A chemical reaction $A\,\,\leftrightharpoons\,\,B$ is said to be at equilibrium when:

$28\;g$ ${{\rm{N}}_2}$ and $6\;g$ ${H_2}$ were mixed. At equilibrium $17\;g$ ${\rm{N}}{{\rm{H}}_3}$ was formed. The weight of ${{\rm{N}}_2}$ and${H_2}$ of equilibrium are respectively :

Which factor does NOT affect the equilibrium position of a reversible reaction?

For the reaction,

$2{\rm{HI}}\left( {\rm{g}} \right)\leftrightharpoons{{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right) - {\rm{Q }}\;{\rm{kJ}},$ the equilibrium constant depends upon

A catalyst affects the equilibrium of a reversible reaction by:

Which of the following is correct for the reaction?${{\rm{N}}_2}\left( {\rm{g}} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right)$

If the value of ${{K_c}}$ for an equilibrium reaction is${10^{ - 4}}$, then the reaction is in

In a reaction at equilibrium $'X'$ mole of the reactant $A$ decompose to give $1\;mole$ each of $C\;and\;D$ . If the fraction of $A$ decomposed at equilibrium is independent of initial concentration of $A$, then the value of $'X'$ is :

At equilibrium, the Gibbs free energy change ($\Delta G$) is:

The equilibrium constant $K$ for the reaction $2{\rm{HI}}\left( {\rm{g}} \right)\leftrightharpoons{{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right)$ at room temperature $300\;{\rm{K}}$ is ${\rm{2}}{\rm{.85}}$ and at $698\;{\rm{K}}\,\;1.84 imes {10^{ - 2}}.$ Hence the reason that ${\rm{HI}}$ exists as a stable compound at room temperature is because:

The equilibrium constant $K_p$ for the reaction $2NO_2(g) \rightleftharpoons N_2O_4(g)$ is $6.7$ at $298\,\mathrm{K}$. Calculate $\Delta G^{\circ}$ for the reaction. $(R = 8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\mathrm{K}^{-1})$