Chemical Equilibrium - AP Chemistry

Card 0 of 866

Question

What is the difference between the reaction quotient and the equilibrium constant?

Answer

The correct answer gives the accurate definition of both the equilibrium constant and the reaction quotient.

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Question

Which of the following can always be determined based on the equation itself?

Answer

According to the law of mass action, the equilibrium constant expression can always be written given the equation of the reaction itself. The equilibrium-constant expression will be written as the products over the reactants, each raised to their respectively stoichiometric coefficient. The rate law and concentrations can only be determined if there is additional data given.

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Question

The law of mass action gives an expression that is specific for a certain __________.

Answer

Equilibrium expressions are specific for certain temperatures. Physical state, pressure, and volume do not factor into the expression.

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Question

Consider formation of Nitrogen monoxide: $N_2 (g) + O_2 (g) \Leftrightarrow 2 NO (g)$ with $K = 4.2 x 10^{-8}$ . If the initial concentration of N2 was 0.085M and O2 was 0.038 M, what is the concentration of nitrogen monoxide at equilibrium?

Answer

$K = \frac{[NO]^2}{[N_2 ][O_2 ]} =\frac{(2x)^2} {(0.085-x)(0.038-x) }$

$\sim \frac{4x^2}{(0.085)(0.038)} = 4.2 x 10^{-8}$

$x = 5.82 x 10^{-6}$

$[NO] = 2x = 2(6.82 x 10^{-6}) = 1.2 x 10^{-5}$

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Question

Given the ICE table below, what are the signs on the terms that would appear in the “Change” row?

$M + 2L \Leftrightarrow ML2$

Answer

Based on the balanced equation, one would use -x, -2x, and +x.

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Question

Find the equilibrium concentrations for C in the following chemical reaction: A + B -> 2C

K = 9.0 x 10-8

Answer

$K = 9 x 10^{-8} =\frac{ [C]^2} {[A][B]} = \frac{(2x)^2}{(0.5-x)^2} \sim \frac{4x^2}{(0.25) }$

$[C] = 2 (7.5 x10^{-5} ) = 1.5 x 10^{-4}$

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Question

What is the equilibrium concentration for A- in the following reaction if the starting concentration of HA is 0.60 M?

$HA \Leftrightarrow H^+ + A^-$ $K = 2.0 x 10^{-5}$

Answer

$K = \frac{[H^+ ][A^- ]}{[HA]} = \frac{x^2}{0.6-x} = 2.0 x 10^{-5}$

$\frac{x^2}{0.6} \sim 2.0 x 10^{-5}$

$x = 3.5 x 10^{-3}$

$[A^- ] = x = 3.5 x 10^{-3}$

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Question

What is the equilibrium concentration for A- in the following reaction if the starting concentration of HA is 0.30 M?

$HA \Leftrightarrow H^+ + A^-$ $K = 5.0 x 10^{-9}$

Answer

$K =\frac{[H^+ ][A^- ]}{[HA]} =\frac{x^2}{0.3-x} = 2.0 x 10^{-5}$

$\frac{x^2}{0.3} \sim 5.0 x 10^{-9}$

$x = 3.87 x 10^{-5}$

$[A^- ] = x = 3.9 x 10^{-5}$

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Question

Consider the following gasesous, reversible, exothermic reaction:

$A \rightarrow C + D + E$

What could be done to increase the equilibrium concentration of species C?

Answer

According to Le Chatelier's principle, any changes in concentration or pressure to a system at equilibrium will cause the system to readjust to a new equilbrium. The addition of a reactant will cause the reaction to shift to the right, increasing the equilibirum concentration of the products. Thus, adding additional reactant A would increase the equlibrium concentration of product C.

Removing species A or adding species E will drive the reaction to the left, reducing the amount of species C. Since the reaction is exothermic, heat can be considered a product. Thus, increasing the temperature will also shift the reaction to the left.

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Question

The reaction $A+3B\rightleftharpoons2C+2D$ is at equilibrium. You have measured the concentration of and to be

What is the equilibrium constant, ?

Answer

Recall that the rate constant for the equilibrium reaction

$aA+bB\rightleftharpoons cC+dD$

is $K = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$ , always remembering that it is products over reactants.

Since we are at equilibrium, for every mol of we have, we have 3 mols of . Therefore,

Likewise, for every 2 mols of we have, we have 2 mols of . Therefore,

Plugging all of this into the above equation we get

$K = \frac{[0.5M]^{2}[0.5M]^{2}}{[1.2M]^{1}[3.6M]^{3}}=1.116*10^{-3}$

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Question

Consider the following chemical reaction.

$A_{(g)}\rightleftharpoons B_{(g)}+C_{(g)}$

This reaction is allowed to equilibrate at some volume and pressure.

Which of the following actions will not shift the equilibrium toward the products?

Answer

Adding an inert gas will not shift the equilibrium toward the products, but neither will it shift the equilibrium toward the reactants. It has no effect because the partial pressure of each species, , , and , will still remain constant. Since the partial pressures are constant, we know from the ideal gas law that the concentration of each species remains constant. For example, for ,

$[A]=\frac{n_{A}}{V}=\frac{P_{A}}{RT}$

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Question

A reaction at equilibrium releases 25 Joules of heat energy at $37^{o}C$ . What is the entropy change for this same reaction?

