In the synthesis of $HI$,the amounts of $H_{2(g)}$,$I_{2(g)}$,and $HI_{(g)}$ at equilibrium were found to be $0.8 \ mol$,$0.8 \ mol$,and $2.4 \ mol$ respectively in a $10 \ L$ vessel. Calculate the equilibrium constant $(K_c)$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ and the equilibrium constant for the reverse reaction.
- A$K_c = 9, K_c' = 0.11$
- B$K_c = 0.11, K_c' = 9$
- C$K_c = 9, K_c' = 0.09$
- D$K_c = 0.09, K_c' = 9$
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View SolutionThe value of equilibrium constant of the reaction $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2_{(g)}} + \frac{1}{2} I_{2_{(g)}}$ is $8.0$. The equilibrium constant of the reaction $H_{2_{(g)}} + I_{2_{(g)}} \rightleftharpoons 2HI_{(g)}$ will be
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View SolutionFind out the value of $K_{c}$ for each of the following equilibria from the value of $K_{p}$:
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Medium
View SolutionOne mole of $A_{(g)}$ is heated to $T(K)$ until the following equilibrium is obtained:
$A_{(g)} \rightleftharpoons B_{(g)}$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium,$0.5 \ mol$ of $A_{(g)}$ is added and heated. The equilibrium is again established. The value of $\frac{[A]}{[B]}$ is:
$A_{(g)} \rightleftharpoons B_{(g)}$
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