Question

The unit of rate constant obeying the rate expression $\text{r = k}\left( \text{A} \right)'{{\left( \text{B} \right)}^{2/3}}$ is:
A. $\text{mo}{{\text{l}}^{-2/3}}\text{litr}{{\text{e}}^{2/3}}\text{tim}{{\text{e}}^{-1}}$
B. $\text{mo}{{\text{l}}^{2/3}}\text{litr}{{\text{e}}^{-2/3}}\text{tim}{{\text{e}}^{-1}}$
C. $\text{mo}{{\text{l}}^{-5/3}}\text{litr}{{\text{e}}^{5/3}}\text{tim}{{\text{e}}^{-1}}$
D. None of these

Answer 1

Hint: For this problem, we have to study about the rate constant and then we have to identify the correct unit of the expression by putting the value of the unit of each species after which we will get the net unit of the rate constant of the given expression.

Complete step by step answer:
- In the given question, we have to identify the correct unit of the given expression of the rate constant among the given options.
- As we know that rate constant is a coefficient which is denoted by 'K'. Rate constant is generally a constant of the proportionality of the concentration of the reactant to the concentration of the product.
- So, as we have to find the unit of rate constant which is denoted as 'k' in the question we write the equation as:
$\text{k = }\dfrac{\text{r}}{\left( \text{A} \right)'{{\left( \text{B} \right)}^{2/3}}}$ …. (1)
- Now, the unit of 'r' or reaction rate is mol/L/s and the unit of the concentration of the reactant and product is mol/L.
- So, by putting the units in the equation (1) we will get:
 $\text{k = }\dfrac{\text{mole}}{\text{litre }\times \text{ sec }}\left( \dfrac{\text{Litre}}{\text{mole}} \right){{\left( \dfrac{\text{Litre}}{\text{mole}} \right)}^{2/3}}$

- Now, by solving the above equation we will get the final and precise unit i.e.
$\text{K = mo}{{\text{l}}^{-2/3}}\text{litr}{{\text{e}}^{2/3}}\text{se}{{\text{c}}^{-1}}$
So, the correct answer is “Option A”.

Note: The rate constant is determined by the order of the reaction which is the sum of the exponents of the concentration of the reaction i.e. the reactant and product. So, the unit of rate constant will be different for zero, first and second order.