# Chapter 15 Test...And Then Some

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Quizzes Created: 6 | Total Attempts: 4,798
Questions: 23 | Attempts: 185  Settings  • 1.

### Units for equilibrium constants are conventionally omitted because they vary from reaction to reaction.  If they were to be included, the units on Keq for this reaction would be:

• A.

Atm

• B.

1/atm

• C.
• D.

All units would cancel out

C.
Explanation
Equilibrium constants are calculated by dividing the concentrations of products by the concentrations of reactants, each raised to the power of their respective stoichiometric coefficients. Since the concentrations of reactants and products are expressed in different units (such as moles per liter or molarity), the units on Keq would depend on the specific units used for concentrations. Therefore, including units for equilibrium constants would not be meaningful or consistent across different reactions.

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• 2.

### At 150 degrees Celsius the equilibrium constant for the reaction below has a value of 300.  The reaction was allowed to reach equilibrium in a sealed container and the partial pressure due to IBr (g) was found to be 3 atm.  Which of the following could be the partial pressure due to the Br2 (g) and I2 (g) in the container?

• A.

A

• B.

B

• C.

C

• D.

D

• E.

E

A. A
Explanation
At 150 degrees Celsius, the equilibrium constant for the reaction is 300. This means that the ratio of the products (Br2 and I2) to the reactant (IBr) is 300:1. Since the partial pressure due to IBr is 3 atm, the partial pressure due to Br2 and I2 combined should be 300 times smaller, which is 0.01 atm. Therefore, the partial pressure due to Br2 and I2 in the container could be 0.01 atm.

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• 3.

### Consider a system at equilibrium based on the reaction above.  In such a system, the concentration of oxygen gas is:

• A.

Constant and equal to the concentration of ozone gas

• B.

Constant and less than the concentration of ozone gas

• C.

Constant and greater than the concentration of ozone

• D.

Increasing and the concentration of ozone is increasing

• E.

Increasing and the concentration of ozone is decreasing

C. Constant and greater than the concentration of ozone
Explanation
In a system at equilibrium, the concentrations of reactants and products remain constant over time. The given answer states that the concentration of oxygen gas is constant and greater than the concentration of ozone gas. This suggests that the reaction is not consuming all of the oxygen gas and that there is an excess of oxygen gas present in the system compared to the concentration of ozone gas.

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• 4.

### Given that the Keq for the reaction,  , is 36.  What will be the value for the following reaction:

• A.

36

• B.

18

• C.

6

• D.

1/6

• E.

1/36

D. 1/6
Explanation
The value for the given reaction can be determined by comparing the Keq values of the reactions. Since the Keq for the given reaction is 36, which is larger than 1, it indicates that the reaction favors the forward direction. Therefore, the value for the given reaction would be a positive fraction, and the only positive fraction option provided is 1/6.

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• 5.

### Which of the following concentration vs. time graphs represents the reaction:

• A.

A

• B.

B

• C.

C

• D.

D

A. A
Explanation
Graph a shows a rapid decrease in concentration over time, indicating a fast reaction. This suggests that the reactants are being consumed quickly and converted into products. Graphs b, c, and d show either no change or a slow change in concentration, which suggests a slower or no reaction occurring. Therefore, graph a is the most representative of a reaction.

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• 6.

### Solids are not included in an equilibrium expression because:

• A.

Their concentrations are all 1.0 M and drop out

• B.

Adding more solid does not change the density (concentration of the solid

• C.

Adding more solid does not change the surface area of the solid

• D.

Equilibrium reactions are usually balanced with the lowest hole numbers

• E.

Solids no longer continue to dissolve once the solution because saturated

B. Adding more solid does not change the density (concentration of the solid
• 7.

### When the system is at equilibrium:

• A.

No further reaction occurs in either direction

• B.

The concentrations of reacts and products are equal

• C.

Forward and reverse reactions are still taking place

• D.

All reactants have been consumed

C. Forward and reverse reactions are still taking place
Explanation
When the system is at equilibrium, it means that the forward and reverse reactions are occurring at the same rate. This does not mean that no reaction is happening, but rather that the rates of the forward and reverse reactions are balanced. The concentrations of reactants and products may be equal at equilibrium, but this is not always the case. Additionally, it is not necessary for all reactants to be consumed for the system to be at equilibrium.

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• 8.

B.
• 9.

### Consider the equilibrium reaction: What will be the effect of halving the concentration of sulfur trioxide?

• A.

Sulfur dioxide and oxygen will increase equally

• B.

Sulfur increases more than oxygen

• C.

Sulfur dioxide and oxygen decrease equally

• D.

Sulfur dioxide decreases more than oxygen

• E.

Oxygen decreases more than sulfur dioxide

D. Sulfur dioxide decreases more than oxygen
Explanation
Halving the concentration of sulfur trioxide will result in a decrease in the concentration of both sulfur dioxide and oxygen. However, according to the balanced equation for the equilibrium reaction, the stoichiometric ratio between sulfur trioxide and sulfur dioxide is 1:1, while the ratio between sulfur trioxide and oxygen is 1:0.5. Therefore, a halving of the concentration of sulfur trioxide will have a greater effect on the concentration of sulfur dioxide compared to oxygen. As a result, sulfur dioxide decreases more than oxygen.

