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AP Chemistry : Equilibrium Constant and Reaction Quotient

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Example Questions

Example Question #11 : Equilibrium Constant And Reaction Quotient

What is the correct equilibrium expression for this reaction?

$2ZnO (s) + 2F_{2} (g) \rightarrow 2ZnF_{2} (s)+ O_{2} (g)$

Possible Answers:

$K_{eq} = \frac{[O_{2}]}{[F_{2}]}$

$K_{eq} = \frac{[O_{2}][ZnF_{2}]^{2}}{[ZnO]^{2}[F_{2}]^{2}}$

$K_{eq}=\frac{[O_2]*2[ZnF_2]}{2[ZnO]*2[F_2]}$

$K_{eq} = \frac{[O_{2}]}{[F_{2}]^{2}}$

$K_{eq} = \frac{[O_{2}][ZnF_{2}]}{[ZnO][F_{2}]}$

Correct answer:

$K_{eq} = \frac{[O_{2}]}{[F_{2}]^{2}}$

Explanation:

Solids are not included in any equilibrium expression. The equilibrium expression will contain only aqueous solutions and gases. In addition, the coefficients of a balanced chemical equation correspond to exponents, not coefficients, in the equilibrium expression.

$2ZnO (s) + 2F_{2} (g) \rightarrow 2ZnF_{2} (s)+ O_{2} (g)$

For this reaction, the only relevant product is oxygen gas and the only relevant reactant is fluorine gas.

$K_{eq} = \frac{[O_{2}]}{[F_{2}]^{2}}$

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Example Question #52 : Reactions And Equilibrium

Consider the following balanced reaction.

$\small BaSO_{4}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)}$

Write the equilibrium constant expression for this reaction.

Possible Answers:

$\small K = \frac{[BaSO_{4}]}{[Ba^{2+}][SO_{4}^{2-}]}$

$\small K = [BaSO_{4}]$

$\small K = [Ba^{2+}][SO_{4}^{2-}]$

$\small K = \frac{[Ba^{2+}][SO_{4}^{2-}]}{[BaSO_{4}]}$

Correct answer:

$\small K = [Ba^{2+}][SO_{4}^{2-}]$

Explanation:

A key point to remember is that the equilibrium constant expression only includes compounds that are in aqueous solution or in the gaseous phase. Pure solids and liquids are omitted from the expression. As a result, solid barium sulfate will not be included in the expression.

The equilibrium constant is found by the concentration of the products over the concentration of the reactants, each raised to the power of their coefficients.

$aA+bB\rightleftharpoons cC+dD$

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

In our reaction, the only reactant is a pure solid, so only the products are used in the expression.

$\small BaSO_{4}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO_{4}^{2-}_{(aq)}$

$K_{eq}=[Ba^{2+}][SO_4^{2-}]$

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Example Question #12 : Equilibrium Constant And Reaction Quotient

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, $K_{eq}$ ?

Possible Answers:

$5.8*10^{-2}$

$6.4*10^{-3}$

$3.8*10^{-4}$

$156.25$

$1.6*10^{-1}$

Correct answer:

$6.4*10^{-3}$

Explanation:

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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Example Question #13 : Equilibrium Constant And Reaction Quotient

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

Possible Answers:

Change in free energy = 0.

The change in free energy is greater than zero.

The reaction stops completely.

The change in free energy is less than zero.

Correct answer:

Change in free energy = 0.

Explanation:

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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Example Question #14 : Equilibrium Constant And Reaction Quotient

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

Possible Answers:

80.9

55.6

7.4

100

5.9

Correct answer:

7.4

Explanation:

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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Example Question #15 : Equilibrium Constant And Reaction Quotient

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?

Possible Answers:

K < 1

K > 1

K = 1

Not enough information

All of the above

Correct answer: K > 1

Explanation:

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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Example Question #14 : Equilibrium Constant And Reaction Quotient

For the following reaction, what would be the correct equilibrium expression?

$Mg (s) + 2HCl (aq) \rightarrow MgCl_{2} (aq) + H_{2} (g)$

Possible Answers:

$K_{eq} = \frac{HCl]^{2}}{[MgCl_{2}][H_{2}]}$

$K_{eq} = \frac{[MgCl_{2}]}{[HCl]^{2}}$

$K_{eq} = \frac{[MgCl_{2}][H_{2}]}{[HCl]^{2}}$

$K_{eq} = \frac{[MgCl_{2}][H_{2}]}{2[HCl]}$

$K_{eq} = \frac{[MgCl_{2}][H_{2}]}{[Mg][HCl]^{2}}$

Correct answer:

$K_{eq} = \frac{[MgCl_{2}][H_{2}]}{[HCl]^{2}}$

Explanation:

For any equilibrium expression, solids and liquids are excluded because their concentrations are presumed to be constant during the reaction. The equilibrium expression is defined as the ratio of the concentration of products divided by the concentration of the reactants. Each reactant or product is raised to the power corresponding to its coefficient in the balanced chemical equation.

Using an arbitrary example:

$aA+bB\rightleftharpoons cC+dD;\ K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Compare this to our question. Remember to exclude the solid magnesium!

