Medium
For the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$,the equilibrium constant is $K$. For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,the equilibrium constant is $K'$. Then $K'$ is equal to:
- A$K^2$
- B$K^{1/2}$
- C$1/K^{1/2}$
- D$1/K^2$
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In the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$. At equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:
Medium
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Medium
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