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The unit,\[mol{L^{ - 1}}{s^{ - 1}}\] is meant for the rate constant of the reaction having the order:
A. 0
B. 2
C. 1
D. 3

Answer
VerifiedVerified
162k+ views
Hint: In general, the units of the rate constant of a reaction are \[mo{l^{1 - n}}{\rm{ }}{{\rm{L}}^{n - 1}}{s^{ - 1}}\] where n is the order of the reaction. Comparing the general units of the rate constant with the units given in the question will allow us to find n. This will be the order of the reaction.

Complete Step by Step Solution:
Consider a reaction \[A \to P\]. Let’s assume the order of the reaction to be \[n\]. Thus, the rate-law expression of the reaction will be
\[r = k{[A]^n}\] where \[r\] is the rate of the reaction and \[k\] is the rate constant. The rate-law expression can be rewritten as follows:

\[r = k{[A]^n}\]
\[ \Rightarrow k = \dfrac{r}{{{{[A]}^n}}}\] … (1)

To find out the units of \[k\], we must substitute the units of \[r\] and \[{[A]^n}\] in equation (1).
\[Unit{\rm{s of k = }}\dfrac{{Units{\rm{ of r}}}}{{{{\left( {Unit{\rm{ of [A]}}} \right)}^n}}}\]
\[ \Rightarrow Units{\rm{ of k = }}\dfrac{{mol{\rm{ }}{{\rm{L}}^{ - 1}}{s^{ - 1}}}}{{{{\left( {mol{\rm{ }}{{\rm{L}}^{ - 1}}} \right)}^n}}}\]
\[ \Rightarrow {\rm{Units\ of\ k = }}\dfrac{{{\rm{mol }}{{\rm{L}}^{ - 1}}{s^{ - 1}}}}{{mo{l^n}{\rm{ }}{{\rm{L}}^{ - n}}}}\]
\[ \Rightarrow Units{\rm{ of k = mo}}{{\rm{l}}^{1 - n}}{\rm{ }}{{\rm{L}}^{n - 1}}{s^{ - 1}}\] … (2)
Equation (2) represents the general units of the rate constant \[k\].

The units of rate constant that we have been given are \[mol{L^{ - 1}}{s^{ - 1}}\]. To find out the order from the units of the rate constant, we must compare the given units with those obtained in equation (2).
\[mo{l^{1 - n}}{\rm{ }}{{\rm{L}}^{n - 1}}{s^{ - 1}} = mol{\rm{ }}{{\rm{L}}^{ - 1}}{s^{ - 1}}\]
On equating the exponents of either \[mol\]or \[L\]units, we get
\[n - 1 = - 1\]
\[ \Rightarrow n = 2\]
Thus, the reaction is of the second order.
Thus, option B is correct.

Note: Equation (2) is a very important formula and needs to be memorised because it finds extensive use throughout chemical kinetics, for example, in certain numerical type questions on integrated rate equations, the order of the reaction will not be given directly. Instead, it must be inferred from the units of the given rate constant using equation (2) in the way described here.