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What is the unit for the rate constant of a second order reaction?
A. ${s^{ - 1}}$
B. $mol{L^{ - 1}}$
C. $mol{L^{ - 1}}{s^{ - 1}}$
D. $Lmo{l^{ - 1}}{s^{ - 1}}$
E. $mo{l^2}{L^{ - 2}}{s^{ - 2}}$

Last updated date: 14th Jun 2024
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Hint: In second order reaction the rate of reaction is directly proportional to the product of concentration of reactants and the sum of exponents of reactants is two. Now the constant defining this proportionality is the constant for second order reaction. Form the equation for second order reaction and substitute units of all the variables to get units for constant.

Complete answer: According to rate law, the rate of chemical reaction is given as follows:
$R = k{[A]^x}{[B]^y}$
Where $R$ is the rate of reaction with units $mol{L^{ - 1}}{s^{ - 1}}$, $[A],[B]$ are the concentrations of the reactants each with units $mol{L^{ - 1}}$and $x,y$ are the exponents appropriate for the chemical reaction and dimensionless. Here, $k$ is the constant of reaction. Now when we talk of a second order reaction then, $x + y = 2$, that is the sum of the exponents should be 2. Now generally the exponents are integers but they can be zero if for some reactions, the rate does not depend on concentration of one of the products.
Consider the second order reaction. Then the power over the concentration is two. So we can generally write for units as follows:

$R = k{[C]^2}$ where $C$ is variable with units of concentration. To find units of $k$, we will have following expression:
$k = \dfrac{R}{{{{[C]}^2}}}$
Substituting the units, we get,
$k = \dfrac{{mol{L^{ - 1}}{s^{ - 1}}}}{{{{(mol{L^{ - 1}})}^2}}} = \dfrac{{mol{L^{ - 1}}{s^{ - 1}}}}{{mo{l^2}{L^{ - 2}}}} = mo{l^{ - 1}}L{s^{ - 1}} = Lmo{l^{ - 1}}{s^{ - 1}}$
Comparing the final result with the given choices,
So, the correct answer is “Option D”.

Note: Reaction between nitrogen dioxide and carbon monoxide is second order but the rate of reaction depends only on concentration of nitrogen dioxide and thus exponent over carbon monoxide is zero and nitrogen dioxide is two.