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Rate law and rate constant

For the non-stoichiometric reaction, 2 A+B C+D, the following

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

Name: Video solution 1: 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
Uploaded: 2022-12-22T17:20:19.000Z
Duration: 12 min
Description: Video solution 1: 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

\frac{d c}{d t} = k [A] [B]

\frac{d c}{d t} = k [A]^{2} [B]

\frac{d c}{d t} = k [A] [B]^{2}

\frac{d c}{d t} = k [A]

Views: 5,772 students

Updated on: Dec 22, 2022

Book:

Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2014

Found 6 tutors discussing this question

Camila Discussed

For the non-stoichiometric reaction,

2 A + 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

14 mins ago

Discuss this question LIVE

14 mins ago

Text solutionVerified

Rate law equation is given by

\frac{d c}{d t} = k [A]^{x} [B]^{y}

∴ \frac{d c}{d t}

for

C

at various concentrations are

1.2 \times 1 0^{- 3} = k (0.1)^{x} (0.1)^{y}

1.2 \times 1 0^{- 3} = k (0.1)^{x} (0.2)^{y}

2.4 \times 1 0^{- 3} = k (0.2)^{x} (0.1)^{y}

Solving for

x

[divide Eq. (iii) by Eq. (i)]

\frac{2.4 \times 1 0 ^{- 3}}{1.2 \times 1 0 ^{- 3}} = \frac{k ( 0.2 ) ^{x} ( 0.1 ) ^{y}}{k ( 0.1 ) ^{x} ( 0.1 ) ^{y}}

2 = (2)^{x} \Rightarrow x = 1

Solving for

y

[divide Eq. (iii) by Eq. (i)]

\frac{2.4 \times 1 0 ^{- 3}}{1.2 \times 1 0 ^{- 3}} = \frac{k ( 0.2 ) ^{x} ( 0.1 ) ^{y}}{k ( 0.1 ) ^{x} ( 0.2 ) ^{y}}

2 = (2)^{x} (\frac{0.1}{0.2})^{y}

Put

x = 1

in above Eq., we get

2 = (2)^{'} (\frac{0.1}{0.2})^{y}

1 = (0.5)^{y}

(1)^{0} = (0.5)^{y} \Rightarrow y = 0

Thus,

\frac{d c}{d t} = k [A]^{1} [B]^{0} \Rightarrow \frac{d c}{d t} = k [A]^{1}

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Question Text	For the non-stoichiometric reaction, $2 A + B ⟶ C + D$ , the following kinetic data were obtained in three separate experiments, all at $298 K$ . The rate law for the formation of $C$ is
Updated On	Dec 22, 2022
Topic	Chemical Kinetics
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	121
Avg. Video Duration	12 min

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

Text solutionVerified

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