Medium
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
(N/A) The molar concentration of a pure solid or liquid is constant,meaning it is independent of the amount present. For a substance '$X$',the values $[X_{(s)}]$ and $[X_{(l)}]$ remain constant regardless of the quantity taken.
$\therefore \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume}} = \frac{\text{Mass} / \text{Molar mass}}{\text{Volume}} = \frac{\text{Density}}{\text{Molar mass}}$
Since density and molar mass are constant for a pure substance at a given temperature,the concentration is constant.
Because these values are constant,they are incorporated into the equilibrium constant ($K_c$ or $K_p$). Thus,they are omitted from the expression.
Example: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
The expression is $K_c = [CO_{2}]$.
Here,$[CaCO_{3(s)}]$ and $[CaO_{(s)}]$ are treated as $1$.
$\therefore \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume}} = \frac{\text{Mass} / \text{Molar mass}}{\text{Volume}} = \frac{\text{Density}}{\text{Molar mass}}$
Since density and molar mass are constant for a pure substance at a given temperature,the concentration is constant.
Because these values are constant,they are incorporated into the equilibrium constant ($K_c$ or $K_p$). Thus,they are omitted from the expression.
Example: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
The expression is $K_c = [CO_{2}]$.
Here,$[CaCO_{3(s)}]$ and $[CaO_{(s)}]$ are treated as $1$.
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$4.5 \ mol$ each of hydrogen and iodine are heated in a $10 \ L$ closed vessel. At equilibrium,$3 \ mol$ of $HI$ is formed. The equilibrium constant for the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be:
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View SolutionIf the equilibrium constant for $A \rightleftharpoons B + C$ is $K_{eq}^{(1)}$ and that of $B + C \rightleftharpoons P$ is $K_{eq}^{(2)}$,the equilibrium constant for $A \rightleftharpoons P$ is :-
In the reaction,$A + B \rightleftharpoons 2C$,at equilibrium,the concentration of $A$ and $B$ is $0.20 \ mol \ L^{-1}$ each and that of $C$ was found to be $0.60 \ mol \ L^{-1}$. The equilibrium constant of the reaction is
Medium
View Solution$3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$
For the above reaction at $298 \ K$,$K_c$ is found to be $3.0 \times 10^{-59}$. If the concentration of $O_2$ at equilibrium is $0.040 \ M$,then the concentration of $O_3$ in $M$ is ...... .
For the above reaction at $298 \ K$,$K_c$ is found to be $3.0 \times 10^{-59}$. If the concentration of $O_2$ at equilibrium is $0.040 \ M$,then the concentration of $O_3$ in $M$ is ...... .
MediumNEET 2022
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
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