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 Bihar Board Class 12 Chemistry Solutions Chapter 4 Chemical Kinetics

BSEB Bihar Board Class 12 Chemistry Solutions Chapter 4 Chemical Kinetics

Bihar Board Class 12 Chemistry Chemical Kinetics Text Book Questions and Answers

Question 1.
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:

Question 2.
In a reaction 2A → Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during the reaction.
Answer:
Rate of reaction = rate of disappearance of A

Question 3.
For a reaction, A + B → Product, the rate law is given by r = k [A]1/2 [B]2 what is the order of the reaction?
Answer:
Order of a reaction is the sum of the powers to which the concentrations of the reactants have to be raised in order to specify the rate of the reaction
r = k [A]1/2 [B]2
Order of the reaction = 12 + 2 = 212

Question 4.
The conversion of molecules X to Y follows second order kinetics. If concentration of x is increased to three times, how will it affect the rate of formation of Y?
Answer:
X → Y
Rate = k[X]2 … (i)
If cone of x is increased to 3 times
[X’] = [3X]
new rate will be = k [3X]2
= k.9 [X]2 = 9.k [X]2 … (ii)
From (i) and (ii) new rate will be 9 times.

Question 5.
A first order reaction has a rate constant 1.15 × 10-3 s-1, How long will 5g of this reactant take to reduce to 3g?
Answer:

Question 6.
Time required to decompose SO2 Cl2 to half of its inital amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer:
t1/2 for SO2 Cl2 = 60 min = 60 × 60 sec
Since it is in first order reaction
∴ k = 0.693t1/2 = 0.69360×60
∴ rate constant of the reaction = 1.925 × 10-4 s-1.

Question 7.
What will be the effect of temperature on rate constant?
Answer:
Most of the chemical reactions are accelerated by increase in temperature, whether the reactions are endothermic or exothermic. In a mixture of KMnO4and H2C2O4, potassium permangnate (KMnO4) gets decolourised faster at higher temperature than at a lower temperature. It has been found that for a rise of 10° in temp., the rate constant of of a reaction is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation
k = Ae-Ea/RT
Where A is the Arrhenius or frequency factor. It is also called PRE-EXPONENTIAL factor. It is a constant specific to a reaction.
R = Gas constant; Ea is the activation energy measured in joules mole.

Question 8.
The rate of the chemical reaction doubles for an increase of’ 10 K in absolute temperature from 298 K. Calculate Ea.
Answer:
T1 = 298 K
T2 = 298 + 10 = 308 K
R = 8.314 JK-1 mol-1
If initial rate constant is k1, then for 10° rise in temp.
final rate constant = 2k,
∴ putting these values in Arrhenius equation.

Question 9.
The activation energy for the reaction 2HI (g) → H2 + I2, (g) is 2.09.5 kj mol-1 at 581 K. Calculate the fraction of the molecules of reactants having energy equal to or greater than activation energy.
Answer:
Fraction of molecules with energy equal to or greater than activation energy, x = e-Ea/RT

Bihar Board Class 12 Chemistry Chemical Kinetics Text Book Questions and Answers

Question 1.
From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants.
(i) 3NO(g) → N2(g); Rate = k[NO]2
(ii) H2O2(aq) + 3I–(aq) + 2H+ → 2H2O(I)–3 ; 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:

Question 2.
For the reaction 2A + B → A2B the rate = k[A][B]2 with k = 2.0 x 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–. Calculate the rate of the reaction after [A] is reduced to 0.6 mol-1.
Answer:
(i) Intial 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 L-1 s-1

(ii) When [A] is reduced from 0.1 mol L-1 to 0.06 mol L-1
i.e. (0.1 – 0.06) = 0.04 mol L-1 has reacted, B reacted
12 x 0.04 molL-1
= 0.02 mol L-1
Hence new [B] = (0.2 – 0.02) = 0.18 mol L-1
Now, 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 × 10-9 mol L-1 s-1

Question 3.
The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.50 × 10-4 mol L S-1 ?
Answer:
2NH3 (g) ⇌ N2(g) + 3H2(g)
rate = – 12.Δ(NH3)Δt = Δ(N2)Δt = 13 Δ(H2)Δt
rate = k as it is a zero order reaction
Δ(N2)Δt = 2.50 x 10-4 mol L s-1 and
Δ(H2)Δt = 7.5 x 10-4 mol L s-1

Question 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 constants?
Answer:
The decomposition reaction is CH3OCH3 → CH4 + H2 + CO
In terms of concentrations
Units of rate = mol L-1 min-1

Question 5.
Mention the factors^ that affect the rate of a chemical reaction.
Answer:
There are a number of factors which influence the rate of:
The important factors are –

1. Nature of the reacting species.
2. Concentration of the reacting species.
3. Temperature at which a reaction proceeds.
4. Surface area of the reactants.
5. Exposure to radiations.
6. Presence of a catalyst.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of the reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Answer:
Rate = k [A]2 = ka2
if [A] = 2a, rate = k[2a]2 = 4ka2 = 4 times
if [A] = 12a; rate = k(12)² = 14ka2 = 14th
Thus the rate of a II order reaction becomes 4 times when the concentration of the reactant is doubled and it becomes 14th when the concentration of reactant is halved.

Question 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 of a reaction increases with increase in temperature and becomes nearly double for every 10° rise of temperature. The effect can be represented quantitatively by Arrhenius equation.
k = Ae-Ea/RT
where A = constant called frequency factor. It is related to number of binary molecular collisions per second per litre.
Ea = Energy of activation
T = Absolute temperature
N = The integrated form
R = Gas constant
log k2kr = Ea2.303R[ [T2−T1T1T2] ]
where k1 and k2 are the rate constants for the reaction between two different temperatures T1 and T2 respectively.

Question 8.
In a pseudo first order hydrolysis of ester in water the following results were obtained –

(i) Calculate the average rate of the reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Answer:
Average rate during the interval 30 – 60 sec.

Question 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 concentration of both A and B are doubled?
Answer:
(i) dxdt = k[A][B]2

(ii) rate = kab2. If [B] is tripled.
Rate = ka.(3b)2 = 9 kab2; rate increases 9 times.

(iii) If both [A] and [B] are doubled
then Rate = k (2a) (2b)2 = 8 kab2 = 8 times.
Rate of the reaction increases 8 times.

Question 10.
In a reaction between A and B, the initial rate of the reaction (r0) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B?
Answer:

Question 11.
The following results have been obtained during the kinetic studies of the reaction 2A + B → C + D.

Determine the rate law and the rate constant for the reaction.
Answer:
From Experiment I and IV, it may be noted that [B] is same, but [A] has been made 4 times.
The rate of the reaction also has become four times
∴ Rate w.r.t. A
Rate α [A] … (1)
From experiments II and III, it may be noted that [A] is kept same, and [B] has been doubled, the rate of the reaction has become 4 times. This mean w.r.t. B,
Rate α [B]² … (2)
Combining (1) and (2), we get the rate law of the reaction
2A + B → C + D
Rate Law is Rate = k [A] [B]2
Over – all order of the reaction = 1 + 2 = 3
To calculate rate constant
k =  Rate [A][B]2 = 6.0×10−3(0.1)(0.1)2 = 6.0
hence rate constant k = 6.0 M-2 m-1

Question 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:

Answer:
The reaction between A and B is first order with respect to A and zero order with respect to B.
∴ rate of the reaction w.r.t. A = k [A]
rate of the reaction w.r.t.B = k[B]0
∴ overall rate of the reaction = k [A] [B]0
It is of 1st (1 + 0) order.
From I Experiment,
∴ k =  Rate [A][B]∘ = 2.0×10−2[0.1]
= 2.o x 10-1
Now, in Experiment II, the cone, of A will be

The given artefact is 1845 years old.

Question 15.
The experimental data for the decomposition of N2O5 [2N2O5 → 4NO2 + O2] in gas phase at 318 K are given below.

(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:
(a) Plot [N2O5] against t

The different values of k calculated are 4.96 × 10-4 s-1,
4.82 × 10-4 s-1, 5.0 × 10-4 s-1, 5.44 × 10-4 s-1, 5.44 × 10-4 s-1,
5.47 × 10-4 s-1, 5.91 × 10-4 s-1, 5.81 × 10-4 s-1 and 5.71 × 10-4 s-1.
Since the value of k is almost constant, the rate equation is
γ = k[N2O5]
The mean value of rate constant is 5.2873 × 10-4 s-1 order of the reaction = 1.
Half life period t1/2 = 0.693k = 0.6935.2873×10−4 = 1310 s.

Question 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 1/16 th value?
Answer:
It is a first order reaction
k = 60 s-1
Let the initial cone, of the reactant = a M L-1
Let t be the time (in seconds) during which its initial concentration reduces to 1/16 of the value.
∴ conc. after time t = a16
Putting the values in the 1st order square

Hence after 4.6 × 10-2 second, the initial concentration of the reactant reduces to 1/16.

Question 17.
During nuclear explosion, one of the product is 90Sr with half life of 28.1 years. If 1 pg 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:
Half life period of 90Sr = 28.1 years
decay constant (k) = 0.693/28.1
Amount of 90Sr absorbed in the bones of the newly born baby = 1 μg = 10-6gm.
After time t = 10 years, the amount of radioactive 90Sr = N

∴ Amount of 90Sr left after 10 years, 60 years respectively is 0.7812

Question 18.
For a first order reaction show that the time required for 99% completion is twice the time requiered for the competion of 90% of reaction.
Answer:

Question 19.
A first order reaction takes 40 min for 30%decomposition. Calculation tt/2.
Answer:

Question 20.
For the composition of azoisopropane to hexane and nitrogen at 543 k, the following data are obtained.

