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Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared
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the JEE exam syllabus. Information about Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam.
Find important definitions, questions, meanings, examples, exercises and tests below for Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer?.
Solutions for Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Not all chemical reactions proceed to a stage at which the concentrations of the reactants become vanishingly small. Here we consider the kinetics of such reactions.Let a reaction be represented in general terms by the schemewhere k., and k_., represent the rate constants for the forwards and reverse reactions, respectively. The equilibrium constant for this reaction may be written asK = [B] /[A] = K1/K1 --(1)where the subscript 00 refers to a time t, sufficiently long to establish equilibrium at the given temperature. The initial concentration of species A is [A]O, and that of B is [B]O . After a time t, let the concentration of species A be [A/t and that of B be [Bat. The total rate of change of [At is given byd[A]t/dt= - K1[A]t + k-1[B]tIf, as is usual, [B]O is initially zero, it follows from a mass balance that at any time t,[Bat = [A]0 - [A]t, whenced[A]t/dt = K1[A]t + k-1([A]O - [A]t)ord[A]t/dt = (K1 + k-1)For the reaction (having both 1ts order reactions), the concentration as a function of time are given for a certain experimental run along with a tangent to the graph at the origin. The ratio of the magnitude of the slopes of the graph of [A/ and [C/ at the origin would be :a)1b)3/4c)4/3d)None of these.Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.