Question 50.
For a decomposition reaction, the values of k at two different temperatures are given below:
k1 = 2.15 × 10-8 L mol-1 s-1 at 650 K k2 = 2.39 × 10-7 L mol-1 s-1 at 700 K
Calculate the value of activation energy for this reaction.
(Log 11.11 = 1.046) (R = 8.314 J K-1 mol-1) (Comptt. All India 2014) Answer:
Given: k1 = 2.15 × 10-8 L mol-1 s-1, T1 = 650 K k2 = 2.39 × 10-7 L mol-1 s-1, T2 = 700 K
R = 8.314 J K-1 mol-1 Ea =?
Question 51.
The rate constant of a reaction at 500 K and 700 K are 0.02 s-1 and 0.07 s-
1 respectively. Calculate the value of activation energy, En (R = 8.314 J K-1 mol-1) (Comptt. Delhi 2015)
Answer:
Given : k2 = 0.07 s-1, k1, = 0.02 s-1, T1 = 500 K, T2 = 700 K, Ea = ?
Question 52.
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 l/10th value? (Comptt. All India 2015)
Answer:
Given : k = 60 s-1, t = ? If initial concentration is [A0]
Question 53.
The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.2 – 1.0×104TK
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 J K-1 mol-1) (Delhi 2016) Answer:
Given: t1/2 = 200 min Ea = ?, T = ? Using Arrhenius equation
Question 54.
For the first order thermal decomposition reaction, the following data were obtained:
C2H5Cl(g) → C2H4(g) + HCl(g) Time/sec Total pressure/atm 0 0.30 300 0.50
Calculate the rate constant (Given: log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021) (All India 2016)
Answer:
Question 55.
If the half-life period of a first order reaction in A is 2 minutes, how long will it take [A] to reach 25% of its initial concentration? (Comptt. Delhi 2016)
Answer:
Question 56.
The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. [R = 8.314 JK-1 mol-1] (Comptt. All India 2016)
Answer:
Question 57.
Following data are obtained for the reaction:
N2O5 → 2NO2 + 12O2
t/s 0 300 600
[N205]/mol L-1 1.6 × 10-2 0.8 × 10-2 0.4 × 10-2
(a) Show that it follows first order reaction.
(b) Calculate the half-life.
(Given log 2 = 0.3010 log 4 = 0.6021) (Delhi 2016) Answer:
(a) For first order reaction:
Question 58.
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.
(Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) (All India 2016) Answer:
Given:
t = 20 min, A0 = 100%, A = 100 – 25 = 75%, k = ?
Question 59.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume :
Experiment Time(s) Total pressure (atm)
1 0 0.4
2 100 0.7
Calculate the rate constant (k).
[Given : log 2 = 0.3010; log 4 = 0.6021] (Comptt. Delhi 2016) Answer:
SO2Cl2 (g) → SO2 (g) + Cl2(g)
Using formula, K = 2.303tlogP02P0−Pt
When t = 100 s
Question 60.
For the first order decomposition of azoisopropane to hexane and nitrogen at 543 K , the following data were obtained :
Calculate the rate constant. The equation for the reaction is :
Experiment Time(s) Total (mmHg)
1 0 35.0
2 720 63.0
(CH3)2CHN = NCH(CH3)2 C6H14 (g) + N2 (g)
[Given : log 3 = 0.4771; log 5 = 0.6990] (Comptt. Delhi 2016) Answer:
(CH3)2CHN = NCH(CH3)2 → N2 + C6H14
Initial pressure P0 O O
After time t P0 – P P P
Total Pressure after time t(Pt) = (P0 – P) + P + P P = Pt – P0
a ∝ P0
(a – x) ∝ P0 – P
Substituting the value of P
a – x ∝ P0 – (Pt – P0) or, (a – x) ∝ P0 – Pt
As decomposition of azoisopropane is a first order reaction
Question 61.
For a first order reaction, show that time required for 99% completion is twice the time required for completion of 90% reaction. (Comptt. All India 2016)
Answer:
∴ Ea = 2.303 × 8.314 (JK-1 mol-1) × 4250 K
= 81.375 J mol-1 or 81.375 kJ mol-1 Question 62.
Half-life for a first order reaction 693 s. Calculate the time required for 90%
completion of this reaction. (Comptt. All India 2016)
Answer:
Chemical Kinetics Class 12 Important Questions Long Answer Type (LA) Question 63.
(a) Explain the following terms : (i) Rate of a reaction
(ii) Activation energy of a reaction (b) The decomposition of phosphine, PH3, proceeds according to the following equation:
4 PH3 (g) → P4 (g) + 6 H2 (g)
It is found that the reaction follows the following rate equation : Rate = K [PH3].
The half-life of PH3 is 37.9 s at 120° C.
(i) How much time is required for 3/4th of PH3 to decompose?
(it) What fraction of the original sample of PH3 remains behind after 1 minute? (All India 2010)
Answer:
(a) (i) Rate of a reaction: The change in the concentration of any one of the reactants or products per unit time is called rate of reaction.
(ii) The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.
The activation energy of the reaction decreases by the use of catalyst.
(b) (i) According to the formula :
Question 64.
(a) Explain the following terms : (i) Order of a reaction
(ii) Molecularity of a reaction
(b) The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 J K-1 mol-1) (All India 2016)
Answer:
(a) (i) Order of a reaction: It is the sum of powers of the molar concentrations of reacting species in the rate equation of the reaction.
