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The concentrations of a species A undergoing thereaction A--->P is 1.0,0.5,0.33,0.25 mol/dm3 at t=0,1,2and 3 seconds, respectively. The order of the reaction is?
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The concentrations of a species A undergoing thereaction A--->P is 1.0...
Solution:

Given, the concentrations of species A at different times are:

- t=0 s: [A] = 1.0 mol/dm3
- t=1 s: [A] = 0.5 mol/dm3
- t=2 s: [A] = 0.33 mol/dm3
- t=3 s: [A] = 0.25 mol/dm3

We need to determine the order of the reaction from these concentration values.

Definition of Reaction Order:

The order of a reaction is defined as the sum of the exponents of the concentration terms in the rate law equation. The rate law equation is an experimentally determined equation that relates the rate of a reaction to the concentrations of the reactants.

For a general reaction,

aA + bB --> cC + dD

the rate law equation can be written as:

rate = k[A]x[B]y

where k is the rate constant, [A] and [B] are the concentrations of the reactants A and B, and x and y are the orders of the reaction with respect to A and B, respectively.

Method to determine order of reaction:

We can determine the order of a reaction experimentally by measuring the rate of the reaction at different concentrations of the reactants. If the rate of the reaction changes by a factor of n when the concentration of a reactant changes by a factor of m, then the order of the reaction with respect to that reactant is given by:

order = log(n) / log(m)

We can apply this method to determine the order of the reaction in this problem.

Calculation of Order of Reaction:

The given reaction is A --> P.

Since there is only one reactant, the rate law equation is:

rate = k[A]x

The rate of the reaction can be determined from the change in concentration of A with time. Using the concentration values given in the problem, we can calculate the rate of the reaction at different times.

- At t=0 s, [A] = 1.0 mol/dm3, so the initial rate is:

rate0 = k[1.0]x = k

- At t=1 s, [A] = 0.5 mol/dm3, so the rate is:

rate1 = k[0.5]x

- At t=2 s, [A] = 0.33 mol/dm3, so the rate is:

rate2 = k[0.33]x

- At t=3 s, [A] = 0.25 mol/dm3, so the rate is:

rate3 = k[0.25]x

Now, we need to compare the rates at different times to determine the order of the reaction.

- If x = 0, then the rate is independent of [A]. In this case, the rate should remain constant over time. However, we see that the rate decreases over time, which rules out the possibility of zero order reaction.

- If x = 1, then the rate is directly proportional to [A]. In this case, the rate should decrease by a factor of 2 when [A] decreases by a factor of 2. We see that the rate decreases by a factor of 2 when [A] decreases from 1.0 mol/dm3 to 0.
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The concentrations of a species A undergoing thereaction A--->P is 1.0,0.5,0.33,0.25 mol/dm3 at t=0,1,2and 3 seconds, respectively. The order of the reaction is?
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The concentrations of a species A undergoing thereaction A--->P is 1.0,0.5,0.33,0.25 mol/dm3 at t=0,1,2and 3 seconds, respectively. The order of the reaction is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The concentrations of a species A undergoing thereaction A--->P is 1.0,0.5,0.33,0.25 mol/dm3 at t=0,1,2and 3 seconds, respectively. The order of the reaction is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The concentrations of a species A undergoing thereaction A--->P is 1.0,0.5,0.33,0.25 mol/dm3 at t=0,1,2and 3 seconds, respectively. The order of the reaction is?.
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