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  • Unit of rate constant(k) for the zero order reaction is:Option: 1 <img alt="\mathrm{mol} \mathrm{\: L}^{-1} \mathrm{t}^{-1}" src="https://learn.caree

Unit of rate constant(k) for the zero order reaction is:

Option: 1

\mathrm{mol} \mathrm{\: L}^{-1} \mathrm{t}^{-1}


Option: 2

\mathrm{mol} \mathrm{\: L}^{2} \mathrm{t}^{-1}


Option: 3

\mathrm t^{-1}


Option: 4

None of above


Answers (1)

As we have learnt,

The differential rate expression for nth order reaction is as follows:
                       -\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}(\mathrm{a}-\mathrm{x})^{\mathrm{n}}
\text { or } \quad \mathrm{k}=-\frac{\mathrm{d} \mathrm{x}}{(\mathrm{a}-\mathrm{x})^{\mathrm{n}} \mathrm{dt}}

\text{Thus, units of K }=\frac{(\text { concentration })}{(\text { concentration })^{\mathrm{n}} \text { time }}=(\text { conc. })^{1-\mathrm{n}} \text { time }^{-1}
\begin{array}{l}{\mathrm{k}=\left(\text { mole } \mathrm{L}^{-1}\right)^{1-\mathrm{n}} t ^{-1}} \\ {\text {For zero order reaction, } \mathrm{n}=0 \text { and hence, } \mathrm{k}=\mathrm{mole\: L}^{-1} \mathrm{t}^{-1}}\end{array}

Therefore, option(1) is correct

Posted by

Sumit Saini

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