The equilibrium constant for the reversible reaction,$N_2 + 3H_2 \rightleftharpoons 2NH_3$ is $K$ and for the reaction $\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3$ the equilibrium constant is $K'$. $K$ and $K'$ will be related as

For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$ at $495 \ K$,$\Delta_{r}G^{\circ} = -9.478 \ kJ \ mol^{-1}$. If we start the reaction in a closed container at $495 \ K$ with $22 \ mmol$ of $A$,the amount of $B$ in the equilibrium mixture is $x \ mmol$. Find $x$ (Round off to the nearest integer). $[R = 8.314 \ J \ mol^{-1} \ K^{-1}; \ln 10 = 2.303]$

One mole $H_2O_{(g)}$ and one mole $CO_{(g)}$ are taken in a $1 \ L$ flask and heated to $725 \ K$. At equilibrium,$40 \%$ of water reacted with $CO_{(g)}$ as follows:
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$
Its $K_c$ value is:

For the reaction $H_2 + I_2 \rightleftharpoons 2HI$,the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $8.0 \ mol \ L^{-1}$,$3.0 \ mol \ L^{-1}$,and $28.0 \ mol \ L^{-1}$ respectively. The equilibrium constant $(K_c)$ of the reaction is: (in $.66$)

For which of the following reactions will ${K_p} > {K_c}$?

Explain why the partial pressure $p$ of a gas is proportional to its concentration $c$.