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.
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If the equilibrium constant for the reaction ${N_2}_{(g)} + {O_2}_{(g)} \rightleftharpoons 2NO_{(g)}$ is $K$,then what is the equilibrium constant for the reaction $\frac{1}{2}{N_2}_{(g)} + \frac{1}{2}{O_2}_{(g)} \rightleftharpoons NO_{(g)}$?
Medium
View SolutionIn the figure shown below,reactant $A$ (represented by square) is in equilibrium with product $B$ (represented by circle). The equilibrium constant is:
The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:
Medium
View SolutionThe following equilibrium constants are given:
$N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$
$N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$
The equilibrium constant for the oxidation of $2 \text{ mole}$ of $NH_{3}$ to give $NO$ is
$N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$
$N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$
The equilibrium constant for the oxidation of $2 \text{ mole}$ of $NH_{3}$ to give $NO$ is
MediumWBJEE 2018
View SolutionFor the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$,the equilibrium mixture is given. If $1 \ mol$ of $H_2S$,$0.2 \ mol$ of $H_2$,and $0.8 \ mol$ of $S_2$ are taken in a $2 \ L$ vessel,find the value of $K_c$.
Difficult
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