The solubility product ($K_{sp}$) of $AgCl$ in water at 25°C is $1.8 imes 10^{-10}$. A solution of $AgCl$ in water is in equilibrium with solid $AgCl$. If a small amount of $NaCl$ is added to this saturated solution, what will happen to the solubility ($s$) and $K_{sp}$ of $AgCl$?

The reaction quotient $\left( Q \right)$ for the reaction is given by: ${{\rm{N}}_2}\left( {\rm{g}} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{N}}{{\rm{H}}_3};Q = \frac{{{{\left[ {{\rm{N}}{{\rm{H}}_3}} \right]}^2}}}{{\left[ {{{\rm{N}}_2}} \right]{{\left[ {{{\rm{H}}_2}} \right]}^3}}}$The reaction will proceed from right to left if:

According to Le Chatelier's principle, increasing the pressure on the following equilibrium system: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ will:

Consider the following reaction at equilibrium${N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)$if small amount of Helium gas is added to the reaction mixture at constant pressure then

The equilibrium, $2{\rm{S}}{{\rm{O}}_2}\left( {\rm{g}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)\;\leftrightharpoons\;2{\rm{S}}{{\rm{O}}_3}\left( {\rm{g}} \right)$ shifts forward if:

According to Le-Chatelier principle, adding heat a solid and liquid in equilibrium will cause the

For the gaseous reaction, ${{\rm{C}}_2}{{\rm{H}}_4} + {{\rm{H}}_2}\leftrightharpoons{{\rm{C}}_2}{{\rm{H}}_6},\;\Delta H = - 130\;{\rm{kJ}}\;{\rm{ mo}}{{\rm{l}}^{ - 1}}$ carried in a closed vessel, the equilibrium concentration of the ${{\rm{C}}_2}{{\rm{H}}_6}$ can definitely be increased by

For the reaction ${{\rm{N}}_2}\left( {\rm{g}} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right)\;\leftrightharpoons\;2{\rm{N}}{{\rm{H}}_3}\left( {\rm{g}} \right);\;\;\;\;\;\;\;\Delta H = - 93.6\;{\rm{kJ}}\;{\rm{ mo}}{{\rm{l}}^{ - 1}}$, the concentration of ${\rm{N}}{{\rm{H}}_3}$ at equilibrium can be increased by(I) lowering the temperature(II)low pressure(III)excessof ${{\rm{N}}_2}$(IV)excess of ${{\rm{H}}_2}$

Consider the reaction equilibrium

$2{\rm{S}}{{\rm{O}}_2}\left( {\rm{g}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)\;\leftrightharpoons\;2{\rm{S}}{{\rm{O}}_3}\left( {\rm{g}} \right),\;\;\Delta H{\rm{^\circ }} = - 198\;\;{\rm{kJ}}$

On the basis of Le-Chatelier’s principle, the condition favourable for the forward reaction is

In lime kiln, the reversible reaction. $CaC{O_3}\left( s \right)\;\leftrightharpoons\;CaO\left( s \right){\rm{ }} + {\rm{ }}C{O_2}\left( g \right)$ proceeds to completion because :

On applying pressure to the equilibrium, $ice\;\leftrightharpoons\;water$

Which phenomenon will happen?