(C) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.For $K_p = K_c$,the change in the number of moles of gaseous pro

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$

(C) The relationship between $K_p$ and $K_c$ is given by $K_p = K_c(RT)^{\Delta n_g}$.For $K_p = K_c$,the change in the number of moles of gaseous products and reactants must be zero,i.e.,$\Delta n_g = 0$.In option $(C)$,the reaction is $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.Here,$\Delta n_g = (2) - (1 + 1) = 0$.Therefore,$K_p = K_c$ for this reaction.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

Medium

For which of the following reactions will $K_p = K_c$?

A
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
B
$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$
C
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$
D
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

6-1.Equilibrium (Chemical Equilibrium)

Subtopic

Kp and Kc Relationship

When one mole of $A$ and one mole of $B$ were heated in a one-litre flask at $T \ K$,$0.5 \ mole$ of $C$ was formed at equilibrium for the reaction $A + B \rightleftharpoons C + D$. The equilibrium constant,$K_C$,is:

EasyTS EAMCET 2015

View Solution

What is the value of the equilibrium constant $K_p$ for the reaction ${H_2O_{(l)}} \rightleftharpoons {H_2O_{(g)}}$ at $60^{\circ}C$? (The vapor pressure of water at $60^{\circ}C$ is $0.185 \ bar$.)

Medium

View Solution

For which of the following equilibria are $K_P$ and $K_C$ different?

Difficult

View Solution

In the reversible reaction $A + B \rightleftharpoons C + D$,the concentration of each $C$ and $D$ at equilibrium was $0.8 \ mol/L$. If the initial concentration of $A$ and $B$ was $1 \ mol/L$ each,then the equilibrium constant $K_c$ will be:

Medium

View Solution

Partial pressures of $A$,$B$,$C$,and $D$ for the gaseous system $A + 2B \rightleftharpoons C + 3D$ are $A = 0.20 \ atm$,$B = 0.10 \ atm$,$C = 0.30 \ atm$,and $D = 0.50 \ atm$. The numerical value of the equilibrium constant $(K_p)$ is:

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

For which of the following reactions will $K_p = K_c$?

Similar Questions

When one mole of $A$ and one mole of $B$ were heated in a one-litre flask at $T \ K$,$0.5 \ mole$ of $C$ was formed at equilibrium for the reaction $A + B \rightleftharpoons C + D$. The equilibrium constant,$K_C$,is:

What is the value of the equilibrium constant $K_p$ for the reaction ${H_2O_{(l)}} \rightleftharpoons {H_2O_{(g)}}$ at $60^{\circ}C$? (The vapor pressure of water at $60^{\circ}C$ is $0.185 \ bar$.)

For which of the following equilibria are $K_P$ and $K_C$ different?

In the reversible reaction $A + B \rightleftharpoons C + D$,the concentration of each $C$ and $D$ at equilibrium was $0.8 \ mol/L$. If the initial concentration of $A$ and $B$ was $1 \ mol/L$ each,then the equilibrium constant $K_c$ will be:

Partial pressures of $A$,$B$,$C$,and $D$ for the gaseous system $A + 2B \rightleftharpoons C + 3D$ are $A = 0.20 \ atm$,$B = 0.10 \ atm$,$C = 0.30 \ atm$,and $D = 0.50 \ atm$. The numerical value of the equilibrium constant $(K_p)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module