(ii) Molecularity of a reaction :
• It is the total number of reacting species (molecules, atoms or ions) which bring the chemical change.
• It is always a whole number.
• It is a theoretical concept.
• It is meaningful only for simple reactions or individual steps of a complex reaction. It is meaningless for overall complex reaction.
(b) Given : T1 = 300 K T2 = 320 K K1 = K (Consider)
K2 = 4 K R = 8.314 Ea = ?
Substituting these values in the formulae,
∴ Energy of activation, Ea = 55327.46 = 55.3 KJ mol-1 Question 65.
(a) With the help of a labelled diagram explain the role of activated complex in a reaction.
(b) A first order reaction is 15% completed in 20 minutes. How long will it take to complete 60% of the reaction ? (Comptt. Delhi 2012)
Answer:
(a) In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram
This diagram is obained by plotting potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products.
(b) For the first order reaction
Question 66.
(a) What is the physical significance of energy of activation ? Explain with diagram.
(b) In general, it is observed that the rate of a chemical reaction doubles with every 10 degree rise in temperature. If the generalization holds good for the reaction in the temperature range of 295 K to 305 K, what would be the value of activation energy for this reaction ?
[R = 8.314 J mol-1 K-1] (Comptt. Delhi 2012) Answer:
(a) The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy. Less is the activation energy, faster is the reaction or greater is the activation energy, slower is the reaction
Question 67.
(a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation,
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 for this reaction. (Given log 1.428 = 0.1548) (Delhi 2013)
Answer:
(a) (i) Differential rate equation : dxdt = K [A]2[B]
(ii) When concentration of A is increased to three times, the rate of reaction becomes 9 times
r = K[3A]2B ∴ r = 9KA2B i.e. = 9 times
(iii) r = K[2A]2[2B] ∴ r = 8KA2B i.e. = 8 times (b) Given : Time, t = 40 minutes, t =?
Let a = 100, ∴ x = 30% of 100 = 30
Using the formula :
Question 68.
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. .
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – Ea2.303R(1T)
where Ea is the activation energy. When a graph is plotted for log k vs. 1T¯¯¯¯,a straight line with a slope of – 4250 K is obtained. Calculate ‘Ea‘ for the reaction. (R = 8.314 JK-1 mol-1) (Delhi 2013)
Answer:
∴ Ea = 2.303 × 8.314 (JK-1 mol-1) × 4250 K
= 81.375 J mol-1 or 81.375 kJ mol-1 Question 69.
(a) The decomposition of A into products has a 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 × 104 s-1? (b) (i) If half life period of a first order reaction is x and 3/4,th life period of the same reaction is y, how are x and y related to each other?
(ii) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why? (Comptt. Delhi 2013) Answer:
(a) Given : K1 = 4.5 × 103 s-1, T1 = 10K + 273K = 283K K2 = 1.5 × 104 s-1, T2 = ? Ea = 60 KJ mol-1
Using formula :
∴ Temperature, T2 will be = 297° – 273° = 24° C (b) (i) t1/2 = 0.693K (For first order reaction) t3/4 = K ⇒ t3/4 = 1.3864K
According to condition
(The value 1.3864 is double of 0.693) From the above equation it is clear that t3/4 = 2t1/2 ∴ y = 2X
(ii) It is due to improper orientation of the colliding molecules at the time of collision.
Question 70.
(a) A first order reaction takes 100 minutes for completion of 60% of the reaction.
Find the time when 90% of the reaction will be completed.
(b) With the help of diagram explain the role of activated complex in a reaction.
(Comptt. Delhi 2013) Answer:
(a) For the first order reaction,
(b) In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram
This diagram is obained by plotting potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products.
Question 71.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :
t/s 0 30 60
[CH3COOCH3]/mol L-1 0.60 0.30 0.15
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021) (Delhi 2015) Answer:
As k is constant in both the readings, hence it is a pseudo first order reaction.
(ii) Rate = – Δ[R] /Δt, Average rate between 30 to 60 seconds
= −(0.15−0.30)60−30=0.1530
= 0.5 × 10-2 mol L-1 sec-1 Question 72.
(a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010) (Delhi 2015) Answer:
(a) For the reaction A + B → P rate is given by Rate = k[A]1[B]2
(i) r1 = k[A]1 [B]2 r2 = k[ A]1[2B]2 =
r2 = k[A]1 [2B]2=4k[A]1 [B]2
r1 = 4r2, rate will increase four times of actual rate.
(ii) When A is present in large amount, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2, second order reaction.
Question 73.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:
t/s 0 10 20
[CH3COOCH3]/mol L-1 0.10 0.05 0.025
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 10 to 20 seconds.
(Given : log 2 = 0.3010, log 4 = 0.6021) (All India 2015) Answer:
As k1 and k2 are equal, hence pseudo rate constant is same.
It follows the pseudo first order reaction.
(ii) Average rate of reaction between 10 to 20 seconds
= −Δ[R]Δt=−(0.025−0.05)(20−10)=0.02510
= 0.0025 mol lit-1 sec-1 Question 74.
(a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (All India 2015)
Answer:
(a) For the reaction A + B → P rate is given by Rate = k[A]1[B]2 (i) r1 = k[A]1 [B]2
r2= k[A]1 [2B]2 = 4k[A]1 [B]2
r1 = 4r2 (rate of reaction becomes 4 times)
(ii) When A is present in large amounts, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2
