
What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) $ {L^{ - 1}}mol{s^{ - 1}} $
(ii) $ Lmo{l^{ - 1}}{s^{ - 1}} $
Answer
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Hint: Order of the reaction is the power dependence of the rate on the concentration of each reactant present in the reaction.
Rate law formula:
Rate= $ k{[A]^x}{[B]^y} $
Where, $ k $ is the proportionality constant, $ A $ and $ B $ are the reactants and $ x $ and $ y $ are the powers of the reactants of $ A $ and $ B $ respectively.
Order of reaction,
Order= $ x + y $
Where, $ x $ and $ y $ are the powers of the reactants.
To find the order of the reaction from the units of the reaction rate constant:
$ k = {M^{(1 - n)}}{t^{ - 1}} = {(mol{L^{ - 1}})^{(1 - n)}}{t^{ - 1}} $
Where, $ k $ is the unit of the reaction rate constant, $ M $ is molarity, $ t $ is time and $ n $ is the order of the reaction.
Complete step by step solution:
First of all, we start with the definition of the order:
Order of a reaction: Order of the reaction is the power dependence of the rate on the concentration of each reactant present in the reaction. It is defined as the sum of the exponents of the reactants taking part in the reaction.
Now, finding the units of the reaction rate constant,
(i) $ {L^{ - 1}}mol{s^{ - 1}} $
Here, first of all we change the unit in the form of $ mol{L^{ - 1}} $ : $ {(mol{L^{ - 1}})^1} $
Now, equating the power of the $ {(mol{L^{ - 1}})^{(1 - n)}}{t^{ - 1}} $ with the $ {L^{ - 1}}mol{s^{ - 1}} $ :
$ 1 - n = 1 $
$ \Rightarrow n = 0 $
Hence, the order of reaction is zero order reaction.
(ii) $ Lmo{l^{ - 1}}{s^{ - 1}} $
Here, first of all we change unit in the form of $ mol{L^{ - 1}} $ : $ {(mol{L^{ - 1}})^{ - 1}} $
Now, equating the power of the $ {(mol{L^{ - 1}})^{(1 - n)}}{t^{ - 1}} $ with the $ Lmo{l^{ - 1}}{s^{ - 1}} $ :
$ 1 - n = - 1 $
$ \Rightarrow n = 2 $
Hence, the order of reaction is second order reaction.
Note:
Reaction rate constant is the proportionality constant of the rate of the reaction between the reactants taking part in the reaction. It is a proportionality constant in the rate law equation.
Rate law formula:
Rate= $ k{[A]^x}{[B]^y} $
Where, $ k $ is the proportionality constant, $ A $ and $ B $ are the reactants and $ x $ and $ y $ are the powers of the reactants of $ A $ and $ B $ respectively.
Order of reaction,
Order= $ x + y $
Where, $ x $ and $ y $ are the powers of the reactants.
To find the order of the reaction from the units of the reaction rate constant:
$ k = {M^{(1 - n)}}{t^{ - 1}} = {(mol{L^{ - 1}})^{(1 - n)}}{t^{ - 1}} $
Where, $ k $ is the unit of the reaction rate constant, $ M $ is molarity, $ t $ is time and $ n $ is the order of the reaction.
Complete step by step solution:
First of all, we start with the definition of the order:
Order of a reaction: Order of the reaction is the power dependence of the rate on the concentration of each reactant present in the reaction. It is defined as the sum of the exponents of the reactants taking part in the reaction.
Now, finding the units of the reaction rate constant,
(i) $ {L^{ - 1}}mol{s^{ - 1}} $
Here, first of all we change the unit in the form of $ mol{L^{ - 1}} $ : $ {(mol{L^{ - 1}})^1} $
Now, equating the power of the $ {(mol{L^{ - 1}})^{(1 - n)}}{t^{ - 1}} $ with the $ {L^{ - 1}}mol{s^{ - 1}} $ :
$ 1 - n = 1 $
$ \Rightarrow n = 0 $
Hence, the order of reaction is zero order reaction.
(ii) $ Lmo{l^{ - 1}}{s^{ - 1}} $
Here, first of all we change unit in the form of $ mol{L^{ - 1}} $ : $ {(mol{L^{ - 1}})^{ - 1}} $
Now, equating the power of the $ {(mol{L^{ - 1}})^{(1 - n)}}{t^{ - 1}} $ with the $ Lmo{l^{ - 1}}{s^{ - 1}} $ :
$ 1 - n = - 1 $
$ \Rightarrow n = 2 $
Hence, the order of reaction is second order reaction.
Note:
Reaction rate constant is the proportionality constant of the rate of the reaction between the reactants taking part in the reaction. It is a proportionality constant in the rate law equation.
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