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  • The unit of rate constant for the second order reaction will be:Option: 1 <img alt="\mathrm{L}^{2} \mathrm{mol}^{-2} \mathrm{t}^{-1}" src="https://le

The unit of rate constant for the second order reaction will be:

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

\mathrm{t}^{-1}


Answers (1)

best_answer

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}}=\frac{(\text { concentration })}{(\text { concentration })^{\mathrm{n}} \text { time }}=(\text { conc. })^{1-\mathrm{n}} \text { time }^{-1}
If concentration be expressed in mole L-1 and time in minutes, then
\mathrm{k}=\left(\text { mole } \mathrm{L}^{-1}\right)^{1-\mathrm{n}} \mathrm{min}^{-1}

The unit of rate constant for the second order reaction will be: \mathrm{L} \mathrm{\: mol}^{-1} \mathrm{t}^{-1}

Therefore, option(2) is correct

Posted by

Ritika Kankaria

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