For the following rate law determine the unit of rate constant. Rate $=-\frac{d[ R ]}{d t}=k[ A ]^{\frac{1}{2}}[ B ]^{2}$

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The total order of reaction $n=\frac{1}{2}+2=\frac{5}{2}=2.5$ Rate $k[\mathrm{~A}]^{\frac{1}{2}}[\mathrm{~B}]^{2}=[\mathrm{R}]^{\frac{5}{2}}$

$\therefore k=\frac{\text { Rate }}{[\mathrm{R}]^{5 / 2}}$

$\therefore$ unit of $k=\frac{\text { unit of rate }}{\text { (unit of concentration) }^{5 / 2}}$

$=\frac{\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{1} \mathrm{~s}^{-1}}{\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)^{\frac{5}{2}}}$

$=\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{1-\frac{5}{2}} \mathrm{~s}^{-1}$

$=\left(\mathrm{mol} \mathrm{L}^{-1}\right)^{-\frac{3}{2}} \mathrm{~s}^{-1}$

$=(\mathrm{mol})^{\frac{-3}{2}}\left(\mathrm{~L}^{-1}\right)^{\frac{-3}{2}} \mathrm{~s}^{-1}$

$=\mathrm{mol}^{\frac{-3}{2}} \mathrm{~L}^{\frac{+3}{2}} \mathrm{~s}^{-1}$

If the order of reaction $=\frac{5}{2}$ then unit of rate constant $k$ is $\mathrm{L}^{\frac{+3}{2}} \mathrm{~mol}^{\frac{-3}{2}} \mathrm{~s}^{-1}$.

Similar Questions

The data for the reaction $A + B \to C$ isThe rate law corresponds to the above data is

Exp.

$[A]_0$

$[B]_0$

Initial rate

$(1)$

$0.012$

$0.035$

$0.10$

$(2)$

$0.024$

$0.070$

$0.80$

$(3)$

$0.024$

$0.035$

$0.10$

$(4)$

$0.012$

$0.070$

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  • [AIPMT 1994]

For a reaction $A \to B$, the rate of reaction quadrupled when the concentration of $A$  is doubled. The rate expression of the reaction is $r = K{(A)^n}$. when the value of $n$ is

For the reaction:

$2 A + B \rightarrow A _{2} B $

the rate $=k[ A ][ B ]^{2}$ with $k =2.0 \times 10^{-6} \,mol ^{-2}\, L ^{2} \,s ^{-1}$. Calculate the initial rate of the reaction when $[ A ]=0.1 \,mol \,L ^{-1},[ B ]=0.2\, mol \,L ^{-1}$. Calculate the rate of reaction after $[A] $ is reduced to $0.06 \,mol\, L ^{-1}$

For a certain reaction : $(A)(g)  \to  B(g)$ Half life for different initial pressures of $A$  is given below

 $\begin{array}{|l|l|l|} \hline {P_{{A_0}}}(atm) & 0.1 & 0.025 \\ \hline {t_{1/2}}(\sec\,\,) & 100 & 50 \\ \hline \end{array}$

The correct statement about order of reaction is

Consider following two reaction,

$A \to {\text{Product ;}}\,\, - \frac{{d[A]}}{{dt}} = {k_1}{[A]^o}$

$B \to {\text{Product ;}}\,\, - \frac{{d[B]}}{{dt}} = {k_2}{[B]}$

Units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as