Question

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 temperature an increase in the amount of NH4HS in the reaction vessel all of the above

Answer 1

Given question is incomplete. The complete question is as follows.

The decomposition of $NH_{4}HS$ is endothermic:

     $NH_{4}HS(s) \rightarrow NH_{3}(g) + H_{2}S(g)$

Which change to an equilibrium mixture of this reaction results in the formation of more $H_{2}S$?

a decrease in the volume of the reaction vessel (at constant temperature), an increase in temperature, an increase in the amount of $NH_{4}HS$ in the reaction vessel, all of the above.

Explanation:

As per Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.

When volume is decreased then there will occur increase in pressure because according to Boyle's law pressure is inversely proportional to volume.

Hence, equilibrium will shift in the direction where there is less pressure. Therefore, in the given reaction a decrease in the volume of the reaction vessel (at constant temperature) will shift the equilibrium in the backward direction.

When we increase the temperature then more number of collisions will take place and product formation will be more. Hence, reaction will shift in the forward direction.

When amount of $NH_{4}HS$ increases then equilibrium this disturbance will shift the equilibrium in forward direction.

Thus, we can conclude that following change to an equilibrium mixture of this reaction results in the formation of more $H_{2}S$:

• an increase in temperature.

• an increase in the amount of $NH_{4}HS$ in the reaction vessel.