The relation between equilibrium constant $K_p$ and $K_c$ is

  • A
    $K_c = K_p (RT)^{\Delta n}$
  • B
    $K_p = K_c (RT)^{\Delta n}$
  • C
    $K_p = \left( \frac{K_c}{RT} \right)^{\Delta n}$
  • D
    $K_p - K_c = (RT)^{\Delta n}$

Explore More

Similar Questions

The values of $K_p/K_c$ for the following reactions at $300 \ K$ are respectively (At $300 \ K, RT = 24.62 \ dm^3 \ atm \ mol^{-1}$):
$(i) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$
$(ii) \ N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
$(iii) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_p = 41$ at $400 \ K$. Calculate $K_c$ for the following reactions at $400 \ K$:
$(a)$ $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$
$(b)$ $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$(c)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$

Difficult
View Solution

For the reaction $Mg(HCO_3)_{2(s)} \rightleftharpoons MgCO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$,the equilibrium constant $K_p = 64 \ atm^2$. Calculate the total pressure at equilibrium.

Difficult
View Solution

$A + B \rightleftharpoons C + D$. If the final equilibrium concentrations of $A$ and $B$ are equal,and the equilibrium concentration of $D$ is twice that of $A$,what is the equilibrium constant $(K_c)$ of the reaction?

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

Difficult
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