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QuestionWhat happens to the equilibrium when H2 S\mathrm{H}_{2} \mathrm{~S} gas is removed from this reaction?
NH4HS(s)⇌NH3( g)+H2 S( g) \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})
A. Shift left, NH3\mathrm{NH}_{3} decreases.
B. Shift right, NH3\mathrm{NH}_{3} increases.
C. Shift left, NH3\mathrm{NH}_{3} increases.
D. Shift right, NH3\mathrm{NH}_{3} decreases.

Studdy Solution

STEP 1

Assumptions1. The system was initially at equilibrium. . H\mathrm{H}_{} \mathrm{} gas is removed from the system.
3. The system will adjust to reestablish equilibrium according to Le Chatelier's principle.

STEP 2

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. In this case, the removal of H2\mathrm{H}_{2} \mathrm{} gas from the system reduces its concentration.

STEP 3

To counteract the change and reestablish equilibrium, the system will shift in the direction that increases the concentration of H2\mathrm{H}_{2} \mathrm{}. Looking at the chemical equation, we see that H2\mathrm{H}_{2} \mathrm{} is a product of the reaction. Therefore, the reaction will shift to the right (towards the products).

STEP 4

As the reaction shifts to the right, the concentration of the other product, NH3\mathrm{NH}_{3}, will also increase. This is because the forward reaction (from left to right) is being favored to produce more products.

STEP 5

Therefore, the correct answer isB. The reaction shifts to the right (products) and the concentration of NH3\mathrm{NH}_{3} increases.

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