Medium
For the reaction ${N_2}_{(g)} + 2{O_2}_{(g)} \rightleftharpoons 2{NO_2}_{(g)}$,the equilibrium constant is $100$. Find the equilibrium constant for the following reactions:
$(1)$ $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$
$(2)$ ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$
$(1)$ $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$
$(2)$ ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$
(N/A) Given reaction: ${N_2}_{(g)} + 2{O_2}_{(g)} \rightleftharpoons 2{NO_2}_{(g)}$,$K = 100$.
$(1)$ The reaction $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$ is the reverse of the given reaction.
Therefore,$K_1 = \frac{1}{K} = \frac{1}{100} = 0.01$.
$(2)$ The reaction ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$ is half of the reverse reaction.
Therefore,$K_2 = (K_1)^{1/2} = (0.01)^{1/2} = 0.1$.
$(1)$ The reaction $2{NO_2}_{(g)} \rightleftharpoons {N_2}_{(g)} + 2{O_2}_{(g)}$ is the reverse of the given reaction.
Therefore,$K_1 = \frac{1}{K} = \frac{1}{100} = 0.01$.
$(2)$ The reaction ${NO_2}_{(g)} \rightleftharpoons \frac{1}{2}{N_2}_{(g)} + {O_2}_{(g)}$ is half of the reverse reaction.
Therefore,$K_2 = (K_1)^{1/2} = (0.01)^{1/2} = 0.1$.
Explore More
Similar Questions
In the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$. At equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:
Medium
View SolutionFor the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$,the equilibrium constant expression $K_c$ is:
Easy
View SolutionFor the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the concentrations of $N_2O_4$ and $NO_2$ at equilibrium are $4.8 \times 10^{-2} \, mol \, L^{-1}$ and $1.2 \times 10^{-2} \, mol \, L^{-1}$ respectively. Calculate the value of $K_c$.
Easy
View SolutionFor the reaction equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$,the concentrations of $N_2O_4$ and $NO_2$ at equilibrium are $4.8 \times 10^{-2} \ mol \ L^{-1}$ and $1.2 \times 10^{-2} \ mol \ L^{-1}$ respectively. The value of $K_c$ for the reaction is:
MediumAIEEE 2003
View Solution$4$ moles of $A$ are mixed with $4$ moles of $B$. At equilibrium for the reaction $A + B \rightleftharpoons C + D$,$2$ moles of $C$ and $D$ are formed. The equilibrium constant for the reaction will be
Easy
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