Difficult
State the law of chemical equilibrium for an equilibrium process.
(N/A) For a general reversible reaction $aA + bB \rightleftharpoons cC + dD$,where $A$ and $B$ are reactants and $C$ and $D$ are products,the law of chemical equilibrium states that at a given temperature,the product of the concentrations of the products raised to their respective stoichiometric coefficients,divided by the product of the concentrations of the reactants raised to their respective stoichiometric coefficients,is a constant value known as the equilibrium constant $(K_c)$.
The mathematical expression is given by:
$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
Here,$[A], [B], [C],$ and $[D]$ represent the molar concentrations of the species at equilibrium. This relationship is also referred to as the law of mass action.
The mathematical expression is given by:
$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
Here,$[A], [B], [C],$ and $[D]$ represent the molar concentrations of the species at equilibrium. This relationship is also referred to as the law of mass action.
Explore More
Similar Questions
For the reactions:
$A \rightleftharpoons B$ $K_C = 2$
$B \rightleftharpoons C$ $K_C = 3$
$C \rightleftharpoons D + E$ $K_C = 5$
$K_C$ for the reaction $A \rightleftharpoons D + E$ is:
$A \rightleftharpoons B$ $K_C = 2$
$B \rightleftharpoons C$ $K_C = 3$
$C \rightleftharpoons D + E$ $K_C = 5$
$K_C$ for the reaction $A \rightleftharpoons D + E$ is:
Medium
View Solution$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}; K_1$
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_2$
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightleftharpoons NH_{3(g)}; K_3$
$2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}; K_4$
If $K_1 = K_2^x = K_3^y = K_4^z$,then the correct values of $x, y,$ and $z$ are respectively:
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_2$
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightleftharpoons NH_{3(g)}; K_3$
$2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}; K_4$
If $K_1 = K_2^x = K_3^y = K_4^z$,then the correct values of $x, y,$ and $z$ are respectively:
Medium
View SolutionFor the gaseous reaction,equilibrium constant is given:
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
The equilibrium constant for the reaction:
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
The equilibrium constant for the reaction:
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be
Medium
View SolutionAt $T \ K$,the equilibrium constant for the reaction $a A_{(g)} \rightleftharpoons b B_{(g)}$ is $K_c$. If the reaction takes place in the following form $2a A_{(g)} \rightleftharpoons 2b B_{(g)}$,its equilibrium constant is $K_c^{\prime}$. The correct relationship between $K_c$ and $K_c^{\prime}$ is
For the reaction ${N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$,the equilibrium constant ${K_p = 35}$ at a given temperature. Calculate the values of ${K_p}$ for the following reactions at the same temperature:
$(i) \ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$
$(ii) \ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)$
$(i) \ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$
$(ii) \ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)$
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo