Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The value of $\dfrac{{{K}_{p}}}{{{K}_{c}}}$for the following reactions at 300K are, respectively:
\[{{N}_{2}}+{{O}_{2}}\leftrightarrows2NO(g)\]
\[{{N}_{2}}{{O}_{4}}(g)\leftrightarrows2N{{O}_{2}}(g)\]
\[{{N}_{2}}+3{{H}_{2}}\leftrightarrows2N{{H}_{3}}(g)\]
(At 300K, RT=$24.62d{{m}^{3}}atmmo{{l}^{-1}}$)
A- 1, 24.61$d{{m}^{3}}atmmo{{l}^{-1}}$,$606d{{m}^{6}}at{{m}^{2}}mo{{l}^{-2}}$
B-1,$4.1\text{x1}{{\text{0}}^{-2}}d{{m}^{3}}at{{m}^{-1}}mo{{l}^{-1}}$, $606d{{m}^{6}}at{{m}^{2}}mo{{l}^{2}}$
C-$606d{{m}^{6}}at{{m}^{6}}mo{{l}^{2}}$,$1.65\text{x1}{{\text{0}}^{-3}}d{{m}^{3}}at{{m}^{2}}mo{{l}^{-1}}$
D-1,$24.62d{{m}^{3}}atmmo{{l}^{-1}}$,$1.65\text{x1}{{\text{0}}^{-3}}d{{m}^{-6}}at{{m}^{-2}}mo{{l}^{2}}$


seo-qna
Last updated date: 20th Jun 2024
Total views: 394.8k
Views today: 5.94k
Answer
VerifiedVerified
394.8k+ views
Hint: Find out the relation between the gas equilibrium constant and the other data is given in question. The gas equilibrium constant helps to find the relation between the products, and reactants of a chemical reaction.

Complete step by step solution:
-First let us define the relation between the gas equilibrium constant i.e ${{K}_{p}}\text{ and }{{\text{K}}_{c}}$. Here ${{K}_{c}}$ is related to the molar concentrations, whereas ${{K}_{p}}$is related to the partial pressure of the gases. The mathematical relation is given below-
\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\] , where $\Delta {{n}_{g}}$= number of moles of product-number of moles of reactants
-It should be kept in mind that it is applicable only to the gaseous state.
-So, now coming back to the calculation, we will first write the reaction for the calculation of $\Delta {{n}_{g}}$
-\[{{N}_{2}}(g)+{{O}_{2}}(g)\leftrightarrows2NO(g)\]
Here, $\Delta {{n}_{g}}$= number of moles of product-number of moles of reactants = 2-2=0
-Therefore, $\dfrac{{{K}_{p}}}{{{K}_{c}}}={{(RT)}^{\Delta {{n}_{g}}}}$=1
Now, we will write the second reaction which is given below-
\[{{N}_{2}}{{O}_{4}}(g)\leftrightarrows2N{{O}_{2}}(g)\]
Here, $\Delta {{n}_{g}}$= number of moles of product-number of moles of reactants = 2-1=1
-Therefore, $\dfrac{{{K}_{p}}}{{{K}_{c}}}={{(RT)}^{\Delta {{n}_{g}}}}$= (RT) = $24.62d{{m}^{3}}atmmo{{l}^{-1}}$(It is already given in the question)
${{K}_{p}}$ and ${{K}_{c}}$are both different terms. The difference lies in the concentration. ${{K}_{c}}$ is defined by molar concentrations whereas ${{K}_{p}}$ is defined by partial pressure developed by the gases in a closed vessel. The former one is expressed in terms of molarity whereas the latter one is expressed in terms of atmospheric pressure.
Let us consider the reaction-
$2A(g)+B(g)\rightleftarrows 2C(g)$
All are in the gas phase.
Thus, the ${{K}_{p}}$ is given by-
${{K}_{p}}=\dfrac{{{P}_{c}}^{2}}{{{P}_{A}}^{2}{{P}_{B}}} (1)$
Now, as per Ideal Gas equation-PV=nRT
With the rearrangement of terms, we get-
\[P=\dfrac{n}{V}RT\]
Substituting in (1), we have
\[\begin{align}
& {{K}_{P}}=\dfrac{{{[C]}^{2}}{{(RT)}^{2}}}{{{[A]}^{2}}{{(RT)}^{2}}[B]{{(RT)}^{2}}} \\
& {{K}_{p}}=\dfrac{{{[C]}^{2}}}{{{[A]}^{2}}[B]}\times \dfrac{{{(RT)}^{2}}}{{{(RT)}^{2}}(RT)} \\
& {{K}_{p}}=\dfrac{{{K}_{c}}}{RT} \\
& {{K}_{p}}={{K}_{c}}{{(RT)}^{-1}} \\
\end{align}\]
Generally, it is mentioned as-${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$ ,where, $\Delta n$=number of moles of product-number of moles of reactant.
Now, for the third reaction which is given below-
\[{{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g)\]
Here, $\Delta {{n}_{g}}$= number of moles of product-number of moles of reactants = 2-4=-2
- Therefore, $\dfrac{{{K}_{p}}}{{{K}_{c}}}={{(RT)}^{\Delta {{n}_{g}}}}$=${{(RT)}^{-2}}$=$1.65\text{x1}{{\text{0}}^{-3}}d{{m}^{-6}}at{{m}^{-2}}mo{{l}^{2}}$

-Now from the above calculation, it is clear that the correct option is-(D)

Additional information:
The formula of gas equilibrium constants is only applicable to the gaseous state, as its name can suggest only, and these terms are related to the gaseous factors. These both factors show their dependency on the reactants, and products.

Note: There is the difference in reaction quotient (Q) and (${{K}_{p}}$ and ${{K}_{c}}$) together. It should be noted that though the formula is the same, the reaction may not necessarily be at equilibrium. Also,
When (i) K(ii) K>Q- The reaction is favoured towards the product side
(iii) K=Q- The reaction is at equilibrium