માટે સંતુલન પુરોગામી દિશામાં .......... દ્વારા ખસે.
equilibrium. This change in the direction of equilibrium is governed by Le-Chatelier's principle. According to Le-Chatellier's principle, equilibrium shifts in die opposite direction to undo the change. \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{Heat}\)
(a) Increasing the concentration of \(N H_{3}(g):\) On increasing the concentration of \(N H_{3}(g)\). the equilibrium shifts in the backward direction where concentration of \(N H_{3}(g)\) decreases
(b) Decreasing the pressure: since, \(p \propto n\) (number of moles), therefore, equilibrium shifts in the backward direction where number of moles are increasing.
(c) Decreasing the concentration of \(N_{2}(g)\) and \(H_{2}(g)\) Equilibrium shifts in the backward direction when concentration of \(H_{2}(g)\) and \(H_{2}(g)\) decreases.
(d) Increasing pressure and decreasing temperature: On increasing pressure, equilibrium shifts in the forward direction where number of moles decreases. It is an example of exothermic reaction therefore decreasing temperature favours the forward direction
$Co{O_{2\left( g \right)}} + C{O_{\left( g \right)}} \rightleftharpoons Co{O_{\left( s \right)}} + C{O_{2\left( g \right)}}\,\,;\,K = 490$
તો નીચેની પ્રક્રિયા માટે સંતુલન અચળાંક .... થશે.
$C{O_{2\left( g \right)}} + {H_{2\left( g \right)}} \rightleftharpoons C{O_{\left( g \right)}} + {H_2}{O_{\left( g \right)}}$
${A_2}(g)\, + \,{B_2}(g)\,\overset {{K_1}} \leftrightarrows \,2AB(g)\,\,\,......(1)$
$6AB\,(g)\,\,\overset {{K_2}} \leftrightarrows \,\,3{A_2}(g)\, + \,3{B_2}(g)......(2)$
તો $K_1$ અને $K_2$ વચ્ચેનો સંબંધ શું થાય?
(અહીં : $SrCO_{3(s)} \rightleftharpoons SrO_{(s)}+ CO_{2(g)} \,, K_p=1.6\,atm$)