Question

For a gaseous reaction $2B\to A$ , the equilibrium constant ${{K}_{p}}$ is _____ than ${{K}_{c}}$ at $298K$ .

Answer 1

Hint: Before talking about the answer, you should know what is equilibrium constant, ${{K}_{p}}$ and ${{K}_{c}}$. Equilibrium constant is the ratio of the amount of products to amount of reactants. ${{K}_{p}}$ is the equilibrium constant for partial pressure and ${{K}_{c}}$ is the equilibrium constant for concentration.

Formula used:${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}$
where, ${{K}_{p}}$ is the equilibrium constant at pressure, ${{K}_{c}}$ is the equilibrium constant at volume, $R$ is the universal gas constant, $T$ is the temperature and $\Delta {{n}_{g}}$ is the change in the number of moles in gaseous form.


Complete step by step answer:
For a gaseous reaction,
$2B\rightleftharpoons A$
The change in the number of gaseous moles,
$\Delta {{n}_{g}}$ = Total number of moles in product – Total number of moles in reactant
$ \Delta {{n}_{g}}={{n}_{P}}-{{n}_{R}}$
$ \Delta {{n}_{g}}=1-2$
$ \Delta {{n}_{g}}=-1 \\
According to the formula, ${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}$
Substituting the given values, we get,
${{K}_{p}}={{K}_{c}}{{(RT)}^{-1}}$
Therefore, ${{K}_{p}}$ is less than ${{K}_{c}}$ .

Hence, for a gaseous reaction $2B\to A$ , the equilibrium constant ${{K}_{p}}$ is less than ${{K}_{c}}$ at $298K$ .

Additional information:
${{K}_{p}}$ is defined as the equilibrium constant for the partial pressure of the reaction. ${{K}_{p}}$ is used to express the relationship between partial pressure of products and partial pressure of reactants.
${{K}_{c}}$ is defined as the ratio of concentration of products to that of reactants, each raised to the power equals to stoichiometric coefficients.
Equilibrium constant is defined as the ratio of the products and reactants.
There are several characteristics of equilibrium constant and they are as follows:
The catalyst changes the rate of reaction, but it does not change the value of equilibrium constant.
If the reaction is reversed, then the value of equilibrium constant is the reciprocal of the one in the forward reaction.

Note: The change in number of moles can be calculated from the difference of moles of products and the moles of reactants.
${{K}_{p}}$ and ${{K}_{c}}$ both are unitless quantities.