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• 10.

### At 2000 K, the value of Kc is  for the reaction:

• A.

There is more product present than the reactant

• B.

There is more reactant present than product

• C.

There is a 50/50 mix of reactant and product

• D.

A. There is more product present than the reactant
Explanation
At 2000 K, the value of Kc is greater than 1, indicating that the concentration of the products is higher compared to the concentration of the reactants. This suggests that the reaction strongly favors the formation of products at this temperature.

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• 11.

### Raising the temperature of a system which has reached equilibrium:

• A.

Changes the value of K

• B.

Increases the kinetic energy of the system

• C.

Causes an increase in concentration of reactants for an exothermic reaction

• D.

All of the above

• E.

More than one of the above, but not all of the above

A. Changes the value of K
Explanation
When the temperature of a system at equilibrium is raised, it causes a shift in the equilibrium position according to Le Chatelier's principle. This shift can favor either the forward or the reverse reaction, depending on whether the reaction is exothermic or endothermic. As a result, the concentrations of reactants and products may change, which in turn affects the equilibrium constant, K. Therefore, raising the temperature of a system at equilibrium changes the value of K.

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• 12.

### If the equilibrium constant for this reaction is 100, what will happen to a mixture in which all species are initially 0.01 M?

• A.

Reaction goes tot he right

• B.

Reaction goes left

• C.

No changes occur, reaction is at equilibrium

• D.

Cannot predict

B. Reaction goes left
Explanation
If the equilibrium constant for the reaction is 100 and all species are initially at a concentration of 0.01 M, it means that the reactants are initially in excess compared to the products. This suggests that the reaction will shift to the left in order to reach equilibrium and establish a balance between reactants and products. Therefore, the answer "reaction goes left" is correct.

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• 13.

### Which must be true of the equilibrium concentrations when 0.10 mole of each reactant are added to an empty 1.0 L container and allowed to reach equilibrium?

• A.

[NO]

• B.

[NO]>[O2]

• C.

[NO]=[O2]

• D.

[NO]=[NO2]

• E.

[NO2]=2[O2]

A. [NO]
Explanation
When 0.10 mole of each reactant are added to an empty 1.0 L container and allowed to reach equilibrium, the concentration of [NO] must be the same as the concentration of [NO] in order to satisfy the equilibrium condition. Therefore, the equilibrium concentrations of [NO] must be equal.

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• 14.

### Consider the reaction: the yield of PCl5 could be increased by.

• A.

Adding Ne gas at constant total pressure

• B.

Increasing the temperature

• C.

Removing the Cl2 from the system

• D.

• E.

Reducing the volume of the system

E. Reducing the volume of the system
Explanation
Reducing the volume of the system would increase the pressure, which would favor the forward reaction according to Le Chatelier's principle. Since the formation of PCl5 is an exothermic reaction, increasing the pressure would shift the equilibrium towards the side with fewer moles of gas, which in this case is the formation of PCl5. Therefore, reducing the volume of the system would increase the yield of PCl5.

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• 15.

### For the reaction: 1.0 mole of Fe and 0.50 moles of oxygen are mixed together, reacted and allowed to come to equilibrium in a 1.0 L container.  At equilibrium, 0.20 moles of Fe2O3 have formed.  What is the Keq at this temperature?

D.
Explanation
The reaction can be represented as:

4Fe + 3O2 ⇌ 2Fe2O3

Since 0.20 moles of Fe2O3 have formed at equilibrium, it means that 0.10 moles of Fe have reacted. This implies that 0.10 moles of O2 have also reacted, as the reaction stoichiometry is 4:3.

The initial moles of Fe and O2 were 1.0 and 0.50 respectively. So, at equilibrium, the moles of Fe and O2 remaining are 1.0 - 0.10 = 0.90 and 0.50 - 0.10 = 0.40 respectively.

Using the law of mass action, Keq can be calculated as [Fe2O3]^2 / ([Fe]^4 * [O2]^3), where [Fe2O3], [Fe], and [O2] represent the molar concentrations at equilibrium.

Plugging in the values, Keq = (0.20^2) / ((0.90^4) * (0.40^3))

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• 16.

### The graph shows the variation of concentration with time for the reaction at 25 degrees Celsius. What is the value of the equilibrium constant, K, at this temperature?

• A.

0.667

• B.

0.845

• C.

1.19

• D.

2.67

C. 1.19
Explanation
Based on the graph, the equilibrium constant (K) can be determined by finding the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. In this case, the ratio is approximately 1.19, which means that the concentration of the products is 1.19 times greater than the concentration of the reactants at equilibrium. Therefore, the value of the equilibrium constant (K) at this temperature is 1.19.

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• 17.

### The quantities which are equal in an equilibrium system are

• A.

Forward and reverse reaction rates

• B.

Amounts of products and reactants

• C.

