /RT
corresponds to
the fraction of molecules that have kinetic energy greater than Ea Taking natural logarithm of both sides of equation (3 18)
ln k = –
Ea
RT + ln A
(3 19)
The plot of ln k vs 1/T gives a straight line according to the equation
(3 |
1 | 3366-3369 | Taking natural logarithm of both sides of equation (3 18)
ln k = –
Ea
RT + ln A
(3 19)
The plot of ln k vs 1/T gives a straight line according to the equation
(3 19) as shown in Fig |
1 | 3367-3370 | 18)
ln k = –
Ea
RT + ln A
(3 19)
The plot of ln k vs 1/T gives a straight line according to the equation
(3 19) as shown in Fig 3 |
1 | 3368-3371 | 19)
The plot of ln k vs 1/T gives a straight line according to the equation
(3 19) as shown in Fig 3 10 |
1 | 3369-3372 | 19) as shown in Fig 3 10 Thus, it has been found from Arrhenius equation (3 |
1 | 3370-3373 | 3 10 Thus, it has been found from Arrhenius equation (3 18) that
increasing the temperature or decreasing the activation energy will
result in an increase in the rate of the reaction and an exponential
increase in the rate constant |
1 | 3371-3374 | 10 Thus, it has been found from Arrhenius equation (3 18) that
increasing the temperature or decreasing the activation energy will
result in an increase in the rate of the reaction and an exponential
increase in the rate constant Rationalised 2023-24
81
Chemical Kinetics
The rate constants of a reaction at 500K and 700K are 0 |
1 | 3372-3375 | Thus, it has been found from Arrhenius equation (3 18) that
increasing the temperature or decreasing the activation energy will
result in an increase in the rate of the reaction and an exponential
increase in the rate constant Rationalised 2023-24
81
Chemical Kinetics
The rate constants of a reaction at 500K and 700K are 0 02s–1 and
0 |
1 | 3373-3376 | 18) that
increasing the temperature or decreasing the activation energy will
result in an increase in the rate of the reaction and an exponential
increase in the rate constant Rationalised 2023-24
81
Chemical Kinetics
The rate constants of a reaction at 500K and 700K are 0 02s–1 and
0 07s–1 respectively |
1 | 3374-3377 | Rationalised 2023-24
81
Chemical Kinetics
The rate constants of a reaction at 500K and 700K are 0 02s–1 and
0 07s–1 respectively Calculate the values of Ea and A |
1 | 3375-3378 | 02s–1 and
0 07s–1 respectively Calculate the values of Ea and A 2
1
log k
k
=
2
1
a
1
2
2 |
1 | 3376-3379 | 07s–1 respectively Calculate the values of Ea and A 2
1
log k
k
=
2
1
a
1
2
2 303
T
T
E
T T
R
0 |
1 | 3377-3380 | Calculate the values of Ea and A 2
1
log k
k
=
2
1
a
1
2
2 303
T
T
E
T T
R
0 07
log
0 |
1 | 3378-3381 | 2
1
log k
k
=
2
1
a
1
2
2 303
T
T
E
T T
R
0 07
log
0 02
=
a
1
1
700
500
2 |
1 | 3379-3382 | 303
T
T
E
T T
R
0 07
log
0 02
=
a
1
1
700
500
2 303
8 |
1 | 3380-3383 | 07
log
0 02
=
a
1
1
700
500
2 303
8 314 J
mol
700
500
E
K
0 |
1 | 3381-3384 | 02
=
a
1
1
700
500
2 303
8 314 J
mol
700
500
E
K
0 544 = Ea × 5 |
1 | 3382-3385 | 303
8 314 J
mol
700
500
E
K
0 544 = Ea × 5 714 × 10-4/19 |
1 | 3383-3386 | 314 J
