(C) Given the equilibrium reaction: $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$The equilibrium constant expression is $K_c = \frac{[Z]}{[X][Y]}$Giv

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(C) Given the equilibrium reaction: $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$The equilibrium constant expression is $K_c = \frac{[Z]}{[X][Y]}$Given the relations: $[X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]$From this,we can express $[X]$ and $[Y]$ in terms of $[Z]$:$[X] = \frac{1}{2}[Z]$$[Y] = [Z]$Substituting these into the $K_c$ expression:$10^{-4} = \frac{[Z]}{(\frac{1}{2}[Z])([Z])}$$10^{-4} = \frac{[Z]}{\frac{1}{2}[Z]^2}$$10^{-4} = \frac{2}{[Z]}$$[Z] = \frac{2}{10^{-4}} = 2 \times 10^4 \ M$

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

Medium

For the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$ at $550 \ K$,the value of $K_c$ is $10^{-4} \ mol^{-1} \ L$. If at equilibrium $[X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]$,then the value of $[Z]$ at equilibrium will be:

A
$2 \times 10^{-4} \ M$
B
$1 \times 10^{-4} \ M$
C
$2 \times 10^{4} \ M$
D
$1 \times 10^{4} \ M$

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Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

For a reaction $X + Y \rightleftharpoons 2 Z$,$1.0 \ mol$ of $X$,$1.5 \ mol$ of $Y$ and $0.5 \ mol$ of $Z$ were taken in a $1 \ L$ vessel and allowed to react. At equilibrium,the concentration of $Z$ was $1.0 \ mol \ L^{-1}$. The equilibrium constant of the reaction is $-\frac{x}{15}$. The value of $x$ is........

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Two equilibria,$AB \rightleftharpoons A^{+} + B^{-}$ and $AB + B^{-} \rightleftharpoons AB_2^-$,are simultaneously maintained in a solution with equilibrium constants $K_1$ and $K_2$ respectively. The ratio of $[A^{+}]$ to $[AB_2^-]$ in the solution is

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In an equilibrium mixture at $1 \, atm$ pressure and $25 \, ^oC$,the partial pressures of $N_2O_4$ and $NO_2$ are $0.70 \, atm$ and $0.30 \, atm$ respectively. The partial pressure of $N_2O_4$ in the equilibrium mixture at $9 \, atm$ pressure and $25 \, ^oC$ will be ........... $atm$.

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At $T(K)$,the $K_p$ for the reaction $A_2B_{6(g)} \rightleftharpoons A_2B_{4(g)} + B_{2(g)}$ is $0.04 \text{ atm}$. The equilibrium pressure (in $\text{atm}$) of $A_2B_{6(g)}$ when it is placed in a flask at $4 \text{ atm}$ pressure and allowed to come to equilibrium is:

MediumAP EAMCET 2023

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One mole of $N_2O_4(g)$ is taken in a closed vessel at $300 \, K$ under $1 \, atm$ pressure. When it is heated to $600 \, K$,$20 \%$ of $N_2O_4(g)$ dissociates into $NO_2(g)$. What is the final pressure in $atm$?

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For the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$ at $550 \ K$,the value of $K_c$ is $10^{-4} \ mol^{-1} \ L$. If at equilibrium $[X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]$,then the value of $[Z]$ at equilibrium will be:

Similar Questions

Two equilibria,$AB \rightleftharpoons A^{+} + B^{-}$ and $AB + B^{-} \rightleftharpoons AB_2^-$,are simultaneously maintained in a solution with equilibrium constants $K_1$ and $K_2$ respectively. The ratio of $[A^{+}]$ to $[AB_2^-]$ in the solution is

In an equilibrium mixture at $1 \, atm$ pressure and $25 \, ^oC$,the partial pressures of $N_2O_4$ and $NO_2$ are $0.70 \, atm$ and $0.30 \, atm$ respectively. The partial pressure of $N_2O_4$ in the equilibrium mixture at $9 \, atm$ pressure and $25 \, ^oC$ will be ........... $atm$.

At $T(K)$,the $K_p$ for the reaction $A_2B_{6(g)} \rightleftharpoons A_2B_{4(g)} + B_{2(g)}$ is $0.04 \text{ atm}$. The equilibrium pressure (in $\text{atm}$) of $A_2B_{6(g)}$ when it is placed in a flask at $4 \text{ atm}$ pressure and allowed to come to equilibrium is:

One mole of $N_2O_4(g)$ is taken in a closed vessel at $300 \, K$ under $1 \, atm$ pressure. When it is heated to $600 \, K$,$20 \%$ of $N_2O_4(g)$ dissociates into $NO_2(g)$. What is the final pressure in $atm$?

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