
Identify the order of a reaction from each of the following units of rate constants \[\left( k \right)\]
(a) \[{\text{mol}}\,{{\text{L}}^{ - 1}}\,{\sec ^1}\]
(b) \[{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{L}}\,{\sec ^1}\]
Answer
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Hint: The order of reaction is determined experimentally. The reaction order is applicable in all the chemical reactions . Order of any reaction indicates the relationship between the rate of a chemical reaction and the concentration of the species taking part in a particular reaction
Complete step by step answer:
(a) Given that, unit of rate constants \[\left( k \right)\] is \[{\text{mol}}\,{{\text{L}}^{ - 1}}\,{\sec ^1}\], which can be also be written as, \[{\text{Mse}}{{\text{c}}^{ - 1}}\]
We know that, for this, we can write the rate as,
\[{\text{rate}}\, = \,{\text{k}}{\left[ {\text{A}} \right]^{\text{n}}}\]
Here, \[{\text{k}}\] = rate constant, A is the rate constant and \[{\text{n}}\] is the exponential factor.
And \[{\text{M}}\,{\text{se}}{{\text{c}}^{ - 1}} = \,{\left[ {\text{A}} \right]^{\text{n}}}\]
So, now we can say that,
\[
{{\text{M}}^{\text{0}}} = \,{\left[ {\text{A}} \right]^{\text{n}}} \\
{\text{n}}\, = \,{\text{0}} \\
\]
Hence we can conclude that the order of the reaction having unit of rate \[{\text{mol}}\,{{\text{L}}^{ - 1}}\,{\sec ^1}\] is 0.
(b) Given that, unit of rate constants \[\left( k \right)\] is\[{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{L}}\,{\sec ^1}\], which can be also be written as, \[{{\text{M}}^{ - 1}}{\text{se}}{{\text{c}}^{ - 1}}\]
We know that, for this, we can write the rate as,
\[{\text{rate}}\, = \,{\text{k}}{\left[ {\text{A}} \right]^{\text{n}}}\]
Here, \[{\text{k}}\]= rate constant, A is the rate constant and \[{\text{n}}\] is the exponential factor
And \[{\text{M}}\,{\text{se}}{{\text{c}}^{ - 1}} = \,{{\text{M}}^{ - 1}}{\sec ^{ - 1}}{\left[ {\text{A}} \right]^{\text{n}}}\]
So, now we can say that,
\[
{{\text{M}}^2} = \,{\left[ {\text{A}} \right]^{\text{n}}} \\
{\text{n}}\, = \,2 \\
\]
Hence we can conclude that the order of the reaction having unit of rate \[{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{L}}\,{\sec ^1}\] is 2.
Note: We know that a reaction can be zero order, first order, pseudo-first order or second order.
In zero order reaction, we know that the rate of reaction does not depend on the concentration of the reactants.
In the first order reaction, we know that the rates depend on the concentration of only one reactant, i.e. the order of reaction is 1.
We know that, in second order reactions, there may exist multiple reactants but out of them only one reactant will be of first-order concentration and the rest of the reactants will have zero-order concentration.
We know that, in a pseudo-first order reaction, the concentration of one reactant remains constant.
We know that the reaction is said to be a second-order reaction when the order of a reaction is 2.
Complete step by step answer:
(a) Given that, unit of rate constants \[\left( k \right)\] is \[{\text{mol}}\,{{\text{L}}^{ - 1}}\,{\sec ^1}\], which can be also be written as, \[{\text{Mse}}{{\text{c}}^{ - 1}}\]
We know that, for this, we can write the rate as,
\[{\text{rate}}\, = \,{\text{k}}{\left[ {\text{A}} \right]^{\text{n}}}\]
Here, \[{\text{k}}\] = rate constant, A is the rate constant and \[{\text{n}}\] is the exponential factor.
And \[{\text{M}}\,{\text{se}}{{\text{c}}^{ - 1}} = \,{\left[ {\text{A}} \right]^{\text{n}}}\]
So, now we can say that,
\[
{{\text{M}}^{\text{0}}} = \,{\left[ {\text{A}} \right]^{\text{n}}} \\
{\text{n}}\, = \,{\text{0}} \\
\]
Hence we can conclude that the order of the reaction having unit of rate \[{\text{mol}}\,{{\text{L}}^{ - 1}}\,{\sec ^1}\] is 0.
(b) Given that, unit of rate constants \[\left( k \right)\] is\[{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{L}}\,{\sec ^1}\], which can be also be written as, \[{{\text{M}}^{ - 1}}{\text{se}}{{\text{c}}^{ - 1}}\]
We know that, for this, we can write the rate as,
\[{\text{rate}}\, = \,{\text{k}}{\left[ {\text{A}} \right]^{\text{n}}}\]
Here, \[{\text{k}}\]= rate constant, A is the rate constant and \[{\text{n}}\] is the exponential factor
And \[{\text{M}}\,{\text{se}}{{\text{c}}^{ - 1}} = \,{{\text{M}}^{ - 1}}{\sec ^{ - 1}}{\left[ {\text{A}} \right]^{\text{n}}}\]
So, now we can say that,
\[
{{\text{M}}^2} = \,{\left[ {\text{A}} \right]^{\text{n}}} \\
{\text{n}}\, = \,2 \\
\]
Hence we can conclude that the order of the reaction having unit of rate \[{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{L}}\,{\sec ^1}\] is 2.
Note: We know that a reaction can be zero order, first order, pseudo-first order or second order.
In zero order reaction, we know that the rate of reaction does not depend on the concentration of the reactants.
In the first order reaction, we know that the rates depend on the concentration of only one reactant, i.e. the order of reaction is 1.
We know that, in second order reactions, there may exist multiple reactants but out of them only one reactant will be of first-order concentration and the rest of the reactants will have zero-order concentration.
We know that, in a pseudo-first order reaction, the concentration of one reactant remains constant.
We know that the reaction is said to be a second-order reaction when the order of a reaction is 2.
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