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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Compounds ‘A’ and ‘B’ react according to the following chemical equation. A(g)+2B(g)⟶2C(g)Concentration of either ‘A’ or ‘B’ were changed keeping the

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प्रश्न

Compounds ‘A’ and ‘B’ react according to the following chemical equation.
\[\ce{A(g) + 2B(g) -> 2C(g)}\]
Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.

Experiment Initial
concentration
of [A]/mol L¹
Initial
concentration
of [B]/mol L¹
Initial rate of
formation of
[C]/mol L¹ s¹
1. 0.30 0.30 0.10
2. 0.30 0.60 0.40
3. 0.60 0.30 0.20

विकल्प

  • Rate = k[A]2[B]

  • Rate = k[A][B]2

  • Rate = k[A][B]

  • Rate = k[A]2[B]0 

MCQ
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उत्तर १

Rate = k[A][B]

Explanation:

Rate = k[A]x[B]y

When concentration of B is doubled keeping the concentration of A constant, the rate of formation of C increases by a factor of four. This indicates that the rate of reactions depends upon the square of concentration of B. When concentration of A is doubled, the rate of formation of C also doubles from the initial value. This shows that the rate depends on first power of concentration of A. Hence Rate= k[A]x[B]y.

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उत्तर २

Let order of A and B be x and y.

r = k [A]x [B]y

0.1 = k (0.3)x (0.3)y  ...(1)

0.4 = 0.1 = k (0.3)x (0.6)y   ...(2)

0.2 = 0.1 = k (0.6)x (0.3)y   ...(3)

Divide 2 by 1

`0.4/0.1 = (0.6)^"y"/(0.3)^"y"`

Divide 3 by 1

`0.2/0.1 = (0.6)^"x"/(0.3)^"x"`

Hence Rate law is

r = k [A]1 [B]2

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अध्याय 4: Chemical Kinetics - Exercises [पृष्ठ ५१]

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एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
अध्याय 4 Chemical Kinetics
Exercises | Q I. 17. | पृष्ठ ५१

संबंधित प्रश्न

A → B is a first order reaction with rate 6.6 × 10-5m-s-1. When [A] is 0.6m, rate constant of the reaction is

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  • 9 × 10-5s-1
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(ii) How is the rate affected on increasing the concentration of A three times?

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