For the reaction $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the partial pressures of $CO$ and $CO_2$ at equilibrium are $2.0 \ atm$ and $4.0 \ atm$ respectively. The value of $K_p$ for the reaction is:

  • A
    $0.5$
  • B
    $4$
  • C
    $8$
  • D
    $1$

Explore More

Similar Questions

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 $2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$ is:

$5.1 \ g$ $NH_4SH$ is introduced in a $3.0 \ L$ evacuated flask at $327 \ ^\circ C$. $30\%$ of the solid $NH_4SH$ decomposes into $NH_3$ and $H_2S$ gases. The $K_p$ of the reaction at $327 \ ^\circ C$ is ($R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,molar mass of $S = 32 \ g \ mol^{-1}$,molar mass of $N = 14 \ g \ mol^{-1}$)

For the reaction: $HI_{(g)} \rightleftharpoons \frac{1}{2}H_{2(g)} + \frac{1}{2}I_{2(g)}$,the equilibrium constant is $8$. What is the equilibrium constant for the reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$?

At a certain temperature,only $50\%$ $HI$ is dissociated into $H_2$ and $I_2$ at equilibrium. The equilibrium constant is:

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:

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