For ammonia formation from constituent elements,the expression for $K_{C}$ is

  • A
    $K_{C} = \frac{[NH_3]^2}{[N_2][H_2]^3}$
  • B
    $K_{C} = \frac{[N_2][H_2]^3}{[NH_3]^2}$
  • C
    $K_{C} = \frac{[NH_3]}{[N_2][H_2]}$
  • D
    $K_{C} = [NH_3]^2$

Explore More

Similar Questions

The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$

For the reaction,$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant is $K_1$. The equilibrium constant is $K_2$ for the reaction,$2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)}$. What is $K$ for the reaction,$NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$?

If the equilibrium constant for the formation of $NH_3$ is $K_c$,then the dissociation constant of $NH_3$ at the same temperature is:

Explain the calculation of equilibrium concentrations given the value of the equilibrium constant.

Difficult
View Solution

At $200^{\circ} C$,nitric oxide reacts with oxygen to form nitrogen dioxide as follows: $2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)$,$K_C = 3 \times 10^6$. In a mixture of the three species at equilibrium,we can accurately predict that:

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