Class 12 - Chemistry - Chemical Kinetics
Question 4.1.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3NO (g) → N2O (g) Rate = k [NO] 2
(ii) H2O2 (aq) + 3I– (aq) + 2H+ → 2H2O (l) + I3− Rate = k [H2O2] [I-]
(iii) CH3CHO (g) → CH4 (g) + CO (g) Rate = k [CH3CHO] (3/2)
(iv) C2H5Cl (g) → C2H4 (g) + HCl (g) Rate = k [C2H5Cl]
Answer:
(i) Given rate = k [NO] 2
Therefore, order of the reaction = 2
K= (Rate)/ [NO] 2
Dimension of = (mol L-1 s-1)/ (mol L-1)2
= (mol L-1 s-1)/ (mol2 L-2)
= L mol-1 s-1
(ii) Given rate = k [H2O2] [I−]
Therefore, order of the reaction = 2
k= (Rate)/ [H2O2] [I−]
Dimension of = (mol L-1 s-1)/ ((mol L-1) (mol L-1))
= L mol-1 s-1
(iii) Given rate = k [CH3CHO] (3/2)
Therefore, order of reaction = (3/2)
k= (Rate)/ [CH3CHO] (3/2)
Dimension of = (mol L-1 s-1)/ (mol L-1) (3/2)
= (mol L-1 s-1)/ (mol (3/2) L (3/2))
=L (1/2) mol (1/2) s-1
(iv) Given rate = k [C2H5Cl] Therefore, order of the reaction = 1
k= (Rate)/ [C2H5Cl]
Dimension of = (mol L-1 s-1)/ (mol L-1)
= s-1
Question 4.2.
For the reaction:
2A + B → A2B
The rate = k [A] [B] 2 with k = 2.0 × 10–6 mol–2 L2 s–1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1.
Calculate the rate reaction after [A] is reduced to 0.06 mol L–1
Answer:
The initial rate of the reaction is
Rate = k [A] [B] 2
= (2.0 × 10−6 mol−2 L2 s−1) (0.1 mol L−1) (0.2 mol L−1)2
= 8.0 × 10−9 mol−2 L2 s−1
When [A] is reduced from 0.1 mol L−1 to 0.06 mol−1, the concentration of A reacted = (0.1 − 0.06) mol L−1 = 0.04 mol L−1
Therefore, concentration of B reacted = (1/2) x (0.04) mol L-1 =0.02 mol L−1
Then, concentration of B available, [B] = (0.2 − 0.02) mol L−1
= 0.18 mol L−1
After [A] is reduced to 0.06 mol L−1 the rate of the reaction is given by,
Rate = k [A] [B] 2
= (2.0 × 10−6 mol−2 L2 s−1) (0.06 mol L−1) (0.18 mol L−1)2
= 3.89 mol L−1 s−1
Question 4.3.
The decomposition of NH3 on platinum surface is zero order reaction.
What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s–1?
Answer:
2NH3 (g) → N2 (g) + 3H2 (g)
Rate of zero order reaction is equal to rate constant
I.e. Rate = 2.5×10-4molL-1sec-1.
According to rate law, - d [NH3]/2dt = d [N2]/ (dt)
2.5 × 10-4mol L-1sec-1 = d [N2]/ (dt)
i.e. the rate of production of N2 is 2.5 × 10-4mol L-1 sec-1.
According to rate law,
- d [NH3]/2dt = d [H2]/ (3dt)
d [H2]/(dt) =(-3) x d[NH3]/[2dt]
i.e. rate of formation of H2 is 3 times rate of reaction
= 3 × 2.5 × 10-4mol L-1sec-1
= 7.5 × 10-4mol L-1sec-1
Rate of formation of N2 and H2 is 2.5 × 10-4 mol L-1sec-1 and
7.5 × 10-4 mol L-1sec-1 respectively
Question 4.4.
The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by
Rate = k [CH3OCH3] (3/2)
The rate of reaction is followed by increase in pressure in a closed vessel,
so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
Rate = k (pCH3OCH3) (3/2)
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Answer:
If pressure is measured in bar and time in minutes, then
Unit of rate = bar min−1
Rate = k (pCH3OCH3) (3/2)
Therefore, unit of rate constants (k) = (bar min-1)/ (bar) (3/2)
=bar-(1/2) min-1
Question 4.5.
