At $700 \, K$,the equilibrium constant $K_p$ for the reaction $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ is $1.80 \times 10^{-3}$. The numerical value in $mol \, L^{-1}$ of $K_c$ for this reaction at the same temperature will be $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.
- A$3.09 \times 10^{-7} \, mol \, L^{-1}$
- B$5.07 \times 10^{-8} \, mol \, L^{-1}$
- C$8.18 \times 10^{-9} \, mol \, L^{-1}$
- D$9.24 \times 10^{-10} \, mol \, L^{-1}$
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$3.00 \, mol$ of $PCl_5$ kept in $1 \, L$ closed vessel was allowed to attain equilibrium at $380 \, K$. If $K_C = 1.80$,calculate the equilibrium concentrations of $PCl_3$ and $Cl_2$ respectively.
Difficult
View SolutionFor the reaction $SO_{2(g)} + 1/2 O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1$ at $298 \, K$. For the reaction $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$ at the same temperature,the equilibrium constant is $K_2$. Then,which of the following is correct?
Easy
View SolutionAt $600\ ^{\circ}C$,$NH_4COONH_2\ (s) \rightleftharpoons 2NH_3\ (g) + CO_2\ (g)$ has an equilibrium constant $K_p = 3.2 \times 10^2\ bar^3$. Calculate $K_c$. $\left( R = 0.082\ L\ atm\ K^{-1}\ mol^{-1} \right)$
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:
$A_{(g)} \rightleftharpoons B_{(g)}$
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:
$2 \ mol$ of $PCl_5$ is heated in a closed vessel of $2 \ L$ capacity. When equilibrium is attained,$PCl_5$ is $40\%$ dissociated. The equilibrium constant $K_c$ for the reaction will be ........... $mol/L$.
Difficult
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