Rate Law - Practice Questions & MCQ

The rate of any reaction depends upon the concentration of the reactants in rate law equation that is rate law and rate of reaction depend upon the order of the reaction.
Let us consider this reaction:
$\mathrm{xA}+\mathrm{yB} \rightarrow$ Product
Rate law equation for the reaction can be given as:
                            
                      $\begin{aligned} & \mathrm{R} \propto[\mathrm{A}]^{\mathrm{P}}[\mathrm{B}]^{\mathrm{q}} \\ & \mathrm{R}=\mathrm{K}[\mathrm{A}]^{\mathrm{P}}[\mathrm{B}]^{\mathrm{q}}\end{aligned}$
Order of reaction $=p+q$
Here p, q are experimental quantities which may or may not be equal to the respective stoichiometric coefficients ( x, y).

The differential rate expression for n^th order reaction is as follows:
                       $-\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}(\mathrm{a}-\mathrm{x})^{\mathrm{n}}$
$\mathrm{k}=\frac{\mathrm{dx}}{(\mathrm{a}-\mathrm{x})^{\mathrm{n}} \mathrm{dt}}=\frac{(\text { concentration })}{(\text { concentration })^{\mathrm{n}} \text { time }}=(\text { conc. })^{1-\mathrm{n}}$ time $^{-1}$
If concentration be expressed in mole L^-1 and time in minutes, then$\mathrm{k}=\left(\mathrm{mole}^{-1}\right)^{1-\mathrm{n}} \min ^{-1}$

For zero order reaction, $\mathrm{n}=0$ and hence, $\mathrm{k}=\mathrm{moleL}^{-1} \min ^{-1}$
For first order reaction, $\mathrm{n}=1$ and hence,$
\mathrm{k}=\left(\operatorname{mole} \mathrm{L}^{-1}\right)^0 \min ^{-1}=\min ^{-1}
$
For second order reaction, $\mathrm{n}=2$ and hence,

$
\mathrm{k}=\left(\operatorname{mole~}^{-1}\right)^{-1} \min ^{-1}=\text { mole }^{-1} \mathrm{Lmin}^{-1}
$
 

The rate constant of a first-order reaction has only time in its unit. It has no concentration term in the unit. This means the numerical value of k first-order reaction is independent of the unit in which concentration is expressed. If the concentration unit is changed, the numerical value of k for a first-order reaction will not change. However, it would change with change in time.