For a reaction $A_{(s)} \rightleftharpoons B_{(s)} + C_{(g)}$,the set of all correct statements are:
$(a) \ K$ is independent of $[A]$.
$(b) \ K$ is dependent on partial pressure of $C$ at a given temperature.
$(c) \ \Delta H$ will be independent of temperature.
$(d) \ \Delta H$ is independent of the catalyst addition.

  • A
    $a, b, c, d$
  • B
    $a, b$ only
  • C
    $a, b, d$ only
  • D
    $a, b, c$ only

Explore More

Similar Questions

From the given data of equilibrium constants for the following reactions:
$(1) \ CO_{2(g)} + H_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)} \ ; \ K_1$
$(2) \ CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)} \ ; \ K_2$
Wait,the provided question text has a typo in the reaction equations. Assuming the standard problem format where we relate equilibrium constants for reverse or combined reactions,if the target reaction is the same as reaction $(1)$,the answer is $K_1$. However,based on the options provided,this is likely a question asking for the relationship between $K_1$ and $K_2$ where reaction $(2)$ is the reverse of reaction $(1)$. If reaction $(2)$ is the reverse of reaction $(1)$,then $K_2 = \frac{1}{K_1}$. Given the options,please re-verify the input. Assuming the question asks for the equilibrium constant of a reaction derived from these,if the target reaction is $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$,the answer is $K_1^{-1}$. Given the options,if we assume the target reaction is the reverse of reaction $(1)$,then $K = \frac{1}{K_1}$.

If the volume of the container for the reaction $2NO + O_2 \rightleftharpoons 2NO_2$ is reduced to half of its initial volume,the rate of the reaction will become .......

The equilibrium constants of the following are
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,; \quad K_1$
$N_2 + O_2 \rightleftharpoons 2NO \,; \quad K_2$
$H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \,; \quad K_3$
The equilibrium constant $(K)$ of the reaction:
$2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O$ is:

$CH_3COCH_{3(g)} \rightleftharpoons C_2H_{6(g)} + CO_{(g)}$. The initial pressure of $CH_3COCH_3$ is $100 \ mm$. When equilibrium is set up,the mole fraction of $CO_{(g)}$ is $\frac{1}{4}$. Hence,the partial pressure of $CO$ is:

Observe the following equations:
$Ag^{+} + NH_3 \rightleftharpoons [Ag(NH_3)]^{+}$,$K_1 = 1.6 \times 10^3$
$[Ag(NH_3)]^{+} + NH_3 \rightleftharpoons [Ag(NH_3)_2]^{+}$,$K_2 = 6.8 \times 10^3$
The equilibrium constant for the following reaction,$Ag^{+} + 2 NH_3 \rightleftharpoons [Ag(NH_3)_2]^{+}$ 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