Practice questions (Ch. 4-6)

Question 1

True or false: Chemical equilibrium is achieved when the concentrations of products and reactants are equal, signaling that the forward and reverse reactions are completed.

Answer 1

This statement is false. Equilibrium does NOT mean equality in the traditional sense of the word. The concentrations of the products and reactants are not necessarily equal, but the concentrations are not changing over time. The forward and reverse reactions are still proceeding, but at equal rates, so the concentrations of the reactants and products are not changed.

Question 2

What is the correct expression for Keq for the following reaction?
NH3 (g) + HCl (g) → NH4Cl (s)

Answer 2

Keq = 1/[NH3][HCl]
This answer choice correctly divides products over reactants, while accounting for the fact that ammonium chloride is a solid, therefore omitting it from the final equation.

Question 3

Which of the following modifications to a reaction would change Keq?
A) raising temperature
B) adding a catalyst
C) INcrease the concentration of reactants

Answer 3

A raising temp
Most Keq values are given at standard temperature, 25°C. However, heat can also be operationalized as a “participant” in the reaction and can shift the reaction left or right, which we’ll get to soon. But heat is not accounted for in Keq, so Keq can be impacted by heat. For now, though, just understand that changes in temperature result in changes in Keq.

Question 4

True or False: The primary difference between the equilibrium constant, K, and the reaction quotient, Q, is that Q can be used at any point in the reaction, whereas K is a measure at equilibrium.

Answer 4

This statement is true. The formula for Q is the same as for K, with a key difference being that Q informs us what direction the reaction will proceed to reach equilibrium. If Q is lower in comparison to K, then it shows that reactants currently dominate the mixture, and more products will need to be made, so the reaction will proceed in the forward direction to move toward equilibrium. The converse is true if Q is greater than K: products dominate, so more reactants need to be made, driving the reaction in the reverse direction. When Q = K, it indicates that the reaction is already at equilibrium.

Question 5

A biochemist is conducting a thermodynamic analysis of a biologically relevant reaction. The Keq for this reaction is 5.06 × 10-13. At one moment in time, the biochemist calculates that Q is 7.69 × 10-14. Which of the following statements is true at this moment?
A) The reaction is at equilibrium
B) The reaction will proceed to favor reactants
C) the reaction will proceed to favor products

Answer 5

B is correct. The reaction quotient, Q, is calculated in the same manner as Keq, but can be found for a reaction at any stage, not just at equilibrium. Q is much smaller than Keq in this problem. Watch out for the negative exponents in this problem! Thus, in this moment, we have fewer products present than we would “want” to have at equilibrium, and the reaction will proceed in the forward direction to generate more products.

Question 6

2 SO2 (g) + O2 (g) ⇋ 2 SO3 (g)
Adding O2 to the reaction mixture at equilibrium would have what effect?

Answer 6

Addition of O2 results in overabundance of O2, so the reaction would shift to the right - towards the product - to use up the extra O2. Note how the reaction shifts in the opposite direction of the stressor. Whenever the concentration of a species in an equilibrium mixture is increased, the system always shifts in the opposite direction to consume the extra species.

Question 7

H2 (g) + I2 (g) ⇋ 2 HI (g)

True or False: Removing some H2 from the reaction mixture would result in the concentration of I2 increasing.

Answer 7

This statement is true. Removing a reactant is the same as adding a stress that causes the system to shift toward the reactants to generate more of the lost reactant. Since H2 is lost here, the reaction will shift to the left to make up for that loss and replenish H2. Even though I2 wasn’t removed initially, the concentration of it still increases: as the reaction shifts left, I2 is made along with H2.

Question 8

2 NO2 (g) ⇋ N2O4 (g)
If the pressure is increased, in which direction will the reaction shift?

Answer 8

Pressure is considered a stress. To relieve that stress, the system will shift towards the side with fewer moles of gas: the right side. The right side has 1 mole compared to the left side’s 2 moles.

Question 9

Increasing the volume of a reaction container will shift the system towards the side with fewer moles, regardless of the substances’ states of matter.

Answer 9

This statement is false. Increasing the volume is akin to a reduction in pressure. To regain this pressure back, the system will shift towards the side with more gaseous moles. This is due to the increased pressure exerted by the greater amount of gas on that side. Thus, the system will shift towards the side with greater moles. The other error in this statement is the states of matter of the substances in the reaction. For equilibrium expressions, liquid and solid products or reactants are not considered. Likewise, they are not factored into Le Châtelier’s Principle either, because they have no impact on pressure or volume shifts. Only gaseous moles are considered when determining system shifts as part of Le Châtelier’s Principle.

Question 10

H2 (g)  + I2 (g) ⇋ 2 HI (g) ΔH = +53 kJ
What modification(s) to the reaction chamber will shift the system to the right?

Answer 10

Raising the temperature
the reaction is endothermic. The system takes in 53 kJ of heat, making heat a requirement for the reaction to proceed. Thus, heat can be operationalized as a reactant. Adding MORE heat by raising the temperature would shift the system to the right, just as adding more of a reactant would.

Question 11

What is the temperature change associated with a calorimetry experiment on a 50 Calorie candy cane if the starting temperature of the water is 25°C? Note the specific heat capacity of water is 1 cal/(g•°C) and the amount of water used is 150ml.

Answer 11

333°C

To solve for the temperature change of the water in a calorimetry experiment we need to use the equation Q = mcΔT. The heat energy released is represented by “Q” and measured in calories. The mass of the water is represented by “m” and measured in grams. The specific heat capacity of water represented by “c” and is equal to 1 calorie/(gram•°C). The temperature change of the water is represented by “ΔT” and is measured in degrees Celsius. The first step is to convert given values into the proper units.
50 Calories = 50 x 103 calories
150 ml H2O = 150 g H2O
Next, we need to plug the values provided into our calorimetry equation.

Question 12

How much water would be required for a 100-calorie cracker to increase the temperature of that water by 64°C? Note the specific heat capacity of water is 1 calorie/ g°C.

Answer 12

1.5g
To solve this problem, we need to use our calorimetry equation, Q = mcΔT. The heat energy released is represented by Q and measured in calories. The mass of the water is represented by m and measured in grams. The specific heat capacity of water is represented by c and is equal to 1 calorie/(gram)( °C). The temperature change of the water is represented by ΔT and is measured in degrees Celsius. Plug all known values into the equation and solve for m.
Because we rounded the denominator down, we know the true answer is a little bit less than 1.6 g. Choice B is the only answer choice that is just a little bit less than 1.6 g so it must be the correct answer.

Question 13

What is the specific heat capacity of ethanol if 3.25 calories raise the temperature of 1g of ethanol by 6°C? Note that 1 calorie = 4.184 joules

Answer 13

2.46 J/g°C
Notice how all of the answer choices have different units. This means that your initial approach to this question should be to determine what the correct units are before doing any calculations. The specific heat capacity of a substance describes its ability to hold heat as it’s added, or more technically speaking, how much heat must be added for 1 gram of that substance to increase in temperature by 1 degree Celsius. Specific heat capacity changes depending on the substance being used. Heat energy can be measured in joules or calories. This means that specific heat capacity can be measured in joules/(g°C) or calories/(g°C). This is the only answer choice with the correct units for specific heat capacity so it must be the right answer.

Question 14

Which factors influence enthalpy?

Answer 14

Thermal energy, pressure, volume

Question 15

At what temperature will the formation of dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P) from 2 moles of fructose 1,6-bisphosphate (F 1,6-BP) need to be coupled with ATP hydrolysis in order to be spontaneous? Note the standard change in enthalpy for this reaction is 48.79 kJ/mol and the standard change in entropy is 960 J/K.

F 1,6-BP → DHAP + G3P

Answer 15

If the reaction needs to have something done to it in order to be spontaneous (such as be coupled to ATP hydrolysis) this indicates that the reaction must otherwise be nonspontaneous. Therefore, we can read word this question to be “At what temperature is this reaction nonspontaneous?”. Because enthalpy and entropy are both positive, we need to use the equation for Gibbs free energy to determine when the reaction will be nonspontaneous. Because the reaction could be nonspontaneous at a wide variety of temperatures, the best approach is to determine where the tipping point is. The best way to do this is to set ΔG = 0 J and determine what the temperature is. Once you know the temperature at ΔG = 0 J you know that any temperature greater than that means the reaction is spontaneous (-ΔG) and at any temperature lower than that the reaction is nonspontaneous (+ΔG).

Question 16

Which of the following indicates a reaction that will shift to favor the reactants?
A) Keq = .5, Q = 1
B) Keq= 2.73, Q = 4.01
C) Keq = .8, Q = .62

Answer 16

A & B
While the equilibrium constant Keq is used only at equilibrium, Q (the reaction quotient) may be used for reactant and product concentrations at any point in the reaction. When Q is greater than Keq, the relative proportions of products to reactants is higher than it would like to be at equilibrium, so the reaction will want to proceed in reverse from products toward reactants to achieve equilibrium. The question stem asks us to identify a reaction that will shift towards the reactants, meaning it will run in the reverse direction. Therefore, any reaction where Q > Keq will run in reverse to create more reactants.

Question 17

True or false: Viscosity is the amount of energy needed to overcome the intermolecular attractions between liquid particles that cause it to resist flow.

