(A) The given reaction is: $N_2 + O_2 \rightleftharpoons 2 NO$The equilibrium constant expression is given by:$K_C = \frac{[NO]^2}{[N_2][O_2]}$Given e

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(A) The given reaction is: $N_2 + O_2 \rightleftharpoons 2 NO$The equilibrium constant expression is given by:$K_C = \frac{[NO]^2}{[N_2][O_2]}$Given equilibrium concentrations:$[N_2] = 4 \times 10^{-3} \ M$$[O_2] = 3 \times 10^{-3} \ M$$[NO] = 3 \times 10^{-3} \ M$Substituting the values into the expression:$K_C = \frac{(3 \times 10^{-3})^2}{(4 \times 10^{-3})(3 \times 10^{-3})}$$K_C = \frac{9 \times 10^{-6}}{12 \times 10^{-6}}$$K_C = \frac{9}{12} = 0.75$

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

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What is the equilibrium constant $(K_C)$ for the given reaction?
$N_2 + O_2 \rightleftharpoons 2 NO$
Where the equilibrium concentrations of $N_2$,$O_2$ and $NO$ are found to be $4 \times 10^{-3} \ M$,$3 \times 10^{-3} \ M$ and $3 \times 10^{-3} \ M$ respectively.

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A
$0.75$
B
$0.622$
C
$9 \times 10^{-3}$
D
$12.8 \times 10^{-6}$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

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What is the equilibrium constant $(K_C)$ for the given reaction?$N_2 + O_2 \rightleftharpoons 2 NO$Where the equilibrium concentrations of $N_2$,$O_2$ and $NO$ are found to be $4 \times 10^{-3} \ M$,$3 \times 10^{-3} \ M$ and $3 \times 10^{-3} \ M$ respectively.

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For the reaction $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,the units of $K_c$ are

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 $O_{3} ?$

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What is the equilibrium constant $(K_C)$ for the given reaction?
$N_2 + O_2 \rightleftharpoons 2 NO$
Where the equilibrium concentrations of $N_2$,$O_2$ and $NO$ are found to be $4 \times 10^{-3} \ M$,$3 \times 10^{-3} \ M$ and $3 \times 10^{-3} \ M$ respectively.