The figure shows the change in concentration of species $A$ and $B$ as a function of time. The equilibrium constant $K_C$ for the reaction $2A_{(g)} \rightleftharpoons B_{(g)}$ is

At $T \ K$,the equilibrium constant for the reaction $a A_{(g)} \rightleftharpoons b B_{(g)}$ is $K_c$. If the reaction takes place in the following form $2a A_{(g)} \rightleftharpoons 2b B_{(g)}$,its equilibrium constant is $K_c^{\prime}$. The correct relationship between $K_c$ and $K_c^{\prime}$ is

Given the reaction between $2$ gases represented by $A_2$ and $B_2$ to give the compound $AB_{(g)}$:
$A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$
At equilibrium,the concentrations are $[A_2] = 3.0 \times 10^{-3} \, M$,$[B_2] = 4.2 \times 10^{-3} \, M$,and $[AB] = 2.8 \times 10^{-3} \, M$.
If the reaction takes place in a sealed vessel at $527^{\circ}C$,then the value of $K_c$ will be:

$4.5$ moles each of hydrogen and iodine are heated in a sealed $10 \ L$ vessel. At equilibrium,$3$ moles of $HI$ are found. The equilibrium constant for ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$ is

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. If the equilibrium constant $(K_C)$ for the above reaction at $500 \ K$ is $1.79$ and the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $1.41 \ M$ and $1.59 \ M$,respectively,then the concentration of $Cl_2$ is approximately: (in $M$)

How is the direction of a reaction predicted?