Question

Unit of equilibrium constant for the reversible reaction \[{{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI~\] is
A. $mo{{l}^{-1}} litre$
B. $mo{{l}^{-2}} litre$
C. $mol.lite{{r}^{-1}}$
D. none of these

Answer 1

Hint: To solve this question we have to know about equilibrium constant and unit of equilibrium constant. In an equilibrium reaction the rate constant of the reaction is given as the ratio of concentration of product to the concentration of reactant. Any coefficient of the reactant or product is used as the power of the concentration. The unit of concentration is given as $mol.lite{{r}^{-1}}$.

Complete Step by Step Solution:
The given reaction is \[{{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI~\].
In this reaction one mole of hydrogen gas reacts with one mole of iodine gas to produce two moles of hydrogen iodide gas. It is a unimolecular reaction as the change in number of moles is zero.

It is an equilibrium reaction as the product can also decompose to give back the reactants hydrogen gas and iodine gas. As in this reaction the number of moles of reactant is same as the number of moles of the product. So, it has no unit and dimensionless.
Thus the correct option is D.

Note: An equilibrium reaction is that reaction where there is an equilibrium between the reactants and products of the reaction. Here the reaction can proceed in both the forward and backward directions that means reactants react to give product and again product can also break down into the corresponding reactants.