Difficult
At constant temperature,the equilibrium constant $(K_p)$ for the decomposition reaction $N_2O_4 \rightleftharpoons 2NO_2$ is expressed by $K_p = \frac{4x^2P}{1 - x^2}$,where $P = \text{pressure}$,$x = \text{extent of decomposition}$. Which one of the following statements is true?
- A$K_p$ increases with increase of $P$
- B$K_p$ increases with increase of $x$
- C$K_p$ increases with decrease of $x$
- D$K_p$ remains constant with change in $P$ and $x$
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Similar Questions
Given the following equilibria:
$K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$
$K_2 : N_2 + O_2 \rightleftharpoons 2NO$
$K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$
The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is:
$K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$
$K_2 : N_2 + O_2 \rightleftharpoons 2NO$
$K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$
The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is:
Medium
View SolutionFor the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,the equilibrium constant is $k$. What is the equilibrium constant for the reaction $2N_2 + 6H_2 \rightleftharpoons 4NH_3$?
Medium
View SolutionFrom equations $1$ and $2$,
$CO_2 \rightleftharpoons CO + \frac{1}{2} O_2 \, [K_{C_1} = 9.1 \times 10^{-12} \, \text{at} \, 1000^{\circ} C] \, \text{(Eq. } i\text{)}$
$H_2O \rightleftharpoons H_2 + \frac{1}{2} O_2 \, [K_{C_2} = 7.1 \times 10^{-12} \, \text{at} \, 1000^{\circ} C] \, \text{(Eq. } ii\text{)}$
The equilibrium constant for the reaction,$CO_2 + H_2 \rightleftharpoons CO + H_2O$ at the same temperature,is
$CO_2 \rightleftharpoons CO + \frac{1}{2} O_2 \, [K_{C_1} = 9.1 \times 10^{-12} \, \text{at} \, 1000^{\circ} C] \, \text{(Eq. } i\text{)}$
$H_2O \rightleftharpoons H_2 + \frac{1}{2} O_2 \, [K_{C_2} = 7.1 \times 10^{-12} \, \text{at} \, 1000^{\circ} C] \, \text{(Eq. } ii\text{)}$
The equilibrium constant for the reaction,$CO_2 + H_2 \rightleftharpoons CO + H_2O$ at the same temperature,is
MediumKVPY 2010
View SolutionFor the gaseous reactions $(I)$ and $(II)$,the equilibrium constants are $X$ and $Y$,respectively.
$I. \frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$
$II. 2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$
Using the above reactions,the equilibrium constant $Z$ for the reaction $(III)$ given below is:
$III. N_2O_{4(g)} \rightleftharpoons N_{2(g)} + 2 O_{2(g)}$
$I. \frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$
$II. 2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$
Using the above reactions,the equilibrium constant $Z$ for the reaction $(III)$ given below is:
$III. N_2O_{4(g)} \rightleftharpoons N_{2(g)} + 2 O_{2(g)}$
DifficultAP EAMCET 2018
View SolutionFor the given hypothetical reactions,the equilibrium constants are as follows:
$X \rightleftharpoons Y ; K_1=1.0$
$Y \rightleftharpoons Z ; K_2=2.0$
$Z \rightleftharpoons W ; K_3=4.0$
The equilibrium constant for the reaction $X \rightleftharpoons W$ is (in $.0$)
$X \rightleftharpoons Y ; K_1=1.0$
$Y \rightleftharpoons Z ; K_2=2.0$
$Z \rightleftharpoons W ; K_3=4.0$
The equilibrium constant for the reaction $X \rightleftharpoons W$ is (in $.0$)
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