Q: In the reaction 2N2O5→4NO2+O2 , initial pressure is 500 atm and rate constant K is 3.38×10−5sec−1 . After 10 minutes the final pressure of N2O5 is:

p0=500 atmK=t2.303log10ptp03.38×10−5=10×602.303logpt500 or 0.00880=logpt500⇒1.02500=490 atm

Q: A+2B→C+D. If −dtd[A]=5×10−4 mol l−1s−11 , then −dtd[B] is:

A+2B→C+Ddt−d[A]=5×10−4−21dtd[B]=25×10−4=2.5×10−4 mol−1sec−1

Q: The reaction 2NO(g)+O2( g)⇌2NO2( g) is of first order. If volume of reaction vessel is reduced to 1/3, the rate of reaction would be:

For following reaction, 2NO(g)+O2( g)→2NO2(g) When the volume of vessel change into 31 then concentration of reactant become three times.The rate of reaction for first order reaction ∝ concentration. So rate of reaction will increases three times.

Q: For a reactions A+B→ product, it was found that rate of reaction increases four times if concentration of 'A ' is doubled, but the rate of reaction remains unaffected. If concentration of 'B' is doubled. Hence, the rate law for the reaction is:

Let the rate of reaction depends on x power of [A]. Then r1=k[A]x and r2=k[2A]x ∴r2r1=[2A]x[A]x=41=(21)2(∵r2=4r1) ∴x=2 . As the reaction rate does not depend upon the concentration of B . Hence, the correct rate law will be rate =K[A]2[B]o or =K[A]2

Q: The approach to the following equilibrium was observed kinetically from both directions.PtCI42–+H2O⇌Pt(H2O)CI3–+CI–at 25oCIt was found that–dtd[PtCI]2−=(4.8×10−5)[PtCI4]2––(2.4×10–3)[Pt(H2O)CI3−][CI−]Calculate the equilibrium constant for the complexation of CI– with Pt(II).

At equilibrium dtd[PtCI4]2−=0Hence, (4.8×10−5)[PtCI4]2–=(2.4×10–3)[Pt(H2O)CI3−][CI−]K=[Pt(H2O)CI3−][CI−]d[PtCI4]2−=4.8×10−52.4×10–3=50

Question 1

Which of these does not influence the rate of reaction?

Accepted Answer

Molecularity of the reaction does not influence the rate of reaction.

Question 2

The rate law for reaction A+2B=C+2D will be:

Accepted Answer

The rate law for an reaction can be given by; rate =K(A)(B)2 , i.e. the powers are raised which are given as coefficient of reactant.

Question 3

In the reaction 2N2​O5​→4NO2​+O2​ , initial pressure is 500 atm and rate constant K is 3.38×10−5sec−1 . After 10 minutes the final pressure of N2​O5​ is:

Accepted Answer

​p0​=500 atmK=t2.303​log10​pt​p0​​3.38×10−5=10×602.303​logpt​500​​ or 0.00880=logpt​500​⇒1.02500​=490 atm

Question 4

The rate law for the reaction Sucrose + Water ⟶[H+]​ Glucose + Fructose is given by:

Accepted Answer

It is a pseudo-unimolecular reaction.

Question 5

A+2B→C+D. If −dtd[A]​=5×10−4 mol l−1s−11 , then −dtd[B]​ is:

Accepted Answer

​A+2B→C+Ddt−d[A]​=5×10−4−21​dtd[B]​=25×10−4​=2.5×10−4 mol−1sec−1​

Question 6

The rate law corresponds to the above data is:

Accepted Answer

Correct answer is A

Question 7

The experimental data for the reaction 2A+B2​ightarrow2AB is:The rate equation for the above data is:

Accepted Answer

Correct answer is A

Question 8

The reaction 2NO(g)+O2​( g)⇌2NO2​( g) is of first order. If volume of reaction vessel is reduced to 1/3, the rate of reaction would be:

Accepted Answer

For following reaction, 2NO(g)​+O2​( g)→2NO2​(g) When the volume of vessel change into 31​ then concentration of reactant become three times.The rate of reaction for first order reaction ∝ concentration. So rate of reaction will increases three times.

Question 9

For a reaction 2A+B→ Products, doubling the initial concentration of both the reactants increases the rate by a factor of 8 , and doubling the concentration of B alone doubles the rate. The rate law for the reaction is:

Accepted Answer

2A+B→ ProductsAccording to question: Rate of reaction of ' A′∝[B] as increase in rate is double when [B] is doubled.Rate of reaction ∝[A][B] as increase in rate is 8 times when concentration of both reactant is doubled. It means that order of reaction is 3 and overall rate reaction should be r=K[A]2[B]

Question 10

For a reactions A+B→ product, it was found that rate of reaction increases four times if concentration of 'A ' is doubled, but the rate of reaction remains unaffected. If concentration of 'B' is doubled. Hence, the rate law for the reaction is:

Accepted Answer

Let the rate of reaction depends on x power of [A]. Then r1​=k[A]x and r2​=k[2A]x ∴r2​r1​​=[2A]x[A]x​=41​=(21​)2(∵r2​=4r1​) ∴x=2 . As the reaction rate does not depend upon the concentration of B . Hence, the correct rate law will be rate =K[A]2[B]o or =K[A]2

Question 11

For the elementary reaction  mA+nB→products A is present in excess and when the concentration of B is changed from 0.01 to 0.02, the rate of reaction becomes 4 times, then the rate law determined experimentally can be

Accepted Answer

A is in excess, rate ∝  [A]0On doubling the concentration of B, rate becomes 4 times,rate ∝ [B]2.