Answer 17

This statement is false. Viscosity is the resistance of liquid to flow. Liquids with higher viscosity will flow more slowly than liquids of lower viscosity.

Question 18

Complete the following statement: Capillary action is due to the ______________

Answer 18

Capillary action refers to the movement of a liquid up the sides of a narrow tube against the force of gravity. This occurs due to the adhesion of liquid molecules to the sides of the tube, which then pull lower liquid molecules upward by cohesion.

Question 19

True or false: The phase diagram for water has a solid/liquid boundary that slants upwards towards the right due to the decrease in density from solid to liquid.

Answer 19

This statement is false. Unlike most substances, the phase diagram for water has a solid/liquid boundary that slants up and to the left (instead of right) due to water being less dense in its solid form than in its liquid form.

Question 20

Under the kinetic molecular theory, the pressure of a sample of gas comes from collisions between gas particles and ______

Answer 20

he kinetic molecular theory asserts that the pressure on the walls of a container is a result of elastic collisions of the particles with those walls.

Question 21

After a birthday party, a child takes a balloon filled with 1 L of helium from inside (where it was 25°C) into a cold car (where it is 0°C). How will the volume of the balloon change?

Answer 21

Charles’ law states that gas temperature and volume are directly proportional. In this setup, the boy takes a balloon from 298K to 273K. Thus, the volume will decrease by (273 K / 298 K), which is much less than a ½ reduction.

Question 22

If held at the same temperature and pressure, what will be the difference in volume between a red balloon holding 2 moles of gas and a blue balloon holding 4 moles of gas.

A

Answer 22

According to Avogadro’s law, the relationship between moles of gas and volume is directly proportional. Therefore, with twice the number of moles, the blue balloon should have twice the volume of the red balloon.

Question 23

How many liters does one mole of gas occupy at STP?

Answer 23

At STP, a mole of gas occupies 22.4 L.

Question 24

3.75 moles of nitrogen gas is held in a rubber balloon with a current volume of 1 L. The balloon is placed in a room with an initial pressure of 1 atm and temperature of 25°C. If the pressure suddenly drops to 0.75 atm, what change in volume is necessary if the temperature is to remain the same?

Answer 24

While we could plug all of the given values into the ideal gas law, we can make this problem much simpler by using Boyle’s law, P1V1 = P2V2. Since initial pressure is 1 atm and initial volume is 1 L, we get (1 atm)(1 L) = (0.75 atm)(x L). The new volume must be approximately 1.33 L.

Question 25

A sample of gas held at 298K and 10 atm occupies 1 L. How many moles must this sample contain? (Note: R = 0.08 L atm / mol K)

Answer 25

To solve this problem, simply plug the values into the ideal gas law and solve. PV=nRT can be rearranged to give n = PV/RT. The result is roughly 0.4 moles.

Question 26

2 moles of oxygen gas are held in a container at 1 atm and 15°C. For the temperature of this gas to increase to 83°C, what increase in pressure is necessary? Assume that the volume of the container cannot change.

Answer 26

Remember, temperatures must be changed to Kelvin when dealing with the ideal gas law. This gas begins with a temperature of 288 K and must increase about 1.25-fold to reach 356 K. According to the equation PV = nRT, if all other factors remain constant, pressure must change by this amount as well. A 125% change in pressure results in a new value of 1.25 atm. Note that it is also perfectly fine to plug in all values to the ideal gas law to find this answer.

Question 27

A biologist is making a variety of measurements for a sample of CO2 generated by a colony of aerobic bacteria. The sample contains exactly 1 mole of gas and is being held in a 0.05 L container. When the biologist calculates the PV/ RT ratio of the gas, he finds that it is 1.05. How could this finding be explained?

Answer 27

Since the sample contains 1 mole of gas, we can rearrange the ideal gas law to find our expected PV/RT ratio: 1. The given value is higher, implying that either pressure or volume is larger than expected. For the ideal gas law to work properly, two main assumptions are necessary: that the volume of particles themselves is negligible and that no intermolecular forces are present between particles. However, these assumptions become especially untrue at low temperature and high pressure. Classically, we think of low temperatures systems being significantly moved away from ideal gas behavior by intermolecular forces, and high pressure systems as being influenced primarily by intermolecular collisions.

Question 28

A sealed flask is filled with three gases: 0.5 mol ammonia, 0.25 mol oxygen, and 0.75 mol helium. If the partial pressure of NH3 is 415 torr, what is the total pressure in the flask?

