For the reversible reaction in equilibrium
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
If the rate constant for the forward reaction is $k_1 = 2.1 \times 10^{-3} \ s^{-1}$ and for the backward reaction is $k_2 = 4.2 \times 10^{-4} \ s^{-1}$,then the equilibrium constant $K_c$ for the above reaction is:
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
If the rate constant for the forward reaction is $k_1 = 2.1 \times 10^{-3} \ s^{-1}$ and for the backward reaction is $k_2 = 4.2 \times 10^{-4} \ s^{-1}$,then the equilibrium constant $K_c$ for the above reaction is:
- A$5.0$
- B$2.0$
- C$0.5$
- D$2.5$
Explore More
Similar Questions
For a reversible reaction,the rate constants for the forward and backward reactions are $2.38 \times 10^{-4}$ and $8.15 \times 10^{-5}$ respectively. The equilibrium constant for the reaction is
Easy
View SolutionThe $K_{p}$ value at equilibrium of $SO_3$ formation reaction from $SO_{2(g)}$ and $O_{2(g)}$ is $5 \ atm^{-1}$. What is the equilibrium partial pressure of $O_2$ if the equilibrium pressure of $SO_2$ and $SO_3$ are equal (in $atm$)?
$9.2 \ g$ of $N_2O_{4(g)}$ is taken in a closed $1 \ L$ vessel and heated until the following equilibrium is reached: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,$50\%$ of $N_2O_{4(g)}$ is dissociated. What is the equilibrium constant $K_c$ (in $mol \ L^{-1}$)? (Molecular weight of $N_2O_4 = 92$)
Medium
View SolutionOne mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:
For the reversible reaction in equilibrium:
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
Given $C_0 = C e^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0 = C' e^{-4.2 \times 10^{-4}t}$ for the backward reaction,calculate the equilibrium constant $K_c$ for the above reaction.
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
Given $C_0 = C e^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0 = C' e^{-4.2 \times 10^{-4}t}$ for the backward reaction,calculate the equilibrium constant $K_c$ for the above reaction.
Difficult
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