Calculate the rate constant.
Answer:
Azoisopropane decompose according to the equation.

The decomposition is found out to be of the first order.
Initial pressure P0 = 35.0 mm of Hg.
Decrease of pressure of azoisopropane after time t = P = p
∴ increase in pr. of N2 = p
and increase in pr. of hexane = p.
Total pressure of the mixture = pt = PA + PN2 + Pc16H14
pt = (P0 – P) + P + P = p0 + P
P = Pt – P0
p = Pt – P0

Question 21.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at constant volume.
SO2Cl2(g) → SO2 (g) + Cl2(g)

Calculate the rate of the reaction when total pressure is 0.65 atm.
Answer:
1. As in question 4.20

= 2.23 × 10-3 s-1
Hence rate constant of the 1st order reaction = 2.23 × lO-3 s-1.

2. To calculate rate of the reaction when total pressure is 0.65 atm.
dxdt = k(a – x) = (2P0 – Pt) × k
= (1.00 – .65) × 2.23 × lO-3
= 0.35 × 2.23 × lO-3
rate of reaction = 7.8 × 10-4 atm s-1

Question 22.
The rate constant for the decomposition of N2O5 at vsrious temparature is given below:

Draw a graph between k and 1/T and calculate the value of A and Ea. Predict the rate constant at 30°C and 50°
Answer:
From the given data, log k is calculated for each value of k.

On plotting log k versus 1T, the slope of the line is found to be – 5300. [Students are advised to plot the actual graph on graph paper]
∵ Slope = −Ea2.303R
∴ Ea = – 2.303 × 8.314 (-5300)
= 101479.85 JK-1 mol-1
Ea = 101.5 KJ k-1 mol-1

(ii) To calculate the rate constant at 30°C = 303 K.
If T1 = 273 K; k1 = 0.0787 × 10-5
T2 = 303 K K2 = ?

Similarly the rate constant at 323 K is 8.14 x 10-4 s-1.

Question 23.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
Answer:
Ea = 179.9 = 179900 J
k = 2.418 × 10-5
and T = 546 K

Question 24.
Consider a certain reaction A → Products with k = 2.0 × 10-2 s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L-1.
Answer:
The reaction A → Products is a first order reaction ask = 2.0 × 10-2 second L-1.
[The value of k in second-1 indicates it is 1st order reaction]

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with T1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Answer:
Since it follows rate law of first order

fraction of sample of sucrose remaining = 0.1576 M

Question 26.
The decomposition of hydrocarbon follows the equation k = 4.5 × 1011 s-1 e28000k/t Calculate Ea.
Answer:
log k = log A – −Ea2.303R … (i)
The decomposition of the hydrocarbon follows the equation.
k = 4.5 × 1011 s-1 e28000k/t
Applying logs
log k = log 4.5 × 1011 – 28000 T
log10 k = log10 4.5 × 1011 – 280002.303T K … (ii)
Comparing it with equation no. (i)
−Ea2.303R = 280002.303T
∴ Ea – 28,000 x 8.314 J
= 28 x 8.314 kJ mol-1
Ea = 232.79 kJ mol-1.

Question 27.
The rate constant for the first order decomposition of H2O2</su-2 is given by the following equation:
logk = 14.34 – 1.25 x 104 KT
Calculate Ea for the reaction and at what temperature will its half-period be 256 minutes?
Answer:
(i) log k = log A – −Ea2.303R
The given equation is
log k = 14.34 – 1.25×104T K
Comparing the two equations
−Ea2.303R = 1.25 x 104
or Ea = 1.25 x 104 x 2.303 x 8.314 J mol-1
= 239.34 k] mol-1

(ii) t1/2 =256 min = 256 x 60 seconds.

Question 28.
The decomposition of A into product has value of k as 4.5 × 10-3 s-1 at 10°C and energy 60kJ mol-1. At what temparature would k be 1.5 × 104 s-1 ?
Answer:

Question 29.
The time required for 10% completion of a first order reaction at 298 k is equal to that required for its 25% completion at 308 k. if the value of the A is 4 × 1010 s-1, calculate k at 318 k and Ea
Answer:

Putting A = 4 x 1010 s-1; R = 8.314 x 10-3 x 10-3 JK-1 mol-1,
Ea kJ mol-1, T = 318 k in the above equation,

Question 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:
T1 = 293;
T2 = 313 K
T2 – T1 = 20°
k2 = 4k1