The unit of rate constant K for a third-order | vedclass

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Question

(A) The general formula for the unit of the rate constant $K$ for a reaction of order $n$ is given by: $Unit = \left(\frac{mol}{L}\right)^{1-n} \cdot

Vedclass · Accepted Answer

$\left(\frac{L}{mol}\right)^2 \cdot s^{-1}$

Vedclass · Answer

(A) The general formula for the unit of the rate constant $K$ for a reaction of order $n$ is given by: $Unit = \left(\frac{mol}{L}\right)^{1-n} \cdot s^{-1}$ For a third-order reaction,$n = 3$. Substituting $n = 3$ into the formula: $Unit = \left(\frac{mol}{L}\right)^{1-3} \cdot s^{-1} = \left(\frac{mol}{L}\right)^{-2} \cdot s^{-1}$ This can also be written as: $\left(\frac{L}{mol}\right)^2 \cdot s^{-1}$ Comparing this with the given options,option $A$ is $\left(\frac{L}{mol}\right)^2 \cdot s^{-1}$,which matches our derived result. Therefore,the correct option is $A$.

$[A] \ (mol/L)$	$[B] \ (mol/L)$	Rate $(mol/L \cdot s)$
$0.05$	$0.05$	$1.2 \times 10^{-3}$
$0.10$	$0.05$	$2.4 \times 10^{-3}$
$0.05$	$0.10$	$1.2 \times 10^{-3}$

Experiment	$[Cl_2] \ (M)$	$[NO] \ (M)$	Rate $(mol \ L^{-1} \sec^{-1})$
$I$	$0.05$	$0.05$	$1 \times 10^{-3}$
$II$	$0.15$	$0.05$	$3 \times 10^{-3}$
$III$	$0.05$	$0.15$	$9 \times 10^{-3}$

The unit of rate constant $K$ for a third-order reaction is . . . . . . .

Similar Questions

Calculate the order of the reaction with respect to $A$ and $B$ based on the following data:

$[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$

$0.05$ $0.05$ $1.2 \times 10^{-3}$

$0.10$ $0.05$ $2.4 \times 10^{-3}$

$0.05$ $0.10$ $1.2 \times 10^{-3}$

For three reactions of first,second,and third order,the rate constants are numerically equal. If the concentration of the reactant is the same and greater than $1 \ M$,which of the following is true for the rates of these three reactions?

For an elementary reaction,which of the following is correct?

In the reaction $A + B \to \text{Products}$,if $B$ is taken in excess,then it is an example of

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