Practice Problems Chemical Equilibrium

Answer

The Law of Multiple Equilibria states that when reactions are added, the corresponding equilibrium constants are multiplied. Thus, individual reactions are added to give the net reaction and the equilibrium constants of the individual reactions are multiplied to give the equilibrium constant for the net reaction.

Answer

Catalysts increase the rate of a chemical reaction but do not affect the final composition of a reaction. Thus, a catalyst may change the rate at which dynamic equilibrium occurs but does not change the value of the equilibrium constant.

Answer

Use the definition of mass action to write the equilibrium constant expression:

(a)

(b)

(c)

Answer

In each case, first write the balanced chemical reaction and then use the stoichiometric coefficients in the mass action expression to give the equilibrium constant expression.

(a) ½N₂(g) + ½O₂(g) → ← NO(g)

(b) ½N₂(g) + 3 / 2H₂(g) →← NH₃(g)

(c) ½N₂(g) + ½O₂(g) + ½Cl₂(g) →← NOCl(g)

Answer

Write the balanced chemical reaction and then use the definition of mass action to find the equilibrium constant expression.

(a) 2 CO(g) + 2 NO(g) → ← N₂(g) + 2 CO₂(g)

(b) 5 O₂(g) + 4 NH₃(g) → ← 4 NO(g) + 6 H₂O(g)

(c) 2 NaHCO₃(s) → ← Na₂CO₃(s) + H₂O(g) + CO₂(g)

Answer

LeChatelier's Principle says that an equilibrium will shift to compensate for any added stress. In this case, the pressure is raised from 1 atm to 10 atm, so each reaction will shift to alleviate that pressure, i.e., towards the side with fewer moles of gases.

(a) Reactants: 2 moles of gases. Products: 1 mole of gases.

An increase in pressure will cause a shift towards products, therefore there will be more product formation at 10 atm.

(b) Reactants: 2 moles of gases. Products: 2 mole of gases.

Both sides have the same number of moles of gases so there will be no shift in the equilibrium.

(c) Reactants: 1 moles of gases. Products: 1.5 mole of gases.

An increase in pressure will cause a shift towards products, therefore there will be more product formation at 10 atm.

Answer

In each case, find the change in the number of moles of gases between product and reactant and then use the relationship K_p = K_c(RT)^Δn, where T is in K and R = 0.0821 L·atm/mol·K

(a) Δn = [1 + 1] – [1] = 1

K_p = 2.9×10^–2 at 303 K so K_c = K_p/(RT)¹ = 2.9×10^–2/(0.0821×303)¹ = 1.2×10^–3

(b) Δn = [2 + 1] – [2] = 1

K_p = 0.275 at 700 K so K_c = K_p/(RT)¹ = 0.275/(0.0821×700)¹ = 4.8×10^–3

(c) Δn = [1] – [1 + 1] = –1

K_p = 22.5 at 395 °C = 668 K so K_c = K_p/(RT)^–1 = K_p(RT)¹ = 22.5(0.0821×668)¹ = 1.23×10³

Answer

Use the relationship K_p = K_c(RT)^Δn

(a) Δn = 1 – 2 = –1, so K_p = 1.2×10³(0.0821×668)^–1 = 22

(b) Δn = 2 – 3 = –1, so K_p = 1.32×10^–2(0.0821×1000)^–1 = 1.61×10^–4

(c) Δn = 2 – 2 = 0, T = 1000 + 273 = 1273 K so K_p = 2.00(0.0821×1273)⁰ = 2.00

Answer

The given reaction is N₂(g) + O₂(g) → ← 2 NO(g), K_c = 4.08×10^–4

Reversing the reaction gives the proper reactants and products for the target reaction, but with the wrong stoichiometry. Reversing the reaction also means that the new equilibrium constant is the inverse of the original equilibrium constant.

2 NO(g) → ← N₂(g) + O₂(g) K_c = 1/4.08×10^–4 = 2.45×10³

To obtain the correct stoichiometry for the target reaction, all of the stochiometric coefficients are multiplied by ½. This means that the new equilibrium constant is the square root of the old equilibrium constant.

NO(g) → ← ½ N₂(g) + ½ O₂(g) K_c = (2.45×10³)^½ = 49.5

Answer

First, find K_p for the target reaction using the Law of Multiple Equilibria. Then, use the relationship K_p = K_c(RT)^Δn to find the value of K_c.

The target reaction has 2 CH₄(g) as one of the reactants and 1 C₂H₂(g) as a product. This tells us that the first reaction must have all of the stoichiometric coefficients doubled and that the K_p for the given reaction must be squared. The stoichiometric coefficients of the second reaction must all be halved and then the reaction reversed so this changes the K_p to the inverse of the square root of the given reaction.

