What is the relationship between $K_{p}$ and $K_{c}$ when $\Delta n = 0$,$\Delta n > 0$,and $\Delta n < 0$?
The relationship is given by the equation $K_{p} = K_{c}(RT)^{\Delta n}$.
$1$. If $\Delta n = 0$,then $K_{p} = K_{c}(RT)^{0} = K_{c}$.
Example: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,where $\Delta n = 2 - (1+1) = 0$.
$2$. If $\Delta n > 0$,then $K_{p} = K_{c}(RT)^{\text{positive value}}$,which implies $K_{p} > K_{c}$.
Example: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,where $\Delta n = (1+1) - 1 = +1$.
$3$. If $\Delta n < 0$,then $K_{p} = K_{c}(RT)^{\text{negative value}}$,which implies $K_{p} < K_{c}$.
Example: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,where $\Delta n = 2 - (1+3) = -2$.
$1$. If $\Delta n = 0$,then $K_{p} = K_{c}(RT)^{0} = K_{c}$.
Example: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,where $\Delta n = 2 - (1+1) = 0$.
$2$. If $\Delta n > 0$,then $K_{p} = K_{c}(RT)^{\text{positive value}}$,which implies $K_{p} > K_{c}$.
Example: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,where $\Delta n = (1+1) - 1 = +1$.
$3$. If $\Delta n < 0$,then $K_{p} = K_{c}(RT)^{\text{negative value}}$,which implies $K_{p} < K_{c}$.
Example: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,where $\Delta n = 2 - (1+3) = -2$.
Explore More
Similar Questions
Value of $K_{p}$ for the equilibrium reaction $N_{2}O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ at $288 \ K$ is $47.9$. The $K_{C}$ for this reaction at same temperature is $......$ (Nearest integer)
$(R=0.083 \ L \ \text{bar} \ K^{-1} \ mol^{-1})$
$(R=0.083 \ L \ \text{bar} \ K^{-1} \ mol^{-1})$
For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,the equilibrium constant is $K$. What will be the equilibrium constant for the reaction $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$?
Medium
View SolutionExplain why the partial pressure $p$ of a gas is proportional to its concentration $c$.
Medium
View SolutionOne mole of $PCl_{5(g)}$ was heated in a $1 \ L$ closed flask at $500 \ K$. At equilibrium,$0.1 \ mole$ of $Cl_{2(g)}$ was formed. What is its $K_{p}$ (in $atm$)? (Given $R=0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)
Find out the value of $K_{c}$ for each of the following equilibria from the value of $K_{p}$:
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$; $K_{p} = 1.8 \times 10^{-2}$ at $500 \ K$
$(ii)$ $CaCO_{3(s)} \longleftrightarrow CaO_{(s)} + CO_{2(g)}$; $K_{p} = 167$ at $1073 \ K$
$(i)$ $2 NOCl_{(g)} \longleftrightarrow 2 NO_{(g)} + Cl_{2(g)}$; $K_{p} = 1.8 \times 10^{-2}$ at $500 \ K$
$(ii)$ $CaCO_{3(s)} \longleftrightarrow CaO_{(s)} + CO_{2(g)}$; $K_{p} = 167$ at $1073 \ K$
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo