The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be

If the equilibrium constant for the reaction $2AB \rightleftharpoons A_2 + B_2$ is $49$,then the equilibrium constant for the reaction $AB \rightleftharpoons \frac{1}{2}A_2 + \frac{1}{2}B_2$ will be:

The reaction quotient $Q$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ is given by $Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$. The reaction will proceed from right to left when:

For the reaction $2HI \rightleftharpoons H_2 + I_2$,the equilibrium constant is $K_1 = 0.25$. What is the value of $K_2$ for the reaction $H_2 + I_2 \rightleftharpoons 2HI$?

In a closed vessel,$PCl_{5(g)}$ is obtained by the chemical reaction between $PCl_{3(g)}$ and $Cl_{2(g)}$. If the equilibrium concentrations in this vessel of $PCl_3$,$Cl_2$,and $PCl_5$ at $500 \ K$ are $1.59 \ M$,$1.59 \ M$,and $1.41 \ M$ respectively,then find the equilibrium constant $K_c$ for the reaction: $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$

Explain the calculation of equilibrium concentrations given the value of the equilibrium constant.