For the decomposition reaction $N_2O_4 \rightleftharpoons 2NO_2$ at constant temperature,the equilibrium constant is given by $K_p = \frac{4x^2P}{1 - x^2}$,where $P$ is the total pressure and $x$ is the degree of dissociation. Which of the following statements is correct?

  • A
    $K_p$ increases with an increase in $P$.
  • B
    $K_p$ increases with an increase in $x$.
  • C
    $K_p$ decreases with a decrease in $x$.
  • D
    $K_p$ remains constant with changes in $P$ and $x$.

Explore More

Similar Questions

At $717 \ K$,$1.50 \ mol$ each of hydrogen and iodine are placed in a $10 \ L$ closed vessel. At equilibrium,$1.25 \ mol$ of each hydrogen and iodine remain. The value of the equilibrium constant $K_c$ for the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ at $717 \ K$ is .......

For the reactions:
$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$
$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$
Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to:

Difficult
View Solution

At $1100 \ K$ temperature,$CaCO_{3(s)}$ and $CaO_{(s)}$ are in equilibrium. The pressure of $CO_{2(g)}$ is $2.0 \times 10^5 \ Pa$.
Find the equilibrium constant $(K_p)$ for the reaction: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

At $673 \ K$,for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_c = 0.50$. Calculate $K_p$ at this temperature. (Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

The equilibrium constant for the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$ is $K_c = 4.9 \times 10^{-2}$. The value of $K_c$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ will be

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