Question

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

Answer 1

C - An increase in temperature

Step-by-step explanation:

We have the given reaction. Note that NH4HS is a solid! This will be important later.

Using process of elimination:

A - A decrease in volume of the reaction vessel at constant temp.

Since volume is inversely proportional to pressure, decreasing the volume will increase the pressure. To account for this, the system will shift toward the "side" with more moles to decrease the moles in the system and therefore decrease the pressure. In this case, we have 0 mol of gas on the left side (NH4HS is a solid) and 2 mol of gas on the right side. Thus, the reaction will reverse to produce more solid (decreasing mol of gas), resulting in decreasing H2S. A is not correct.

B - An increase in the amount of NH4HS.

Beware of this option! NH4HS is a solid, so its concentration is constant. Increasing the amount will have no effect on equilibrium.

C - An increase in temperature.

We know that the reaction is endothermic, so we can think of heat as a reactant: NH4HS (s) + heat ⇌ NH3 (g) + H2S (g)

Thus, the system will shift to "get rid of" the heat and push in the forward direction to make more product. H2S will increase. Answer C is correct.