Medium
For the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$,the equilibrium constant $K$ is given by:
- A$K = \frac{[H_3O^{+}][H_2O]}{[CH_3COO^{-}][CH_3COOH]}$
- B$K = \frac{[H_3O^{+}][CH_3COO^{-}]}{[H_2O][CH_3COOH]}$
- C$K = \frac{[H_3O^{+}][H_2O]}{[CH_3COOH][CH_3COO^{-}]}$
- D$K = \frac{[H_2O][CH_3COO^{-}]}{[H_2O][CH_3COOH]}$
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The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$
Medium
View SolutionIf the equilibrium constant for the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $K$,what is the equilibrium constant of $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$?
The equilibrium constant expression for a gas reaction is,
$K_{C} = \frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$
Write the balanced chemical equation corresponding to this expression.
$K_{C} = \frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$
Write the balanced chemical equation corresponding to this expression.
Easy
View SolutionFor ammonia formation from constituent elements,the expression for $K_{C}$ is
The 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
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