
A gases catalyst X is added in an equilibrium mixture of ${N_2}$, \[{H_2}\], and \[N{H_3}\] gases at 298K keeping pressure constant. What will be observed at equilibrium reactions?
\[{N_{2{\text{ }}(g)}} + 3{H_{2{\text{ }}(g)}}\; \rightleftarrows 2{\text{ }}N{H_{3{\text{ }}(g)}}\]
A.Equilibrium will be unaffected
B.More $NH_3$ gas will be formed
C.Dissociation of ammonia will be favoured
D.Catalyst will change the value of equilibrium constant.
Answer
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Hint:This question is based on Le-Chatelier’s principle. Le-Chatelier’s principle helps to predict the effect of temperature, concentration, pressure, volume, concentration and catalyst on the position of equilibrium in a chemical reaction.
Complete step by step answer:
-If catalyst X is added to a given reaction there is no change in that position of equilibrium reaction .That is the yield of reaction catalyst only helps to allow the equilibrium to be reached more quickly. Because catalyst only increases the rate of reaction but has no effect on yield. It speeds up both forward and backward reactions equally.
-Now if we kept the pressure of that equilibrium reaction constant then volume will automatically change and Le-Chatelier’s principle, if we add a catalyst to the reaction at constant pressure equilibrium shifts to that side where the number of moles will be higher.
-So in a given reaction there is a total number of moles in reactant side is four while in product side 2.
-Therefore according to Le-Chatelier’s principle equilibrium shifts on reactant side it means ammonia dissociates more that is dissociation of ammonia will be favoured.
Hence, the correct option will be option C.
Note:
Some other factors also affects the equilibrium
-In an endothermic reaction if we increase the temperature of the system the reaction goes forward and the decreasing temperature reaction goes backward. While in exothermic reaction on increasing temperature reaction goes backward while goes forward on decreasing temperature.
-On increasing concentration of reactant equilibrium shift to right hand side while on increasing concentration of product equilibrium shifts to left hand side.
-On increasing pressure equilibrium shifts to that side where lesser number of moles of gas present while on decreasing pressure equilibrium shifts to that side where number of moles are higher.
Complete step by step answer:
-If catalyst X is added to a given reaction there is no change in that position of equilibrium reaction .That is the yield of reaction catalyst only helps to allow the equilibrium to be reached more quickly. Because catalyst only increases the rate of reaction but has no effect on yield. It speeds up both forward and backward reactions equally.
-Now if we kept the pressure of that equilibrium reaction constant then volume will automatically change and Le-Chatelier’s principle, if we add a catalyst to the reaction at constant pressure equilibrium shifts to that side where the number of moles will be higher.
-So in a given reaction there is a total number of moles in reactant side is four while in product side 2.
-Therefore according to Le-Chatelier’s principle equilibrium shifts on reactant side it means ammonia dissociates more that is dissociation of ammonia will be favoured.
Hence, the correct option will be option C.
Note:
Some other factors also affects the equilibrium
-In an endothermic reaction if we increase the temperature of the system the reaction goes forward and the decreasing temperature reaction goes backward. While in exothermic reaction on increasing temperature reaction goes backward while goes forward on decreasing temperature.
-On increasing concentration of reactant equilibrium shift to right hand side while on increasing concentration of product equilibrium shifts to left hand side.
-On increasing pressure equilibrium shifts to that side where lesser number of moles of gas present while on decreasing pressure equilibrium shifts to that side where number of moles are higher.
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