Mention the factors that affect the rate of a chemical reaction.
Answer:
The factors that affect the rate of a reaction are as follows.
(i) Concentration of reactants (pressure in case of gases)
(ii) Temperature
(iii) Presence of a catalyst
Question 4.6.
A reaction is second order with respect to a reactant.
How is the rate of reaction affected if the concentration of the reactant is?
(i) Doubled (ii) reduced to half?
Answer:
Let the concentration of the reactant be [A] = a
Rate of reaction, R = k [A] 2
= ka2
(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R’ = k (2a) 2
= 4ka2
= 4 R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e., [A] = (1/2) a then the rate of the reaction would be
R’’ =k (1/2) (a) 2
= (1/4) ka
= (1/4) R
Therefore, the rate of the reaction would be reduced to
Question 4.7.
What is the effect of temperature on the rate constant of a reaction?
How can this temperature effect on rate constant be represented quantitatively?
Answer:
The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.
The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,
K= A e-(Ea/RT)
Where, k is the rate constant,
A is the Arrhenius factor or the frequency factor,
R is the gas constant,
T is the temperature, and
Ea is the energy of activation for the reaction
Question 4.8.
In a pseudo first order hydrolysis of ester in water, the following results were obtained:
|
(t/s) |
0 |
30 |
60 |
90 |
|
[Ester]/mol L-1 |
0.55 |
0.31 |
0.17 |
0.085 |
(i) Calculate the average rate of reaction between the time intervals 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer:
= [change in concentration]/ [time taken] i.e.
= [0.31 -0.17]/ [60-30]
= (0.14)/ (30)
=0.00467 mol L-1 sec-1
= 4.67 x 10-3 mol L-1 s-1
K = (2.303/t) log (Ci/Ct)
Where K = Rate constant
t =s time taken
Ci = initial concentration
Ct = Concentration at time t.
K1 = (2.303/30) log (0.55/0.31) [at t=30 sec]
= 1.911 x 10-2 s-1
K2 = (2.303/60) log (0.55/0.17) [at t=60 sec]
= 1.957 x 10-2 s-1
K3 = (2.303/90) log (0.55/0.31) [at t=90 sec]
=2.075 x 10-2 s-1
Then, average rate constant k = (k1 +k2 +k3)/ (3) = 1.911 x 10-2 s-1
= (1.911 x 10-2) + (1.957 x 10-2) + (2.075 x 10-2)
⇒ K = 1.98 × 10-2 sec-1
Question 4.9.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
Answer:
(i) The differential rate equation will be
- (d[R])/ (dt) =k ([A] 1[B] 2
(i) If conc. of B is increased to 3 times then:
- (d[R])/ (dt) =k ([A] ([3B]) 2
- (d[R])/ (dt) =k ([A] 9[B] 2
=9 x k ([A] 1[B] 2
Therefore rate increases 9 times
(iii) When conc. of both A and B are doubled
- (d[R])/ (dt) =k [A] [B] 2
=k [2A] [2B] 2
=8 k [A] [B] 2
Therefore, the rate of reaction increases by 8 times.
Question 4.10.
In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
|
A/mol L-1 |
0.20 |
0.20 |
0.40 |
|
B/mol L-1 |
0.30 |
0.10 |
0.05 |
|
r0/mol L-1 |
5.07 x 10-5 |
5.07 x 10-5 |
1.43 x 10-4 |
What is the order of the reaction with respect to A and B?
Answer:
Let the order of the reaction with respect to A be x and with respect to B be y. Therefore,
r0 = k [A]x [B]y
5.07×10−5 = k [0.20] x [0.30] y (1)
5.07×10−5 = k [0.20] x [0.10] y (2)
1.43×10−4 = k [0.40] x [0.05] y (3)
Dividing (i) and (ii)
(5.07×10−5) / (5.07×10−5)
= ([0.20] x [0.30] y)/ ([0.20] x [0.10] y)
Or, 1= ([0.30]y/[0.10]y)
Or, 1y = 3y
Using formula, 1=x0
30=3y
Or, y=0
Dividing (ii) and (iii)
(1.43×10−4) / (5.07×10−5) = ([0.40] x [0.05] y)/ ([0.20] x [0.10] y)
(1.43×10−4) / (5.07×10−5) = ([0.40] x)/ ([0.20] x) (since y=0, [0.05] x= [0.30] y=1)
Or, 2.821= (2) x
Taking log both sides, log (2.821) =log (2) x
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.
Question 4.11.
The following results have been obtained during the kinetic studies of the reaction:
2A + B → C + D
|
Experiment |
[A]/mol L-1 |
[B]/mol L-1 |
Initial rate of formation of [D]/mol L-1 min-1 |
|
I |
0.1 |
0.1 |
6.0 x 10-3 |
|
II |
0.3 |
0.2 |
7.2 x 10-2 |
|
III |
0.3 |
0.4 |
2.88 x 10-1 |
|
IV |
0.4 |
0.1 |
2.40 x 10-2 |
Determine the rate law and the rate constant for the reaction.
Answer:
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
Rate = k [A]x [B]y
According to the question,
6.0 x 10-3 = k [0.1] x [0.1] y (1)
7.2 x 10-2 = k [0.3] x [0.2] y (2)
= k [0.3] x [0.4] y (3)
2.40 x 10-2 =k [0.4] x [0.1] y (4)
Dividing equations (4) by (1),
(2.40 x 10-2) / (6.0 x 10-3) = (k [0.4] x [0.1] y)/ (k [0.1] x [0.1] y)
4 = [0.4]x/[0.1]x
4 = (0.4/0.1)x
(4)1 = 4x
=> x=1
Dividing equation (iii) by (ii), we obtain
(2.88 x 10-1)/ (7.2 x 10-2) = (k [0.3] x [0.4] y)/ (k [0.3] x [0.2] y)
4= (0.4/0.2)y
Or y=2
Therefore, the rate law is
Rate = k [A] [B] 2
k = (Rate)/ [A] [B] 2
From experiment I, we obtain
k= (6.0 x 10-3 molL-1min-1)/ (0.1molL-1) (0.1mol L-1)2
= 6.0 L2 mol−2 min−1
From experiment II, we obtain
k= (7.2 x 10-2 molL-1min-1)/ (0.3molL-1) (0.2mol L-1)2
=6.0 L2 mol−2 min−1
= 6.0 L2 mol−2 min−1
From experiment III, we obtain
k= (2.88 x 10-1 molL-1min-1)/ (0.3molL-1) (0.4mol L-1)2
= 6.0 L2 mol−2 min−1
From experiment IV, we obtain
k= (2.40 x 10-2 molL-1min-1)/ (0.4molL-1) (0.1mol L-1)2
= 6.0 L2 mol−2 min−1
Therefore, rate constant, k = 6.0 L2 mol−2 min−1
Question 4.12.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
|
Experiment |
[A]/mol L-1 |
[B]/mol L-1 |
Initial rate of formation of [D]/mol L-1 min-1 |
|
I |
0.1 |
0.1 |
2.0 x 10-2 |
|
II |
- |
0.2 |
4.0 x 10-2 |
|
III |
0.4 |
0.4 |
- |
|
IV |
- |
0.2 |
2.0 x 10-2 |
Answer:
The given reaction is of the first order with respect to A and of zero order with respect to B.
Therefore, the rate of the reaction is given by,
Rate = k [A] 1[B] 0
⇒ Rate = k [A]
From experiment I, we obtain
2.0 × 10−2 mol L−1 min−1 = k (0.1 mol L−1)
⇒ k = 0.2 min−1
From experiment II, we obtain
(4.0 × 10−2 mol L−1 min−1)= 0.2 min−1 [A]
⇒ [A] = 0.2 mol L−1
From experiment III, we obtain Rate
= (0.2 min−1 × 0.4 mol L−1)
= (0.08 mol L−1 min−1)
From experiment IV, we obtain
(2.0 × 10−2 mol L−1 min−1) = 0.2 min−1 [A]
⇒ [A] = 0.1 mol L−1
Question 4.13.
Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s–1 (ii) 2 min–1 (iii) 4 years –1
Answer:
(i) Half-life, t (½) = (0.693 / k)
= (0.693 / 200) s-1
= 3.47 ×10-3 s (approximately)
(ii) Half-life, t (½) = (0.693 / k)
= (0.693 / 2) min-1
= 0.35 min (approximately)
(iii) Half-life, t (½) = (0.693 / k)
= (0.693 / 4) years-1
= 0.173 years (approximately)
Question 4.14.
The half-life for radioactive decay of 14C is 5730 years.
An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample?
Answer:
k= (0.693)/ (t (1/2))
= (0.693)/ (5730) years-1
Also, t = (2.303)/ (k) log[R]o/[R]
= (2.303)/ ((0.693/5730)) log (100/80)
=1845 years (approx.)
Hence, the age of the sample is 1845 years.
Question 4.15.
The experimental data for decomposition of N2O5
[2N2O5 → 4NO2 + O2] in gas phase at 318K are given below:
|
t/s |
0 |
400 |
800 |
1200 |
1600 |
2000 |
2400 |
2800 |
3200 |
|
102 x [N2O5]/ mol L-1 |
1.63 |
1.36 |
1.14 |
0.93 |
0.78 |
0.64 |
0.53 |
0.43 |
0.35 |
(i) Plot [N2O5] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N2O5] and t.
(iv) What is the rate law?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
Answer:
(i)
(ii)
|
t/s |
0 |
400 |
800 |
1200 |
1600 |
2000 |
2400 |
2800 |
3200 |
|
102 x [N2O5]/ mol L-1 |
1.63 |
1.36 |
1.14 |
0.93 |
0.78 |
0.64 |
0.53 |
0.43 |
0.35 |
|
log [N2O5] |
-1.79 |
-1.87 |
-1.94 |
-2.03 |
-2.11 |
-2.19 |
-2.28 |
-2.37 |
-2.46 |
Therefore, the rate law of the reaction is
Rate =k [N2O5]
Slope = (-2.46)-(-1.79)/(3200-0)
=-(0.67/3200)
Again, slope of the line of the plot log [N2O5] v/s t is given by
(-k/2.303)
Therefore,
(-k/2.303) = - (0.67/3200)
=> k= 4.82 x 10-4 s-1
t (1/2) = (0.693)/(k)
= (0.693)/ (4.82 x 10-4) s
=1.438 x 103 sec
= 1438 sec
This value, 1438 s, is very close to the value that was obtained from the graph.
Question 4.16.
The rate constant for a first order reaction is 60 s–1.
How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
Answer:
It is known that,
t= (2.303/k) log[R] 0/[R]
= (2.303)/ (60 s-1) log ((1)/ (1/16))
= (2.303)/ (60 s-1) log (16)
=4.6 x 10-2 sec (approx.)
Hence, the required time is 4.6 × 10−2 s.
Question 4.17.
During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years.
If 1μg of 90Sr was absorbed in the bones of a newly born baby instead of calcium,
how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Answer:
Initial concentration, [R] 0 = 1μg
Final concentration= [R]
Half-life t (1/2) = 28.1 years
We know, t (1/2) = (0.693/k) Where, k = rate constant
⇒ k = (0.693/ t (1/2))
Or k = (0.693/28.1 years)
k=0.0246 years-1
Also, t = (2.303/k) log[R]o/[R]
If t = 10yrs, then, using the formula, we get,
t = (2.303/k) log[R]o/[R]
10 = (2.303)/ (0.0246) log (1/[R])
10 = (2.303)/ (0.0246) log (- [R])
log [R] = - (10 x 0.693)/(2.303 x 28.1)
[R] = antilog (-0.1071)
=0.7814 μg
Therefore, 0.7814 μg of 90Sr will remain after 10 years.
Again,
t = (2.303/k) log[R]o/[R]
60 = (2.303)/ (0.0246) log (1/[R])
60 = (2.303)/ (0.0246) log (1/ [R])
log [R] = - (60 x 0.693)/(2.303 x 28.1)
[R] = antilog (-0.6425)
=0.2278 μg
Therefore, 0.2278 μg of 90Sr will remain after 60 years.
Question 4.18.
For a first order reaction, show that time required for 99% completion is twice the time
required for the completion of 90% of reaction.
Answer:
For a first order reaction, the time required for 99% completion is
t1= (2.303/k) log [(100)/ (100-99)]
= (2.303/k) log (100)
= 2 x (2.303/k)
For a first order reaction, the time required for 90% completion is
T2= (2.303/k) log [(100)/ (100-90)]
= (2.303/k) log (10)
= (2.303/k)
Therefore, t1 = 2t2
Hence, the time required for 99% completion of a first order reaction is twice the time required for the
completion of 90% of the reaction.
Question 4.19.
A first order reaction takes 40 min for 30% decomposition. Calculate t (1/2).
