The reaction between $N_2$ and $H_2$ to form ammonia has $K_c = 6 \times 10^{-2}$ at the temperature $500 \ ^oC$. The numerical value of $K_p$ for this reaction is

  • A
    $1.5 \times 10^{-5}$
  • B
    $1.5 \times 10^{5}$
  • C
    $1.5 \times 10^{-6}$
  • D
    $1.5 \times 10^{6}$

Explore More

Similar Questions

In a chemical equilibrium,the rate constant of the backward reaction is $7.5 \times 10^{-4}$ and the equilibrium constant is $1.5$. The rate constant of the forward reaction is:

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 $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

For the reaction,$CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$,the ratio $K_p/K_c$ is equal to:

For the reaction $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$,at equilibrium,the mixture of steam and hydrogen contains $40\%$ $H_2$ by volume. Find $K_p$ for the reaction.

$NH_4COONH_{2_{(s)}} \rightleftharpoons 2NH_{3_{(g)}} + CO_{2_{(g)}}$. If the equilibrium pressure is $3 \, atm$ for the above reaction,the $K_p$ for the reaction is:

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