Answer

To solve this problem, we must make use of the following equation:

$\Delta G=\Delta H-T\Delta S$

At equilibrium, the value of $\Delta G$ is 0. Therefore, we can simplify the equation.

$0=\Delta H-T\Delta S$

$T\Delta S=\Delta H$

$\Delta S=\frac{\Delta H}{T}$

We must also convert temperature from degress Celsius into Kelvin.

$37^{o}C+273=310K$

$\Delta S=\frac{\Delta H}{T}=\frac{-25J}{310K}=-0.0806\frac{J}{K}$

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Question

If formate is added to a solution of formic acid in water, which of the following changes would occur?

Answer

Because formic acid is a weak acid, it will exist in equilibrium with its conjugate base, formate, while in water according to the following reaction:

$Formic:Acid\left_{(aq)}+H_{2}O_{(l)}\rightleftharpoons Formate_{(aq)}+H_{3}O^{+}_{(aq)}$

If formate is added to this solution, then according to Le Chatlier's principle, the reaction will shift towards the left. Thus, the $[H_{3}O^{+}]$ in solution will decrease and the pH will consequently increase. Furthermore, the addition of formate has no effect on the pKa.

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Question

For a given chemical reaction, $\Delta G^{o}=-2.5\frac{kJ}{mol}$ . What is the $K_{eq}$ for this reaction at $25^{o}C$ ?

Answer

To answer this question, it is necessary to use following equation:

$\Delta G=-RTln\left ( K_{eq}\right )$

We can solve for by rearranging the equation to the following:

$K_{eq}=e^{\frac{-\Delta G^{o}}{RT}}=e^{-\frac{(-2.5\cdot10^{3}\frac{J}{mol})}{(8.314\frac{J}{molK})(298K)}}=2.74$

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Question

$NH_{4}NO_{3}(s) \rightleftharpoons N_{2}O(g)+2H_{2}O(g)$

Which of the following is the correct equilibrium expression for the given reaction?

Answer

For this question, we're asked to determine the correct equilibrium expression for a given reaction that is in equilibrium.

To begin with, we need to remember that pure solids and liquids are omitted from the equilibrium expression. Only gases and aqueous components of a reaction are included. The reason for this is because in a reaction at equilibrium, a pure solid or liquid will not have any appreciable change in concentration. Because their concentrations remain essentially unchanged, their value gets incorporated into the constant.

Also remember that an equilibrium expression is represented by the products divided by the reactants, each of which is raised to its stoichiometric coefficient.

In the case of the reaction given, we can see that it is already balanced. This means we are ready to set up the expression.

$K=[N_{2}O][H_{2}O]^{2}$

Here, the $[H_{2}O]$ term is being squared because it has a stoichiometric coefficient of in the equilibrium reaction. Also notice that $NH_{4}NO_{3}$ does not show up in the expression because it is a solid.

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Question

$\small Zn \rightarrow Zn^{2+} + 2e^{-}$ $\small \small E^{o} = 0.762 V$

$\small Cu \rightarrow Cu^{2+} + 2e^{-}$ $\small \small E^{o}= -0.340 V$

$\small Zn^{2+} + Cu \rightarrow Zn + Cu^{2+}$

What is $\small \small E_{cell}^{o}$ ? Is reaction 3 spontaneous?

Answer

$\small E^{o}_{cell} = E_{red}^{o} + E_{ox}^{o}$

$\small E_{cell}^{o} = -0.762 V - 0.340 V$

$\small E^{o}_{cell} = -1.1 V$

There are two concepts to consider in this problem. First, the question asks for the $\small E^{o}_{cell}$ . Reaction 3 has $\small Zn^{2+}$ being reduced so the potential for the half-reaction becomes negative. $\small Cu$ half-reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

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Question

If delta G for a certain reaction is found to be -4955 J/mol, what is the equilibrium constant (Keq) for this reaction?

Answer

Given the equation delta G = -RTln(Keq), plugging in the correct values for the variables will lead us to the correct answer. Using e = 2.7, R = 8.314 J/molK, T= 298 K, and G = -4955 J, we can calculate that K = 7.4.

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Question

If cesium metal spontaneously combusts when it comes in contact with room temperature water, which of the following must be true of the equilibrium constant of the reaction at 298 K?

Answer

When K is greater than 1, the products are favored and the reaction proceeds forward spontaneously. This must be true of the reaction, since it spontaneously occurs when the metal comes into contact with the water.

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Question

Which of the following is true when the reaction quotient (Q) equals the equilibrium constant (K)?

Answer

When Q = K, the reaction has reached equilibrium and is in a state of dynamic equilibrium, where the forward and reverse reactions are occurring at the same rate. Since there is no net change in concentration, the free energy is zero.

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Question

Consider the following (arbitrary) reaction:

$A_2O_{4\hspace{1 mm}(aq)}\leftrightharpoons 2AO_{2\hspace{1 mm}(aq)}$

At equilibrium, $[A_2O_4]=0.25\hspace{1 mm}M$ and $[AO_2]=0.04\hspace{1 mm}M$ .

What is the value for the equilibrium constant, $K_{eq}$ ?

Answer

First, consider the formula for the equilibrium constant of arbitrary reaction:

$aA+bB \leftrightharpoons cC+dD$

$K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The formula for the equilibrium constant for this reaction is:

$K_{eq}=\frac{[AO_2]^2}{[A_2O_4]}=\frac{[0.04]^2}{[0.25]}=6.4\times10^{-3}$

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