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# Unit of equilibrium constant is:(A) ${{\left( \dfrac{Mol}{l} \right)}^{1-n}}$ (B) ${{\left( \dfrac{Mol}{l} \right)}^{\Delta n}}$ (C) ${{\left( atm \right)}^{\Delta n}}$ (D) All of the above

Last updated date: 18th Jun 2024
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Hint:Equilibrium constant of the reaction is the ratio of concentration of all products to the concentration of all reactants. To find its unit, we can put units of all concentrations of all species mentioned in a case and simplify it mathematically to get the answer.

First of all we should know about equilibrium constant. We denote it by K. we should know that it expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit. We will understand the concept of equilibrium constant by observing the following reaction.
$bB+cC\rightleftharpoons dD+eE$
We can write the value of equilibrium constant expression as follows:
$K=\dfrac{a_{D}^{d}\cdot a_{E}^{e}}{a_{B}^{b}\cdot a_{C}^{c}}$ .......(1)
We should note that the lower case letters in the balanced equation represent the number of moles of each substance; the upper case letters represent the substance itself. If value of K is greater than one, then the equilibrium will favour products and if the value of K is less than one, then the equilibrium will favour reactants.
- Now, coming back to the mentioned question. If we observe the expression of equilibrium constant that is mentioned above, we will find that equilibrium value constant depends on a number of moles of reactants and products. So, we know that concentration is expressed in $\dfrac{mol}{L}$ unit here. So, we can rewrite equation (1) in terms of units as,
$K=\dfrac{{{\left( \dfrac{mol}{L} \right)}^{d}}\cdot {{\left( \dfrac{mol}{L} \right)}^{e}}}{{{\left( \dfrac{mol}{L} \right)}^{b}}\cdot {{\left( \dfrac{mol}{L} \right)}^{c}}}$
So, we can write this as $K={{\left( \dfrac{mol}{L} \right)}^{(d+e)-(b+c)}}$
But, d and e are moles of products and b and c are moles of reactants. So, we can also write above equation as, $K={{\left( \dfrac{mol}{L} \right)}^{(product)-(reactant)}}$
Now, if we suppose that number of moles of product are ${{n}_{product}}$ and number of moles of reactant are ${{n}_{\text{reactant}}}$ , then we can write the above equation as, $K={{\left( \dfrac{mol}{L} \right)}^{\Delta n}}$

Thus (B) ${{\left( \dfrac{Mol}{l} \right)}^{\Delta n}}$ is the correct answer.

Note:
Remember that equilibrium constant is always a unitless quantity but here, this option is not given. So, by our chosen option, when we will put the available values of moles, we will get the answer that the constant is a unitless quantity.