mol
700
500
E
K
0 544 = Ea × 5 714 × 10-4/19 15
Ea = 0 |
1 | 3384-3387 | 544 = Ea × 5 714 × 10-4/19 15
Ea = 0 544 × 19 |
1 | 3385-3388 | 714 × 10-4/19 15
Ea = 0 544 × 19 15/5 |
1 | 3386-3389 | 15
Ea = 0 544 × 19 15/5 714 × 10–4 = 18230 |
1 | 3387-3390 | 544 × 19 15/5 714 × 10–4 = 18230 8 J
Since
k = Ae-Ea/RT
0 |
1 | 3388-3391 | 15/5 714 × 10–4 = 18230 8 J
Since
k = Ae-Ea/RT
0 02 = Ae-18230 |
1 | 3389-3392 | 714 × 10–4 = 18230 8 J
Since
k = Ae-Ea/RT
0 02 = Ae-18230 8/8 |
1 | 3390-3393 | 8 J
Since
k = Ae-Ea/RT
0 02 = Ae-18230 8/8 314 × 500
A = 0 |
1 | 3391-3394 | 02 = Ae-18230 8/8 314 × 500
A = 0 02/0 |
1 | 3392-3395 | 8/8 314 × 500
A = 0 02/0 012 = 1 |
1 | 3393-3396 | 314 × 500
A = 0 02/0 012 = 1 61
The first order rate constant for the decomposition of ethyl iodide
by the reaction
C2H5I(g) ® C2H4 (g) + HI(g)
at 600K is 1 |
1 | 3394-3397 | 02/0 012 = 1 61
The first order rate constant for the decomposition of ethyl iodide
by the reaction
C2H5I(g) ® C2H4 (g) + HI(g)
at 600K is 1 60 × 10–5 s–1 |
1 | 3395-3398 | 012 = 1 61
The first order rate constant for the decomposition of ethyl iodide
by the reaction
C2H5I(g) ® C2H4 (g) + HI(g)
at 600K is 1 60 × 10–5 s–1 Its energy of activation is 209 kJ/mol |
1 | 3396-3399 | 61
The first order rate constant for the decomposition of ethyl iodide
by the reaction
C2H5I(g) ® C2H4 (g) + HI(g)
at 600K is 1 60 × 10–5 s–1 Its energy of activation is 209 kJ/mol Calculate the rate constant of the reaction at 700K |
1 | 3397-3400 | 60 × 10–5 s–1 Its energy of activation is 209 kJ/mol Calculate the rate constant of the reaction at 700K We know that
log k2 – log k1 =
a
1
2
1
1
2 |
1 | 3398-3401 | Its energy of activation is 209 kJ/mol Calculate the rate constant of the reaction at 700K We know that
log k2 – log k1 =
a
1
2
1
1
2 303
E
T
T
R
Subtracting equation (3 |
1 | 3399-3402 | Calculate the rate constant of the reaction at 700K We know that
log k2 – log k1 =
a
1
2
1
1
2 303
E
T
T
R
Subtracting equation (3 20) from (3 |
1 | 3400-3403 | We know that
log k2 – log k1 =
a
1
2
1
1
2 303
E
T
T
R
Subtracting equation (3 20) from (3 21), we obtain
ln k2 – ln k1 =
a
1
E
RT –
a
2
E
RT
ln k
k
E
R
T
T
2
1
1
2
1
1
=
−
a
log |
1 | 3401-3404 | 303
E
T
T
R
Subtracting equation (3 20) from (3 21), we obtain
ln k2 – ln k1 =
a
1
E
RT –
a
2
E
RT
ln k
k
E
R
T
T
2
1
1
2
1
1
=
−
a
log kk
E
R T
T
2
1
1
2
2 303
1
1
=
−
a
(3 |
1 | 3402-3405 | 20) from (3 21), we obtain
ln k2 – ln k1 =
a
1
E
RT –
a
2
E
RT
ln k
k
E
R
T
T
2
1
1
2
1
1
=
−
a
log kk
E
R T
T
2
1
1
2
2 303
1
1
=
−
a
(3 22)
log |
1 | 3403-3406 | 21), we obtain
ln k2 – ln k1 =
a
1
E
RT –
a
2
E
RT
ln k
k
E
R
T
T
2
1
1
2
1
1
=
−
a
log kk
E
R T
T
2
1
1
2
2 303
1
1
=
−
a
(3 22)
log kk
E
T
T
T T
2
1
2
1
1
2
=2 303
−
a
R
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Example 3 |
1 | 3404-3407 | kk
E
R T
T
2
1
1
2
2 303
1
1
=
−
a
(3 22)
log kk
E
T
T
T T
2
1
2
1
1
2
=2 303
−
a
R
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Example 3 10
Example 3 |
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