According to Le Châtelier's principle, how will an increase in concentration of a reactant affect the equilibrium system?

Shift it toward the products

Shift it toward the reactants

Shift it toward the side with higher total mole concentration

Shift it toward the side with lower total mole concentration

Question

Chemistry

Tess Meyers

9 May, 02:56

According to Le Châtelier's principle, how will an increase in concentration of a reactant affect the equilibrium system?

Shift it toward the products

Shift it toward the reactants

Shift it toward the side with higher total mole concentration

Shift it toward the side with lower total mole concentration

+1

Answers (2)

Know the Answer?

Kenna · Answer 1

The answer is Shift it toward the products

Explanation:

In a system that is in equilibrium, an increase in the concentration of a substance will cause the system to move so that the substance that was added is partially used. On the other hand, if the concentration of a substance in a system that is in equilibrium is decreased, it will cause the system to move in the sense that it partially replaces the substance that was removed.

In the case of exercise, if the concentration of a reagent is increased, the direct sense of the reaction, that is, towards the products, will be favored.

Butterball · Answer 2

Shift it toward the products

Explanation:

Imagine this equilibrium:

2A + B ⇄ A₂B

Let's think the expression for Kc:

Kc = [A₂B] / [A]². [B]

Now, we have this situation:

2A ↑ + B ⇄ A₂B

As we are not in equilibrium, we have to think that Kc would decrease.

↓Kc = [A₂B] / [A]² ↑. [B]

so we have to use Qc (reaction quotient). For this case

Qc < Kc, so the reaction → Product side

We have to make more product to counteract the reagent increase and thus return the system to equilibriumr