For the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$,the equilibrium constant expression $K_c$ is:

At $490\,^{\circ}C$,the equilibrium constant for the synthesis of $HI$ is $50$. The value of $K$ for the dissociation of $HI$ will be:

Explain the reaction quotient and how it is used to predict the direction of a reaction.

Observe the following equilibrium in a $1 \text{ L}$ flask. $A_{(g)} \rightleftharpoons B_{(g)}$. At $T \text{ K}$,the equilibrium concentrations of $A$ and $B$ are $0.5 \text{ M}$ and $0.375 \text{ M}$ respectively. $0.1 \text{ moles}$ of $A$ is added into the flask and heated to $T \text{ K}$ to establish the equilibrium again. The new equilibrium concentrations (in $\text{M}$) of $A$ and $B$ are respectively.

The reaction,$2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$ is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 \ M$. When equilibrium is reached,the concentration of $D$ is measured and found to be $0.25 \ M$. The value for the equilibrium constant for this reaction is given by the expression:

In the reaction $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$,one mole of $AB$ reacts with one mole of $CD$. When equilibrium is established,$3/4$ mole of each $AB$ and $CD$ are converted into $AD$ and $CB$. If the volume remains constant,the equilibrium constant for the reaction will be........