MCAT Physical - Le Chatelier's Principle and Common Ion Effect

Consider the reaction reaction below.

$N_2(g)+3H_2(g)\rightarrow2NH_3(g)$

A student allows the system to reach equilibrium and then removes two moles of hydrogen gas. Which of the following will be a result?

The amount of N2 in the reaction vessel will increase

More NH3 will be produced

The reaction will first shift toward the products, then toward the reactants

The reaction will shift to the side with fewer total moles of gas

No change will occur

Explanation

According to Le Chatelier's principle, when a system at equilibrium is disturbed, the system will react to restore equilibrium. In other words, it will seek to undo the stress. Here, if hydrogen gas is removed, the reaction will shift toward the reactants to re-form it. In the process, more nitrogen will be produced.

In the given reaction, which of the following changes takes place if the temperature of the system is increased?

$CaO + H_{2}O\rightleftharpoons Ca(OH)_{2}+heat$

$[Ca(OH)_2]\ \text{decreases }$

$[CaO]\ \text{decreases }$

$[H_2O]\ \text{decreases }$

None of these changes occurs

More than one of these changes occurs

Explanation

This reaction is exothermic, since heat is released in the reaction. In exothermic reactions, decreasing the temperature favors the forward (exothermic) reaction, while increasing the temperature favors the reverse (endothermic) reaction. Similarly, for an endothermic reaction decreasing the temperature favors the reverse (exothermic) reaction, while increasing the temperature favors the forwards (endothermic) reaction.

In this question, increasing the temperature will favor the reverse reaction, increasing the reactant concentration and decreasing the product concentration.

Consider the following saturated solution. Assume it is at equilibrium.

$PbCrO_{4}_{(s)}\ \leftrightharpoons\ Pb^{2+}_{(aq)}\ + CrO_{4}^{2-}_{(aq)}$

$K_{sp} = 1.8 \cdot 10^{-14}$

Adding sodium chromate to the above solution would ___________ the solubility of lead chromate due to ___________.

decrease . . . an increase of chromate ion concentration

increase . . . an increase of chromate ion concentration

increase . . . an increase of sodium ion concentration

decrease . . . an increase of sodium ion concentration

Explanation

Adding sodium chromate increases the concentration of chromate ion in the solution, which shifts the reaction to the left due to the common ion effect. Thus, the solubility of lead chromate would decrease, as there would be an increased amount of solid lead chromate.

$2A_{(s)}+3B_{(g)}\rightleftharpoons 3C_{(g)}+4D_{(g)} +heat$

Which of the following adjustments would shift the equilibrium towards the left?

I. Placing the reaction vessel into an ice bath

II. Increasing the pressure in the reaction vessel

III. Increasing the amount of C and D

IV. Decreasing the amount of A and B

II, III, and IV

I, and III

I, II, and IV

III, and IV

Explanation

The following will shift the equilibrium to the left:

decreased volume (increased pressure)
increased temperature
decreased amounts of A and B
increased amounts of D and C

Of the choices, all are correct except for Roman numeral I. Placing the vessel into an ice bath decreases the temperature of the equilibrium, and thus the system would want to shift towards the side that produces heat to replace the heat lost.

Calcium carbonate dissolves in water based on the following reaction:

$CaCO_{3}_{(s)} \rightarrow Ca^{2+}_{(aq)} + CO_{3}^{2-}_{(aq)}$

Which of the following will decrease the solubility of the salt?

Add calcium chloride to the solution

Remove calcium ions from the solution

Add more calcium carbonate

Add more water to the solution

Explanation

When thinking about the solubility of a salt, it helps to use Le Chatelier's principle. The solubility of a salt is dependent on the amount of ions that are created by the precipitate in solution. As a result, we decrease the solubility by increasing the amount of ions from the salt in solution.

$CaCO_{3}_{(s)} \rightarrow Ca^{2+}_{(aq)} + CO_{3}^{2-}_{(aq)}$

In this case, increasing the amount of calcium or carbonate ions will shift the reaction to the left, decreasing solubility.

The common-ion effect tells us that when an ion made by the salt is increased by another substance, the solubility of the salt will decrease. Calcium chloride will dissolve completely in solution, and will increase the amount of calcium ions. This will shift the reaction to the left, thus reducing the solubility of the precipitate.

Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.

$HA \leftrightharpoons H^{+} + A^{-}$

All of the bases proceed in a similar fashion.

$BOH \leftrightharpoons B^{+} + OH^{-}$

The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.

In studying the acid , a scientist finds that heat is released when the acid dissociates in solution. If this scientist raises the temperature in the vessel after the reaction has reached equilibrium, which of the following is most likely true?

The concentration of will increase

The concentration of will decrease

The concentration of will not change

The concentration of will increase

The concentration of will not change

Explanation

In exothermic process heat is released during the reaction.