Answer:
For a first order reaction,
t= (2.303/k) log [R] 0/[R]
k= (2.303/40 min) log [(100)/ (100-30)]
= (2.303/40 min) log (10/7)
=8.918 x 10-3 min-1
Therefore t (1/2) of the decomposition reaction is
t (1/2) =(0.693/k)
= (0.6913)/ (8.918 x 10-3) min
=77.7 approx.
Question 4.20.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
|
t(sec) |
P (mm of Hg) |
|
0 |
35.0 |
|
360 |
54.0 |
|
720 |
63.0 |
Calculate the rate constant.
Answer:
The decomposition of azoisopropane to hexane and nitrogen at 543 K is represented by the following equation.
(CH3)2CHN=NCH (CH3)2(g) à N2 (g) + C6H14 (g)
At t=0 P0 0 0
At t= t (P0 –p) 0 0
When time t = t, the total partial pressure is Pt = P0 + p
(P0-p) = (Pt-2p), but by the above equation, we know p = (Pt-P0)
Hence, (P0-p) = Pt-2(Pt-P0)
Thus, (P0-p) = 2P0 – Pt
We know that time t= (2.303/k) log[R] 0/[R]
Where, k=rate constant
[R]0 =Initial concentration of reactant
[R]=Concentration of reactant at time‘t’
Here concentration can be replaced by the corresponding partial pressures.
Hence, the equation becomes, t= (2.303/k) log [(P0)/ (P0 – p)] (equation 1)
At time t = 360 s, Pt = 54 mm of Hg and P0 = 30 mm of Hg,
Substituting in equation 1,
⇒ 360 = (2.303/k) log [(30)/ (2x30) – 54)]
⇒ k = 2.175 × 10-3 s-1
At time t = 720 s, Pt = 63 mm of Hg and P0 = 30 mm of Hg,
Substituting in equation 1,
720 = (2.303/k) log [(30)/ (2x30) – 63)]
Thus, k= (2.235 x 10-3 s-1 + 2.175 x 10-3 s-1)/ (2)
Therefore k = 2.21 x 10-3 s-1
Question 4.21.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.
SO2Cl2 (g) -->SO2 (g) + Cl2 (g)
|
Experiment |
Time/s-1 |
Total pressure/atm |
|
1 |
0 |
0.5 |
|
2 |
100 |
0.6 |
Calculate the rate of the reaction when total pressure is 0.65 atm.
Answer:
When t = 0, the total partial pressure is P0 = 0.5 atm
SO2Cl2 (g) --> SO2 (g) + Cl2 (g)
At t=0 P0 0 0
At t= t (P0 –p) 0 0
When time t = t, the total partial pressure is Pt = (P0 + p)
(P0-p) = (Pt-2p), but by the above equation, we know p = (Pt-P0)
Hence, (P0-p) = (Pt)-(2(Pt-P0))
Thus, (P0-p) = (2P0 – Pt)
We know that time, t= (2.303)/ (k) log[R] 0/[R]
Where, k=rate constant
[R]° =Initial concentration of reactant
[R]-Concentration of reactant at time‘t’
Here concentration can be replaced by the corresponding partial pressures.
Hence, the equation becomes, t= (2.303)/ (k) log (Po)/ (P0-p)
⇒t= (2.303)/ (k) log (Po)/ (2P0-Pt) (equation 1)
At time t = 100 s, Pt = 0.6 atm and P0 = 0.5 atm,
Substituting in equation 1,
100 = (2.303)/ (k) log (0.5)/ ((2 x 0.5) -0.6)
Thus, k = 2.231 × 10-3 s-1
The rate of reaction R = (k × P PS02Cl2)
When total pressure Pt = 0.65 atm and P0 = 0.5 atm, then
PS02Cl2 = (2P0-Pt)
Thus, substituting the values, PS02Cl2 = (2(0.5)-0.6) = 0.35 atm
R =k × PS02Cl2 = (2.231 × 10-3 s–1 × 0.35)
Rate of the reaction R = 7.8 × 10-4atm s–1
Question 4.22.
The rate constant for the decomposition of N2O5 at various temperatures is given below:
|
T/0C |
0 |
20 |
40 |
60 |
80 |
|
105 x k/s-1 |
0.0787 |
1.70 |
25.7 |
178 |
2140 |
Draw a graph between ln k and 1/T and calculate the values of A and Ea .Predict the rate constant at 30° and 50°C.
