At a certain temperature,only $50\%$ $HI$ is dissociated into $H_2$ and $I_2$ at equilibrium. The equilibrium constant is:

The value of $K_{p}$ for the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is $3.0$ at $1000 \ K$. If initially $P_{CO_{2}} = 0.48 \ bar$ and $P_{CO} = 0 \ bar$ and pure graphite is present,calculate the equilibrium partial pressures of $CO$ and $CO_{2}$.

One mole of $N_2O_4$ in a $1 \ L$ flask decomposes to attain the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At the equilibrium the mole fraction of $NO_2$ is $1/2$. Hence $K_C$ will be:

At $1000 \ K$,the partial pressures of $CO_{2(g)}$ and $CO_{(g)}$ for the reaction $CO_{2(g)} + C_{(s)} \rightleftharpoons 2 CO_{(g)}$ in a closed vessel at equilibrium are $0.15 \ bar$ and $0.60 \ bar$ respectively. The $K_c$ for this reaction at the same temperature is approximately

When one mole of $A$ and one mole of $B$ were heated in a one-litre flask at $T \ K$,$0.5 \ mole$ of $C$ was formed at equilibrium for the reaction $A + B \rightleftharpoons C + D$. The equilibrium constant,$K_C$,is:

Consider the following reaction in a $1 \ L$ closed vessel: $N_2 + 3H_2 \rightleftharpoons 2NH_3$. If all the species $N_2, H_2$ and $NH_3$ are $1 \ mol$ each at the beginning of the reaction and equilibrium is attained when unreacted $N_2$ is $0.7 \ mol$,what is the value of the equilibrium constant?