Question 12

The approach to the following equilibrium was observed kinetically from both directions.PtCI42–​+H2​O⇌Pt(H2​O)CI3–​+CI–at 25oCIt was found that–dtd[PtCI]2−​=(4.8×10−5)[PtCI4​]2––(2.4×10–3)[Pt(H2​O)CI3−​][CI−]Calculate the equilibrium constant for the complexation of CI– with Pt(II).

Accepted Answer

At equilibrium  dtd[PtCI4​]2−​=0Hence, (4.8×10−5)[PtCI4​]2–=(2.4×10–3)[Pt(H2​O)CI3−​][CI−]K=[Pt(H2​O)CI3−​][CI−]d[PtCI4​]2−​=4.8×10−52.4×10–3​=50

Question 13

The decomposition of a substance follows first order kinetics. Its concentration is reduced to 81​th of its initial value in 24 minutes. The rate constant of the decomposition process is

Accepted Answer

k=t2.303​loga−xa​=242.303​log81​1​=242.303​log18​min−1

Question 14

In the acid hydrolysis reaction A+H2​O+H+→Products,where [H+]=0.1M and H2​O is present in large excess, the apparent rate constant is 1.5×10−5s−1r=K[A][H2​O][H+]=Kapp​[A]The true rate constant (third-order rate constant) of reaction is

Accepted Answer

K=[H2​O][H+]Kapp​​=(181000​M)(0.1M)1.5×10−5s−1​=2.7×10−6M−2s−1

Question 15

Consider the reaction, 2A+B→ product. When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

Accepted Answer

2A+B⟶ Products
[B] is doubled, half-life didn't change.
Half-life is independent of change in concentration of reactant i.e., first order
First order wrt to A
 Unit for the rate constant =conc(1−n)t−1
=(molL)−1⋅S−1
=Lmol−1 s−1

Question 16

For the reaction, H2​+I2​⟶2HI, the differential rate law is

Accepted Answer

H2​+I2​⟶2HI
 Rate of reaction =dt−d[H2​]​=dt−d[I2​]​=21​dtd[HI]​
 or dt−2d[H2​]​=dt−2d[I2​]​=dtd[HI]​

Question 17

Consider the reaction X+Y→ products. If the initial concentration of X is increased by four times, keeping the concentration of Y constant, the rate of reaction increases four times. When the concentrations of both X and Y becomes four times of their original values, the rate of reaction becomes sixteen times of its original value. The observed rate law is

Accepted Answer

On increasing [X] four times, the rate of the reaction becomes four times, hence the order of the reaction with respect to X is one.
Rate ∝[X]1.
But the rate becomes sixteen times, on increasing the concentrations of both X and Y four times. Hence the order is two with respect to X and Y. Thus, the correct rate law is
 Rate =k[X][Y]

Question 18

For the reaction, 2A+B→C, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is

Accepted Answer

Let the rate equation be k[A]x[B]γ From Ist values,
0.045=k[0.05]x[0.05]γ
From 2nd values

0.090=k[0.10]x[0.05]y
From 3rd values,
0.72=k[0.20]x[0.10]y
On dividing equations (i) by (ii), we get

0.090.045​=[0.100.05​]x
[0.100.05​]1=[0.100.05​]x
∴x=1
Similarly on dividing Eq. (ii) by (iii) we get

0.720.09​=[0.20.1​]x[0.100.05​]γ
0.080.01​=0.20.1​[0.10.05​]γ
0.25=[0.100.05​]γ
0.25=[0.5]γ
[0.5]2=[0.5]γ
y =2
Hence, the rate law for the reaction

Rate =k[A][B]2

Question 19

For a reaction, consider the plot of lnk versus 1/T given in the figure. If the rate constant of this reaction at 400 K lnk is 10−5 s−1, then the rate constant at 500 K is

Accepted Answer

The temperature dependence of a chemical reaction is expressed by Arrhenius equation,
k=Ae−Ea^​/RT
Taking natural logarithm on both sides, the Arrhenius equation becomes,
lnk=lnA−RTEa​​
where, −REa​​ is the slope of the plot and lnA gives the intercept.

Eq. (i) at two different temperatures for a reaction becomes,

lnk1​k2​​=REa​​(T1​1​−T2​1​)
⇒ In the given problem,
T1​=400 K,T2​=500 K,k1​=10−5 s−1,k2​=?

−REa​​ (Slope) =−4606
On substituting all the given values in Eq. (ii), we get
ln10−5k2​​=4606(4001​−5001​)

ln10−5k2​​=2.303
10−5k2​​=10⇒k2​=10−4 s−1

Therefore, rate constant for the reaction at 500 K is 10−4 s−1

Question 20

For the non-stoichiometric reaction, 2A+B⟶C+D, the following kinetic data were obtained in three separate experiments, all at 298 K.The rate law for the formation of C is

Accepted Answer

Rate law equation is given by
dtdc​=k[A]x[B]y
∴dtdc​ for C at various concentrations are

1.2×10−3=k(0.1)x(0.1)y
1.2×10−3=k(0.1)x(0.2)y
2.4×10−3=k(0.2)x(0.1)y

Solving for x [divide Eq. (iii) by Eq. (i)]

1.2×10−32.4×10−3​=k(0.1)x(0.1)yk(0.2)x(0.1)y​
2=(2)x⇒x=1

Solving for y [divide Eq. (iii) by Eq. (i)]

1.2×10−32.4×10−3​=k(0.1)x(0.2)yk(0.2)x(0.1)y​

2=(2)x(0.20.1​)y

Put x=1 in above Eq., we get

2=(2)′(0.20.1​)y
1=(0.5)y
(1)0=(0.5)y⇒y=0

Thus, dtdc​=k[A]1[B]0⇒dtdc​=k[A]1

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Solution for Class 12, Chemistry, Chemical Kinetics

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