Answer:
|
T/0C |
0 |
20 |
40 |
60 |
80 |
|
105 x k/s-1 |
0.0787 |
1.70 |
25.7 |
178 |
2140 |
|
T (in Kelvin) |
(0+273) =273 |
(20 + 273)=293 |
(40 + 273)=313 |
(60 + 273) =333 |
(80 + 273)=353 |
|
(1/T) K-1 |
3.66 x 10-3 |
3.41 x 10-3 |
3.19 x 10-3 |
3.0 x 10-3 |
2.83 x 10-3 |
|
ln K |
-7.147 |
-4.075 |
-1.359 |
-0.577 |
3.063 |
The graph is given as:
The Arrhenius equation is given by k = Ae-(Ea/RT)
Where, k= Rate constant
A= Constant
Ea=Activation Energy
R= Gas constant
T=Temperature
Taking natural log on both sides,
ln k = ln A-(Ea/RT) (equation 1)
By plotting a graph, ln K Vs (1/T), we get y-intercept as ln A and Slope is – (Ea/R).
Slope = (y2-y1)/(x2-x1)
By substituting the values, slope = -12.301
⇒ – (Ea/R) = -12.301
But, R = 8.314 JK-1mol-1
⇒ Ea = (8.314 JK-1mol-1 × 12.301 K)
⇒ Ea = 102.27 kJ mol-1
Substituting the values in equation 1 for data at T = 273K
⇒ ln K = ln A – (Ea/RT)
=>-7.147 = ln A – (102.27 x 103)/ (8.314 x 273)
=By taking (T=273 K, ln k =-7.147)
On solving, we get ln A = 37.911
Therefore, A = 2.91×106
When T = 300C, hence T = (30 + 273) = 303K
⇒ ln k = -2.8
⇒ k = 6.08x10-2s-1.
When T = 500C, hence T = (50 + 273) = 323K
Substituting in equation 1,
A = 2.91 × 106, Ea = 102.27 kJ mol-1
Ln K = ln (2.91 × 106) – (102.27)/ (8.314 × 323)
Thus, ln k = -0.5 and k = 0.607s-1.
Question 4.23.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1 at 546 K.
If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
Answer:
k= 2.418 × 10-5 s-1
T= 546 K
Ea= 179.9 kJ mol-1 = 179.9 × 103J mol-1
According to the Arrhenius equation k = Ae-(Ea/RT)
Taking log on both sides,
log k =ln A – (Ea)/(RT)
log k =ln A – (Ea)/(2.303 RT)
log A = log k + (Ea)/(2.303 RT)
=log (2.418 x 10-5 s-1) + (179.9 x 103) Jmol-1/ (2.303x 8.314 Jk-1mol-1 x 546k)
= (0.3835 - 5) + (17.2082)
= 12.5917
Therefore, A = antilog (12.5917)
= 3.9 × 1012 s-1(approximately)
Question 4.24.
Consider a certain reaction A → Products with k = 2.0 × 10–2s–1.
Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L–1.?
Answer:
k = 2.0 × 10–2s-1
Time t = 100s
Concentration [A0] = 1.0 mol L-1
We know, t= (2.303)/ (k) log [A0]/ [A]
On substituting the values, 100 = (2.303)/ (0.02) log [1]/ [A]
log (1/ [A]) = (2.303/2)
Log [A] = - (2.303/2)
[A] =0.135 mol L-1
Question 4.25.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t (1/2) = 3.00 hours.
What fraction of sample of sucrose remains after 8 hours?
Answer:
t (1/2) = 3.00 hours
We know, t (1/2) = (0.693/k)
Therefore, k = (0.693/3)
k = 0.231hrs-1
We know, time t= (2.303)/ (k) log [R0]/[R]
Where, k= rate constant
[R]° =Initial concentration
[R]=Concentration at time‘t’
Thus, substituting the values, 8= (2.303)/ (0.231) log ([R] 0/[R])
log ([R]0/[R]) = 0.8
log ([R]/[R]0) = -0.8
[R]/[R] 0 = 0.158
Hence, 0.158 fraction of sucrose remains.
Question 4.26.
The decomposition of hydrocarbon follows the equation
k = (4.5 × 1011s–1) e (-28000K/T) .Calculate Ea.
