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]

(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

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?

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?

If pressure is measured in bar and time in minutes, then

Unit of rate = bar min−1

Rate = k (pCH3OCH3) (3/2)

• k= (Rate)/ (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.

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?

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?

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.

• Average rate of reaction over interval

= [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

• the pseudo first-order rate constant can be calculated by

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?

(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?

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

• log 2.821 = x log 2
• x=(log 2.821)/(log 2)
• =1.496
• 5 approx.

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.

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

• 4 = 2y
• (2)2 = 2y

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

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 min1

From experiment II, we obtain

(4.0 × 10−2 mol L−1 min−1)= 0.2 min−1 [A]

⇒ [A] = 0.2 mol L1

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

(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?

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).

(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

• Time corresponding to the concentration (1.603 x 102)/ (2) molL-1, is the half-life.
• From the graph, the half-life is obtained as 1450 s.
• The given reaction is of the first order as the plot, log [N2O5] v/s t, is a straight line.

Therefore, the rate law of the reaction is

Rate =k [N2O5]

• From the plot, log [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

• Half-life is given by,

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?

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.

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.

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).

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.

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.

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 1.7 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.

 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?

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.?

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?

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.

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?

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?

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?

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)

• Ea = (2.303 x 314 x298 x 308 x log(2.7296))/(308-298)
• =76640.096 Jmol-1
• =76.64 kJmol-1

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.

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)

• 6021 = (20 x Ea)/( 2.303 x 8.314 x293 x 313)
• Ea = (0.6021 x 2.303 x 8.314 x293 x 313)/(20)
• = ( 52863.33 ) J mol-1
• = 52.86 kJ mol-1

Hence, the required energy of activation is 52.86 kJ mol- 1