$HA\rightleftharpoons H^++A^-+heat$

Heat can be considered a product in this situation, and thus increasing the amount of heat in the vessel after the system has reached equilibrium will drive the reaction to the left. This property is a derivative of Le Chatelier's principle.

A system contains iodine atoms that are in equilibrium with respect to the reaction below:

$2I_{(g)} \rightleftharpoons I_2_{(g)}$

The volume of the system is suddenly reduced, leading to an increase in pressure. What effect will this have on the reaction?

The reaction shifts to the right

The reaction will not be affected

The reaction shifts to the left

The reaction shifts to the right then to the left

Explanation

Since the system is originally at equilibrium and a stress is applied, Le Chatelier's principle is to be considered. By increasing the total pressure, the reaction will move in the direction toward which there are less molecules of gas.

$2I_{(g)} \rightleftharpoons I_2_{(g)}$

Looking at the original reaction there are two moles of gas on the left and only one on the right, indicating that the reaction will shift to the right if the pressure is increased.

The Haber-Bosch process, or simply the Haber process, is a common industrial reaction that generates ammonia from nitrogen and hydrogen gas. A worker in a company generates ammonia from the Haber process. He then dissociates the gaseous ammonia in water to produce an aqueous solution. Since ammonia is a base, it will accept a proton from water, generating and ammonium ion products. The two reactions involved are:

$N_2_(g_)+3H_2_(g_)\rightarrow2NH_3_(g_);; ;;;;;;;;;;;;;;;;;;;;;;;;;(1)$

$NH_3_(g_)+H_2O_(l_)\rightarrow NH_4^+_(aq_)+OH^-_(aq_);;;;;(2)$

The worker could do which of the following to shift the equilibrium towards hydroxide () ions?

I. Add hydrogen gas

II. Add

III. Add

I and III

I only

III only

I and II

Explanation

You need to understand Le Chatelier’s principle to solve this question. Le Chatelier’s principle states that the addition of excess molecules to one side of a reaction will shift the equilibrium toward the other side. This shift makes sure that the reaction remains at equilibrium.

If you add hydrogen gas, reaction 1 will shift to the right and will produce more ammonia. Since more ammonia is produced, reaction 2 will shift to the right to produce more ammonium ions and hydroxide ions. Statement I is correct.

Adding to the solution will increase the concentration of hydroxide ions. This means that the equilibrium of reaction 2 will shift to the left, away from hydroxide ions. While the overall hydroxide ion concentration may increase due to the addition of the base, the equilibrium for the reaction will shift away from the hydroxide ions, making statement II incorrect.

Finally, adding will generate ions. These ions will react with the ions and form water, which will decrease the overall concentration of hydroxide ions. The equilibrium of reaction 2 will shift to the right to produce more ammonium and hydroxide ions. Statement III is correct.

$2A_{(s)}+3B_{(g)}\rightleftharpoons 3C_{(g)}+4D_{(g)} +heat$

Which of the following adjustments would cause the equilibrium to shift towards the right?

I. Adding more substance B

II. Adding more substance D

III. Increasing the temperature

IV. Decreasing the volume

I only

I, II, and III

I and II

I, III, and IV

Explanation

We know that concentration, temperature, medium and catalyst all affect the direction and the rate of the reaction through the Le Chatelier's Principle, where a system tries to relieve stress when stress is being applied. In order to push the equilibrium towards the right, we need to increase volume, decrease temperature, increase concentration of A and B, and decrease the concentrations of C and D. In the choices provided, only roman numeral I provided a correction suggestion. By increasing temperature, the system would want to go to the side that does not produce heat (the left). By increasing the amount of D, it will tend to shift reaction away from the species being added. By decreasing the volume and thus increasing the pressure, the system would want to shift left towards the side with less moles of gas.

Consider the following saturated solution. Assume it is at equilibrium.

$PbCrO_{4}_{(s)}\ \leftrightharpoons\ Pb^{2+}_{(aq)}\ + CrO_{4}^{2-}_{(aq)}$

$K_{sp} = 1.8 \cdot 10^{-14}$

What is the chromate ion concentration after 0.019 moles of lead nitrate is added to 1 L of the above solution?

$9.5 \cdot 10^{-13}M$

$3.4 \cdot 10^{-16}M$

$1.3 \cdot 10^{-7}M$

$5.2\cdot10^{-12}M$

Explanation

Us the solubility constant to calculate the answer to this question. We know that:

$K_{sp} = [Pb^{2+}] \cdot [CrO_{4}^{2-}]$

and

$K_{sp} = 1.8 \cdot 10^{-14}$

Since $K_{sp}$ is a very small number, we can assume that the concentration of chromate, , will be very small compared to the concentration of the added lead, 0.019 M. We can use the two known values ( $K_{sp}$ and lead concentration) to solve for the unknown chromate concentration: