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)

• 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.

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:

• 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?

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

• 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.

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

• 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

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 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

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

• 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?

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 1.7 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.

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 / [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)

• 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.

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)

• 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

.