Answer:
The given equation is k = (4.5 ×1011 s−1) e (−28000 K/T) (i)
Arrhenius equation is given by,
K = Ae-(Ea/RT) (ii)
From equation (i) and (ii), we obtain
(Ea/RT) = (28000K)/ (T)
Ea = (R x 28000 K)
= (8.314 J K−1 mol−1 × 28000 K)
= 232792 J mol−1
= 232.792 kJ mol−1
Question 4.27.
The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.34 – 1.25 × 104K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Answer:
Arrhenius equation is given by,
k= Ae – (Ea/RT)
⇒In k = In A – (Ea/RT)
⇒In k = Log A – (Ea/RT)
⇒ Log k = Log A – (Ea/2.303RT) (i)
The given equation is
Log k = (14.34 - 1.25 104 (K/T)) (ii)
From equation (i) and (ii), we obtain
(Ea/2.303RT) = 1.25 104 (K/T)
⇒ Ea = (1.25 × 104K × 2.303 × R)
= (1.25 × 104K × 2.303 × 8.314 J K - 1mol – 1)
= 239339.3 J mol - 1 (approximately)
= 239.34 kJ mol - 1
Also, when t (1/2) = 256 minutes,
k = (0.693 / t (1/2))
= (0.693 / 256)
= 2.707 × 10 - 3 min - 1
= 4.51 × 10-5s-1
It is also given that, log k= (14.34 - 1.25 × 104(K/T))
=>log (4.51 × 10-5) = (14.34 - 1.25 × 104(K/T))
=> log (0.654 – 05) = (14.34 - 1.25 × 104(K/T))
=> (1.25 x 104 K)/ (T) = 18.686
T= (1.25 x 104 K)/ (18.686)
= 668.95 K
= 669 K (approximately)
Question 4.28.
The decomposition of A into product has value of k as 4.5 × 103 s–1 at 10°C and energy of activation 60 kJ mol–1.
At what temperature would k be 1.5 ×104s–1?
Answer:
From Arrhenius equation, we obtain
log (k2/k1) = (Ea)/ (2.303 R) (T2-T1)/(T1T2)
Also, k1 = 4.5 × 103 s-1
T1 = 273 + 10 = 283 K
k2 = 1.5 × 104 s-1
Ea = 60 kJ mol-1 = 6.0 × 104 J mol-1
Then, log (1.5 x104)/ (4.5 x 104) = (6 x 104)/ (2.303 x 8.314) (T2 -283)/ (283 – T2)
0.5229 = 3133.627 – (T2 -283)/ (283 – T2)
0 .0472T2 = (T2-283)
T2 = 297K or T2 = 240 C
Question 4.29.
The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K.
If the value of A is 4 × 1010 s–1 Calculate k at 318K and Ea?
Answer:
For a first order reaction,
t = (2.303) / (k log a) / (a – x)
At 298 K,
t = (2.303) / (k log 100) / (90)
= (0.1054 / k)
At 308 K,
t' = (2.303) / (k' log 100) / (75)
= (2.2877 / k')
According to the question,
t = t'
⇒ (0.1054 / k) = (2.2877 / k')
⇒ (k' / k) = 2.7296
From Arrhenius equation, we obtain
log (k2/k1) = (Ea)/ (2.303 R) (T2-T1)/(T1T2)
log (2.7296) = (Ea)/ (2.303 x 8.314) (308-298)/ (298 x 308)
To calculate k at 318 K,
It is given that, A = 4 x 1010 s-1, T = 318K
Again, from Arrhenius equation, we obtain
log k = log A – (Ea)/ (2.303 RT)
=log (4 x 1010) – (76.64 x 103)/ (2.303 x 8.314 x 318)
= (0.6021+ 10) – (12.5876)
=-1.9855
Therefore, k = Antilog (-1.9855)
= 1.034 x 10-2 s -1
Question 4.30.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K.
Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answer:
From Arrhenius equation, we obtain
log (k2/k1) = (Ea)/ (2.303 R) (T2-T1)/(T1T2)
It is given that k2 = 4k1
T1 = 293 K
T2 = 313 K
Therefore, log (4k1/k1) = (Ea)/ (2.303 x 8.314) (313-293)/ (293 x 313)
Hence, the required energy of activation is 52.86 kJ mol- 1
.
Email: ExamFearVideos@gmail.com
