The decomposition of NH4HS is endothermic: NH4HS(s)⇌NH3(g)+H2S(g) Part A Which change to an equilibrium mixture of this reaction results in the formation of more H2S? Which change to an equilibrium mixture of this reaction results in the formation of more ? a decrease in the volume of the reaction vessel (at constant temperature) an increase in the amount of NH4HS in the reaction vessel an increase in temperature all of the above

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reaction is as follows.

[tex]NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)[/tex]

(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.

Thus, formation of [tex]H_{2}S[/tex] will increase.

(b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.

Hence, formation of [tex]H_{2}S[/tex] will decrease with decrease in volume.

When we increase the mount of [tex]NH_{4}HS[/tex] then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.

Thus, we can conclude that formation of [tex]H_{2}S[/tex] will increase then.