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

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$[NO]$

$\mathrm{mol} \mathrm{L}^{-1}$

${H}_{2}$

$\mathrm{mol} \mathrm{L}^{-1}$

Rate 

$\mathrm{mol}L^{-1}$ $s^{-1}$

$(A)$ $8 \times 10^{-5}$ $8 \times 10^{-5}$ $7 \times 10^{-9}$
$(B)$ $24 \times 10^{-5}$ $8 \times 10^{-5}$ $2.1 \times 10^{-8}$
$(C)$ $24 \times 10^{-5}$ $32 \times 10^{-5}$ $8.4 \times 10^{-8}$

The order of the reaction with respect to $\mathrm{NO}$ is ..... .

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