Medium
The equilibrium constant for the equilibrium $2HX_{(g)} \rightleftharpoons H_{2(g)} + X_{2(g)}$ is $1 \times 10^{-5}$. What will be the equilibrium concentration of $HX$ if the equilibrium concentrations for $H_2$ and $X_2$ are $1.2 \times 10^{-3} \ M$ and $1.2 \times 10^{-4} \ M$ respectively?
- A$12 \times 10^{-4} \ M$
- B$12 \times 10^{-3} \ M$
- C$12 \times 10^{-2} \ M$
- D$12 \times 10^{-1} \ M$
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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
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View SolutionConsider the following reactions in which all the reactants and products are in gaseous state:
$2PQ \rightleftharpoons P_2 + Q_2\,;\,K_1 = 2.5 \times 10^5$
$PQ + \frac{1}{2}R_2 \rightleftharpoons PQR\,;\,K_2 = 5 \times 10^{-3}$
The value of the equilibrium constant for the reaction:
$\frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 \rightleftharpoons PQR$ is
$2PQ \rightleftharpoons P_2 + Q_2\,;\,K_1 = 2.5 \times 10^5$
$PQ + \frac{1}{2}R_2 \rightleftharpoons PQR\,;\,K_2 = 5 \times 10^{-3}$
The value of the equilibrium constant for the reaction:
$\frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 \rightleftharpoons PQR$ is
Medium
View SolutionFor the reaction $2 H_{2(g)} + 2 NO_{(g)} \rightarrow N_{2(g)} + 2 H_2O_{(g)}$,the observed rate expression is $rate = k_f [NO]^2 [H_2]$. The rate expression of the reverse reaction is:
The equilibrium concentration of $x$,$y$ and $yx_2$ are $4$,$2$ and $2$ respectively for the equilibrium $2x + y \rightleftharpoons yx_2$. The value of equilibrium constant,$K_C$ is
Medium
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