2 CH₄(g) + 2 H₂O(g) → ←2 CO(g) + 6 H₂(g)    K_p = (1.2×10^–25)² = 1.4×10^–50

2 CO(g) + H₂O(g) → ← C₂H₂(g) + 3 / 2O₂(g)     K_p = 1/(1.1×10²)^½ = 9.5×10^–2

The sum of these two reactions is given below, which conveniently eliminates CO(g). K_p for the summed reaction is found by multiplying the two equilibrium constants.

2 CH₄(g) + 3 H₂O(g) → ← C₂H₂(g) + 6 H₂(g) + 3 / 2O₂(g)     K_p = (1.4×10^–50)( 9.5×10^–2) = 1.3×10^–51

This is not the target reaction. To reach the target, the third reaction given above must be used with all of its stoichiometric coefficents multiplied by three. This means that the equilbrium constant must be cubed.

3 H₂(g) + 3 / 2O₂(g) → ← 3 H₂O(g)    K_p = (1.1×10⁴⁰)³ = 1.3×10¹²⁰

Now, adding the last two reactions together gives the target reaction. K_p for the target reaction is found by multiplying equilibrium constants.

2 CH₄(g) → ← C₂H₂(g) + 3 H₂(g)    K_p = (1.3×10^–51)( 1.3×10¹²⁰) = 1.7×10⁶⁹

Finally, to find K_c for the final reaction, Δn = [1 + 3] – [2] = 2 so K_c = K_p/(RT)^Δn = (1.7×10⁶⁹)/(0.0821×298)² = 2.8×10⁶⁶

Answer

Since the equilibrium constant data is given in terms of K_p for the known reactions, first find K_p for the unknown reaction and then, lastly, convert it to K_c.

The target reaction is:

½ N₂(g) + ½ O₂(g) + ½ Cl₂(g) → ← NOCl(g)

Reaction (1) has nitrogen and oxygen as reactants but a wrong product. Reaction (3) can be used to eliminate the nitrogen dioxide product from (1) and also introduces chlorine:

½ N₂(g) + O₂(g) → ← NO₂(g) K_p = 1.0×10^–9 (1)

NO₂(g) + ½ Cl₂(g) → ← NO₂Cl(g) K_p = 0.3 (3)

Adding the two reactions together means that we multiply equilibrium constants:

½N₂(g) + O₂(g) → ← NO₂Cl(g) K_p = (1.0×10^–9)(0.3) = 3×10^–10 (4)

(The NO₂(g) found on both reactants and products sides has been eliminated.)

This has all the correct reactants, but the product is wrong and the stoichiometry is not quite right. If reaction (2) is reversed (meaning K_p is inverted), it can be added to reaction (4) to give the target reaction:

½N₂(g) + O₂(g) → ← NO₂Cl(g) K_p = 3×10^–10 (4)

NO₂Cl(g) → ← NOCl(g) + ½ O₂(g) K_p = (1/1.1×10² = 9.1×10^–3 (2_rev)

Adding (4) and 2_rev gives the desired reaction, both eliminating the NO₂Cl and correcting the stoichiometry.

½ N₂(g) + ½ O₂(g) + ½ Cl₂(g)→ ← NOCl(g)

With K_p = K_p(4)×K_p(2_rev) = (3×10^–10)( 9.1×10^–3) = 3×10^–12

To find K_c, use the relationship K_p = K_c(RT)^Δn

Δn = 1 –(½ + ½ + ½ ) = –½

R = 0.0821 L·atm/mol·K

T = 298 K

so

3×10^–12 = K_c[(0.0821)(298)]^–½

K_c = 3×10^–12[24.5]^+½ = 1×10^–11

Answer

The mass action expression for the reaction is .

The equilibrium concentrations of each component is given, so can be directly put into the mass action expression to give .

Answer

From the balanced reaction and the definition of K_c we can write:

PCl₅(g) → ← PCl₃(g) + Cl₂(g)

To find K_c, then, we need to find the equilibrium concentration of each component in the reaction from the given masses, the volume of the container, and the molar mass of each species.

PCl₅: [30.97 + 5(35.45)] = 208.22 g/mol.

PCl₃: [30.97 + 3(35.45)] = 137.32 g/mol.

Cl₂: [2(35.45)] = 70.90 g/mol.

Plugging each of these concentrations into the expression for K_c gives:

K_p = K_c(RT)^Δn, Δn = (1+1) – 1 = 1; R = 0.0821 L·atm/mol·K; T = 250 + 273 = 523 K

K_p = 4.28×10^–2[(0.0821)(523)]¹ = 1.84

Answer

Set up an ICE table:

NH₄HS(s) → ← NH₃(g) + H₂S(g)

K_p = P_NH3(e)P_H2S(e)

Initial? 0 0

Change–x +x +x

Equilibrium–x x x

In this case x represents the partial pressure of the gases in the container and the total pressure in the container is equal to the sum of the partial pressures (Dalton's Law of Partial Pressures). Thus, 0.658 atm = x + x = 2x, or x = 0.329 atm.