The forward and reverse rate constants

• D.

Forward and reverse activation

• E.

All of the above

A. Forward and reverse reaction rates
Explanation
In an equilibrium system, the forward and reverse reaction rates are equal. This means that the rate at which products are formed from reactants is the same as the rate at which reactants are formed from products. This balance between the forward and reverse reactions is what defines an equilibrium state. The other options listed, such as the amounts of products and reactants, the forward and reverse rate constants, and the forward and reverse activation energies, are not necessarily equal in an equilibrium system.

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• 18.

### As a catalyst is added to a system at equilibrium, the value of the equilibrium constant

• A.

Decreases

• B.

Increases

• C.

Stays the same

C. Stays the same
Explanation
When a catalyst is added to a system at equilibrium, it increases the rate of both the forward and reverse reactions equally. This means that the system will reach a new equilibrium faster, but the concentrations of the reactants and products at the new equilibrium will remain the same. Therefore, the value of the equilibrium constant, which is determined by the concentrations of the reactants and products at equilibrium, will also stay the same.

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• 19.

### The Ksp for PbCl2 is 1.6x106-5.  of the following choices which is the greatest amount of Pb(NO3)2 that can be added to 1.0 L of 0.010 M NaCl solution at constant temperature without causing precipitation to occur.

• A.

0.0004 mol

• B.

0.0015 mol

• C.

0.0080 mol

• D.

0.15 mol

D. 0.15 mol
Explanation
The Ksp value for PbCl2 is given, which represents the solubility product constant. This constant indicates the maximum concentration of the ions in a saturated solution. In this case, the Ksp value indicates that the maximum concentration of Pb2+ and Cl- ions in a saturated solution is 1.6x10^-5 M.

To determine the maximum amount of Pb(NO3)2 that can be added without causing precipitation, we need to consider the reaction between Pb2+ and Cl- ions. Pb(NO3)2 dissociates into Pb2+ and 2NO3- ions. When Pb2+ ions react with Cl- ions, PbCl2 is formed.

Since the concentration of Cl- ions is given as 0.010 M in the NaCl solution, we need to calculate the maximum amount of Pb2+ ions that can be present without exceeding the Ksp value. The stoichiometry of the reaction tells us that for every 1 Pb2+ ion, 2 Cl- ions are required. Therefore, the maximum concentration of Pb2+ ions that can be present without exceeding the Ksp value is 0.010 M / 2 = 0.005 M.

To calculate the amount of Pb(NO3)2, we multiply the concentration of Pb2+ ions by the volume of the solution:
0.005 M * 1.0 L = 0.005 mol

Therefore, the greatest amount of Pb(NO3)2 that can be added without causing precipitation is 0.005 mol, which is not listed as an option. As a result, the correct answer cannot be determined from the given options.

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• 20.

### What is the hybridization of orbitals in the ammonia molecule (NH3)?

• A.

Sp

• B.

Sp2

• C.

Sp3

• D.

Dsp3

• E.

D2sp3

C. Sp3
Explanation
The hybridization of orbitals in the ammonia molecule (NH3) is sp3. This is because the central nitrogen atom is bonded to three hydrogen atoms and has one lone pair of electrons. In order to form four bonds and minimize electron repulsion, the nitrogen atom undergoes hybridization, combining one s orbital and three p orbitals to form four sp3 hybrid orbitals. These hybrid orbitals are then used to form sigma bonds with the hydrogen atoms, resulting in a tetrahedral molecular geometry.

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• 21.

• A.

A

• B.

B

• C.

C

• D.

D

• E.

E

E. E
• 22.

### Consider the chemical bonds found in solid sodium hydrogen carbonate.  For each bond specified, choose the best description of the bond type. a.  ionic bond b.  hydrogen bond c.  single covalent bond d.  double covalent bond e.  resonance covalent bond with bond order between 1 and 2 Sodium/Hydrogen carbonate bond

• A.

A

• B.

B

• C.

C

• D.

D

• E.

E

A. A
Explanation
The correct answer is "a. ionic bond". In solid sodium hydrogen carbonate, the bond between sodium and hydrogen carbonate is an ionic bond. Ionic bonds occur when one atom donates electrons to another atom, resulting in the formation of positive and negative ions that are attracted to each other. In this case, sodium donates an electron to the carbonate ion, forming a positive sodium ion and a negative carbonate ion, which are held together by electrostatic attraction. This type of bond is typically found between a metal and a nonmetal.

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• 23.

### Consider the chemical bonds found in solid sodium hydrogen carbonate.  For each bond specified, choose the best description of the bond type. a.  ionic bond b.  hydrogen bond c.  single covalent bond d.  double covalent bond e.  resonance covalent bond with bond order between 1 and 2 oxygen/hydrogen bond

• A.

A

• B.

B

• C.

C

• D.

D

• E.

E

C. C
Explanation
The bond between oxygen and hydrogen in solid sodium hydrogen carbonate is a single covalent bond. This type of bond involves the sharing of electrons between the oxygen and hydrogen atoms.

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