$Mg (s) + 2HCl (aq) \rightarrow MgCl_{2} (aq) + H_{2} (g)$

$K_{eq}=\frac{[MgCl_2][H_2]}{[HCl]^2}$

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Example Question #1 : Chemical Reactions

Determine the equation for the equilibrium constant of the following unbalanced reaction:

$Ba(OH)_2_{(aq)} + FeCl_2_{(aq)} \rightleftharpoons Ba^{2+}_{(aq)} + Cl^-_{(aq)} + Fe(OH)_{2(s)}$

Possible Answers:

$K_{eq} = \frac{[Ba(OH)_2][FeCl_2]}{[Ba^{2+}][Cl^-]}$

$K_{eq} = \frac{[Ba(OH)_2][FeCl_2]}{[Ba^{2+}][Cl^-]^2}$

$K_{eq} = \frac{[Ba^{2+}][Cl^-]}{[Ba(OH)_2][FeCl_2]}$

$K_{eq} = \frac{[Ba^{2+}][Cl^-]^2}{[Ba(OH)_2][FeCl_2]}$

$K_{eq} = \frac{[Ba^{2+}][Cl^-]^2[Fe(OH)_2)]}{[Ba(OH)_2][FeCl_2]}$

Correct answer:

$K_{eq} = \frac{[Ba^{2+}][Cl^-]^2}{[Ba(OH)_2][FeCl_2]}$

Explanation:

The equilibrium constant, $K_{eq}$ , is the ratio of the concentration of products raised to their coefficients, over the concentration of reactants raised to their coefficients. To find its value, we first need to balance the equation and then consider only the products and reactants that actually have concentrations (i.e. aqueous and gaseous species). Liquids and solids can be omitted from the calculation.

$Ba(OH)_2_{(aq)} + Fe{\color{Red} Cl_2}_{(aq)} \rightleftharpoons Ba^{2+}_{(aq)} + {\color{Red} Cl^-}_{(aq)} + Fe(OH)_{2(s)}$

We note that chlorine only appears once on the right, so we add a 2 coefficient to balance the equation:

$Ba(OH)_2_{(aq)} + FeCl_2_{(aq)} \rightleftharpoons Ba^{2+}_{(aq)} + 2Cl^-_{(aq)} + Fe(OH)_{2(s)}$

Since Fe(OH)_2 is a solid (most hydroxides precipitate, unless they are paired with alkali metals, barium, or calcium) it will not be included in the equilibrium constant calculation.

The chloride ion concentration is raised to the second power because of the 2 coefficient that we added. So, putting products (species on the right side, other than the solid) on the top and reactants (species on the left) on bottom, we get:

$K_{eq} = \frac{[Ba^{2+}][Cl^-]^2}{[Ba(OH)_2][FeCl_2]}$

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Example Question #21 : Chemical Reactions

Consider the following generic reaction:

$AB_{2(aq)} \rightleftharpoons A_{(aq)}+ 2B_{(aq)}$

$K_{eq} = 3.5 * 10^{-5}$

If the equilibrium concentration of AB_2 is 1M, what is the equilibrium concentration of ?

Possible Answers:

$3.0 * 10^{-3}M$

$3.5 * 10^{-5} M$

$2.1 * 10^{-2}M$

$5.9 * 10^{-3} M$

$2.6 * 10^{-2}M$

Correct answer:

$2.1 * 10^{-2}M$

Explanation:

We can set up the equilibrium expression by placing the products, raised to the power of their coefficients, in the numerator and reactants, raised to the power of their coefficients, in the denominator:

$K_{eq} = \frac{[A][B]^2}{[AB_2]}$

We are given the value of the equilibrium constant and the equilibrium concentration of AB_2 . Using stoichiometric coefficients, we can determine that a variable concentration, , of will be present at equilibrium, and of will be present at equilibrium. Plugging in:

$[A]=x,\ [B]=2x,\ [AB_2]=1$

$3.5 * 10^{-5} = \frac{x(2x)^2}{1}$

Simplify:

$3.5 * 10^{-5} = \frac{x*4x^2}{1}$

$3.5 * 10^{-5} = {4x^3}$

Now we can solve for (which is the equilibrium concentration of , as we assigned) by dividing both sides by 4 and then taking the cube root.

$\sqrt[3]{\frac{3.5*10^{-5}}{4}} = x$

x = 0.020606M

Rounding to 2 sig figs (to match the two numbers we were given by the problem), we get:

$x = 2.1 * 10^{-2} M$

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AP Chemistry : Equilibrium Constant and Reaction Quotient

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #11 : Equilibrium Constant And Reaction Quotient

Example Question #52 : Reactions And Equilibrium

Example Question #12 : Equilibrium Constant And Reaction Quotient

Example Question #13 : Equilibrium Constant And Reaction Quotient

Example Question #14 : Equilibrium Constant And Reaction Quotient

Example Question #15 : Equilibrium Constant And Reaction Quotient

Example Question #14 : Equilibrium Constant And Reaction Quotient

Example Question #1 : Chemical Reactions

Example Question #21 : Chemical Reactions

All AP Chemistry Resources

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