Answer 28

Dalton’s law states that the total pressure in a vessel is the sum of the partial pressures of the components. Moreover, the partial pressure of a gas is proportional to its mole fraction in the container. We already know that the total number of moles in the flask is 0.5 + 0.25 + 0.75 = 1.5 moles. Since exactly one-third of this is ammonia, the total pressure is 3*415 torr, or 1245 torr.

Question 29

HCl (aq) + NaHCO3 (aq) → H2O (l) + CO2 (g) + NaCl (aq)

250 mL of 2 M HCl is mixed with excess sodium bicarbonate in a closed container and allowed to react. If the carbon dioxide produced is then transferred to a glass vessel with 0.5 mol water vapor and 0.5 mol gaseous ammonia, what will be the mole fraction of CO2 in the vessel?

Answer 29

Here, mole fraction can be calculated as the number of moles of carbon dioxide present divided by the total moles of gas. 250 mL HCl × (2 mol HCl / 1000 mL) × (1 mol CO2 / 1 mol HCl) = 0.5 mol CO2. To find our answer, take 0.5 mol CO2 / 1.5 total moles in the vessel = 0.33.

Question 30

Gas A, a compound used in industrial cleaning, is observed to diffuse through a room at approximately 0.1 m/s. If helium gas diffuses through the same room at 0.8 m/s, the molar mass of Gas A is:

Answer 30

Graham’s law can be used to compare the rates of diffusion of two gases. This equation can be written as rate 1 / rate 2 = √molar mass 2 / √molar mass 1; in other words, the rate of diffusion of a gas varies with the reciprocal of the square root of its molar mass. Here, let’s give helium “Rate 1” and our unknown gas “Rate 2.” This equation becomes 0.8 m/s / 01. m/s = √molar mass 2 / √4g/mol, and the molar mass of Gas A can be calculated at 256 g/mol.

Question 31

True or false: Oxygen would be expected to effuse more quickly than carbon dioxide.

Answer 31

This statement is true. Graham’s law states that both the effusion rate and the diffusion rate are directly proportional to the reciprocal of the square root of the molar mass of a gas. Thus, as molecular mass increases, the effusion rate decreases. Since oxygen has a smaller molar mass than carbon dioxide, it would be expected to effuse more quickly.

Question 32

Which of the following solvents would most effectively dissolve propane?
A) Acetone
B) Hexane
C) H2O

Answer 32

As the only non-polar solvent among our options, hexane would be the most favorable solvent for a nonpolar molecule such as propane.

Question 33

True or false: A compound must be able to ionize to dissolve in water.

Answer 33

This statement is false. While ionic compounds do tend to readily dissolve in water, and are broken apart into their constituent ions when they do so, covalent molecules are also capable of dissolving in water and do not form ions when they do so. Consider glucose as an example, which dissolves readily in the blood without taking on any ionic form.

Question 34

True or false: The molarity and molality of a solution are always different values.

Answer 34

This statement is false. The molarity and molality of a solution are usually different values, but they can be the same for a very dilute solution with a solvent with a density of 1 (i.e. water).

Question 35

Which of the following unknown solutes would be least soluble in a given solvent?
A) 1 mol of compound A (MW = 58g/mol, Ksp = 1.8x10^-5)
B) o,5 mol Compound B (MW = 80g/mol, ksp = 2.8x10^-9)
C) o.5 mol Compound D (MW = 130g/mol, ksp= 3.1x10^-4)

Answer 35

B

ompound B has the smallest solubility constant, making it the least soluble solute.

Question 36

True or false: Only the concentration of the aqueous product ions are included when calculating the Ksp of a dissolution.

Answer 36

This statement is true. The concentrations of solids, such as the original undissolved solute, and pure liquids, such as the water dissolving the solute, are not included in an equilibrium constant, leaving only the concentration of the product ions.

Question 37

The activation energy is the?

Answer 37

This is the definition of activation energy. For a reaction to start, there needs to be enough energy to get the reactants into the transition state; after this, the reaction will spontaneously proceed to form the products. Remember that high energy compounds are unstable, thus they naturally drop down in energy to more stable compounds

Question 38

True or false: If the products are lower in energy than the reactants, the reaction is not reversible.

Answer 38

Although the reverse reaction (products back into reactants) would require a higher energy input to reach the transition state, the reaction described is definitely reversible. However, it may not necessarily be spontaneous. Remember that spontaneous and reversible are different concepts; if a reaction is not spontaneous it can still be reversible with the right amount of energy!

Question 39

True or false: The order of a reaction can be determined from the stoichiometry of that reaction.

Answer 39

This statement is false. The order of a reaction must be determined experimentally (e.g. by a passage telling you that when you double the amount of a reactant, the reaction rate quadruples). On the MCAT, do not assume the coefficients in a reaction will tell you the rate.