Chemical Kinetics

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Question
CBSEENCH12011433

A first-order reaction has a specific reaction rate of 10–2 s–1. How much time will it take for 20 g of the reactant to reduce to 5 g?

  • 238.6 second

  • 138.6 second

  • 346.5 second

  • 693.0 second

Solution

B.

138.6 second

straight t subscript 1 divided by 2 end subscript space equals space fraction numerator 0.693 over denominator 10 to the power of negative 2 end exponent end fraction
For the reduction of 20 g of reactant to 5 g, two t1/2 is required.
therefore space straight t space equals space 2 space straight x space fraction numerator 0.693 over denominator 10 to the power of negative 2 end exponent end fraction space second
space equals space 138.6 space second

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Question
CBSEENCH12011172

Activation energy (Ea) and rate constants (k1 and k2) of chemical reaction at two different temperatures (T1 and T2) are related by

  • ln space k subscript 2 over k subscript 1 space equals negative E subscript a over R open parentheses 1 over T subscript 1 minus 1 over T subscript 2 close parentheses
  • ln space k subscript 2 over k subscript 1 space equals negative E subscript a over R open parentheses 1 over T subscript 2 minus 1 over T subscript 1 close parentheses
  • ln space k subscript 2 over k subscript 1 space equals negative E subscript a over R open parentheses 1 over T subscript 2 plus 1 over T subscript 1 close parentheses
  • ln space k subscript 2 over k subscript 1 space equals E subscript a over R open parentheses 1 over T subscript 1 minus 1 over T subscript 2 close parentheses

Solution

D.

ln space k subscript 2 over k subscript 1 space equals E subscript a over R open parentheses 1 over T subscript 1 minus 1 over T subscript 2 close parentheses

According to Arrhenius equation, activation energy (Ea) and rate constants (k1 and k2) of chemical reaction at two different temperatures (T1 and T2) are related as,
ln space k subscript 2 over k subscript 1 space equals E subscript a over R open parentheses 1 over T subscript 1 minus 1 over T subscript 2 close parentheses

Question
CBSEENCH12011401

Consider the reaction

N2 (g) + 3 H2 (g) → 2 NH3 (g) 

The equality relationship between  fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction space and space fraction numerator negative straight d left square bracket straight H subscript 2 right square bracket over denominator dt end fraction is:

  • fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction space equals space minus space 1 third fraction numerator straight d left square bracket straight H subscript 2 right square bracket over denominator dt end fraction
  • fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction space equals space minus space 2 over 3 fraction numerator straight d left square bracket straight H subscript 2 right square bracket over denominator dt end fraction
  • plus fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction space equals space minus 3 over 2 fraction numerator straight d left square bracket straight H subscript 2 right square bracket over denominator dt end fraction
  • fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction space equals space minus space fraction numerator straight d left square bracket straight H subscript 2 right square bracket over denominator dt end fraction

Solution

B.

fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction space equals space minus space 2 over 3 fraction numerator straight d left square bracket straight H subscript 2 right square bracket over denominator dt end fraction

For the reaction,
N2 (g) + 3 H2 (g) → 2 NH3 (g)
The space rate space of space reaction space straight w. straight r. straight t space straight N subscript 2 space space equals negative fraction numerator straight d left square bracket straight N subscript 2 right square bracket over denominator dt end fraction
The space rate space of space reaction space straight w. straight r. straight t space straight H subscript 2 equals space fraction numerator negative 1 over denominator 3 end fraction fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction
The space rate space of space reaction space straight w. straight r. straight t space NH subscript 3 space equals space plus 1 half fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction
Hence comma space at space straight a space fixed space time
fraction numerator straight d left square bracket straight N subscript 2 right square bracket over denominator dt end fraction space equals space minus space 1 third fraction numerator straight d left square bracket straight H subscript 2 right square bracket over denominator dt end fraction
space equals space plus 1 half fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction
plus fraction numerator straight d left square bracket NH subscript 3 right square bracket over denominator dt end fraction space equals space minus space 2 over 3 fraction numerator straight d left square bracket straight H subscript 2 right square bracket over denominator dt end fraction
minus fraction numerator 2 straight d left square bracket straight N subscript 2 right square bracket over denominator 2 end fraction

Question
CBSEENCH12011286

During the kinetic study of the reaction, 2A + B --> C+ D,  following results were obtained 


 

[A]/mol L-

[B]/ mol L-

Initial rate of formation of D/ mol L- min-

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 x10-2

 
based on the above data which one of the following is correct?
  • rate = k[A]2[B]

  • rate = k [A][B]

  • rate = k[A]2[B]2

  • rate = k[A][B]2

Solution

D.

rate = k[A][B]2

Let the order of reaction with respect to A is x and with respect to B is y. Thus,
rate = k[A]x[B]y
For the given cases,


(I) rate = k(0.1)x (0.1)y = 6.0 x 10-3
(II) rate = k (0.3)x (0.2)y = 7.2 x 10-2
(III) rate =k(0.3)x (0.40)y = 2.88 x 10-1
(IV) rate = k(0.4)x (0.1)y = 2.40 x 10-2

On dividing eq. (I) and (IV), we get
open parentheses fraction numerator 0.1 over denominator 0.4 end fraction close parentheses to the power of straight x space open parentheses fraction numerator 0.1 over denominator 0.1 end fraction close parentheses to the power of straight y space equals space open parentheses fraction numerator 6.0 space straight x space 10 to the power of negative 3 end exponent over denominator 2.4 space straight x space 10 to the power of negative 2 end exponent end fraction close parentheses

open parentheses 1 fourth close parentheses to the power of straight x space equals space open parentheses 1 fourth close parentheses to the power of 1

straight x equals space 1

on space dividing space eq. space left parenthesis II right parenthesis space by space left parenthesis III right parenthesis comma space we space get
open parentheses fraction numerator 0.3 over denominator 0.3 end fraction close parentheses to the power of straight x space open parentheses fraction numerator 0.2 over denominator 0.4 end fraction close parentheses to the power of straight y space equals space fraction numerator 7.2 space straight x space 10 to the power of negative 2 end exponent over denominator 2.88 space straight x space 10 to the power of negative 1 end exponent end fraction
open parentheses 1 half close parentheses to the power of straight y space equals space open parentheses 1 half close parentheses squared

straight y equals space 2
Thus comma space rate space law space is
rate space equals space straight k left square bracket straight A right square bracket to the power of 1 left square bracket straight B right square bracket squared space equals space straight k left square bracket straight A right square bracket left square bracket straight B right square bracket squared

Question
CBSEENCH12011274

For an endothermic reaction, energy of activation is Ea and enthalpy of reaction is ΔH (both of these in kJ/mol). Minimum value of Ea will be

  • less than ΔH

  • equal to ΔH

  • more than ΔH

  • equal to zero

Solution

C.

more than ΔH

In endothermic reactions, the energy of reactants is less than that of the products. Potential energy diagram for endothermic reactions is,

Where Ea = activation energy of forwarding reaction
Ea' = activation energy of backwards reaction
ΔH = enthalpy of the reaction
From the above diagram,
Ea = Ea' + ΔH
Thus, Ea > ΔH

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