Now, K_p = (x)(x) = x² = (0.329)² = 0.108.

Answer

Convert the mass of butane (abbreviated B) to molarity, then set up a table of concentrations to solve for the equilibrium concentration of isobutane (abbreviated i-B). Then convert this molarity back to a mass of isobutane.

B: C₄H₁₀ molar mass = 4(12.01) + 10(1.01) = 58.14 g/mol

B(g) → ← i-B(g)

Initial6.88×10^–3 0

Change–x +x

Equilibrium6.88×10^–3 – x x

[i-B]_e = 6.11×10^–3 M

i-B: C₄H₁₀ so the molar mass is the same as butane (which must be true if they are isomers) = 58.14 g/mol.

So the mass of isobutane at equilibrium is:

mass = Molarity×Volume×molar mass = 6.11×10^–3 mol/L × 12.5 L × 58.14 g/mol = 4.44 g

Answer

A typical equilibrium problem: write the reaction, write the mass action expression, set up a table of concentrations, then plug into the mass action expression and solve. Assume a 1.00 L reaction vessel.

C(s) + H₂O(g) → ← CO(g) + H₂(g)

Initialxs 0.100 0 0.100

Change  – x +x +x

Equilibrium  0.100 – x x 0.100 + x

This is solved using the quadratic equation to give

x = 0.0436 or x = –0.255

Since x must be a positive number x = [CO] = 0.0436 M = 0.0436 mol in a 1.00 L vessel.

Note: because Δn = +1 for this reaction, the number of moles of CO present at equilibrium depends upon the size of the reaction vessel. For example, if you do the problem assuming a 2.00 L vessel, the amount of CO present at equilibrium is 0.0584 mol.

Answer

Set up a table of concentrations and solve from there.

CO(g) + H₂O(g) → ← CO₂(g) + H₂(g)

Initial0.250 0.250 0 0

Change– x – x +x +x

Equilibrium0.250 – x 0.250 – x x x

Note that the volume of the container does not matter in this problem (this is not always true):

Because the stoichiometry of the reactants and products is the same, the volume terms cancel out.

To solve the problem:

So the number moles of CO₂ = the number of moles of H₂ = 0.207 mol

Answer

First, convert the initial amounts of species from moles to molarity. Then do the usual equilibrium problem. Finally, convert the calculated molarities back to moles to obtain the answers.

PCl₃(g) + Cl₂(g) → ← PCl₅(g)

Initial0.100/6.40 = 0.0156 0.100/6.40 = 0.0156 0.0100/6.40 = 0.00156

Change– x – x +x

Equilibrium0.0156 – x 0.0156 – x 0.00156 + x

Solving with the quadratic equation gives

x = 0.067 or x = 0.0027 (2 significant figures because of K_c)

Since x must be less than 0.0156 (so that the PCl₃ and Cl₂ concentrations are positive at equilibrium), the correct value for x is 0.0027.

Thus, [PCl₃]_e = 0.0156 – 0.0027 = 0.0129 M, [Cl₂]_e = 0.0156 – 0.0027 = 0.0129 M, and [PCl₅]_e = 0.00156 + 0.0027 = 0.0043 M.

The number of moles of each component is found by multiplying each molarity by the volume of the reaction flask:

PCl₃: 0.0129×6.40 = 0.0826 mol

Cl₂: 0.0129×6.40 = 0.0826 mol

PCl₅: 0.0043×6.40 = 0.028 mol

Answer

, i.e., the volumes cancel for this reaction.

Since volume is unimportant for finding the equilibrium constant, suppose the volume of gas chosen for analysis contained 100.0 g of sample. This means that of the 100.0 g, 13.71 g of the sample was C and 86.29 g of the sample was S, based on the analysis. The number of moles of C in the gas phase sample = 13.71 g / 12.011 g/mol = 1.141 mol. All of the carbon in the gas phase is in the carbon disulfide, so the 100.0 g sample must contain 1.141 mol CS₂. The total number of moles of sulfur in the 100.0 g sample = 86.29 g / 32.066 g/mol = 2.691 mol. Of this total, 2(1.141) = 2.282 mol of sulfur is used in CS₂, leaving 2.691 – 2.282 = 0.409 mol sulfur in S₂, or 0.409/2 = 0.205 mol S₂. Thus, at equilibrium, in the 100.0 g sample there is 1.141 mol CS₂ and 0.205 mol S₂. Now K_c can be found:
K_c = (1.141)/(0.205) = 5.57.