$K_{c}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $39.0$. In a closed one litre flask,one mole of $A_{2(g)}$ was heated to $T \ K$. What are the concentrations of $A_{2(g)}$ and $B_{2(g)}$ (in $mol \ L^{-1}$) respectively at equilibrium?
- A$0.025, 0.975$
- B$0.975, 0.025$
- C$0.05, 0.95$
- D$0.02, 0.98$
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Similar Questions
If $2 \ mol$ of $H_2$ and $I_2$ are taken initially in a $1 \ L$ vessel,and the equilibrium concentration of $HI$ is $2 \ mol/L$,find the $K_p$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.
Easy
View SolutionOne mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ of water reacted with $CO_{(g)}$ as follows:
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$
Its $K_c$ value is:
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$
Its $K_c$ value is:
The value of $K_P / K_C$ for the reaction at $T(K)$ is:
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons CO_{2(g)}$
The equilibrium constant $(K_p)$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{-4}$ and at $127^{\circ} C$ is $0.60 \times 10^{-4}$. Calculate the mean heat of formation of ammonia per mole in this temperature range. (in $cal$)
DifficultTS EAMCET 2021
View SolutionAt $1000 \ K$,if the equilibrium constant $K_p$ for the reaction $2 \ NOCl_{(g)} \rightleftharpoons 2 \ NO_{(g)} + Cl_{2(g)}$ is $4.157 \times 10^{-4} \ bar$,the $K_c$ (in $mol \ L^{-1}$) is $(R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1})$
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