Home / AP Chemistry: 3.5 Kinetic Molecular Theory – Exam Style questions with Answer- FRQ

AP Chemistry: 3.5 Kinetic Molecular Theory – Exam Style questions with Answer- FRQ

Question

Elemental sulfur can exist as molecules with the formula \(S_8\) . The \(S_8\) molecule is represented by the incomplete Lewis diagram above.
(a) The diagram of \(S_8\) shows only bonding pairs of electrons. How many lone pairs of electrons does each S atom in the molecule have?
(b) Based on your answer to part (a), determine the expected value of the S–S–S bond angles in the S8 molecule.
(c) Write the electron configuration for the S atom in its ground state.
(d) The complete photoelectron spectrum for the element chlorine is represented below. Peak X in the spectrum corresponds to the binding energy of electrons in a certain orbital of chlorine atoms. The electrons in this orbital of chlorine have a binding energy of 273 MJ/mol, while the electrons in the same orbital of sulfur atoms have a binding energy of 239 MJ/mol.

(i) Identify the orbital and explain the difference between the binding energies in terms of Coulombic forces.
(ii) Peak Y corresponds to the electrons in certain orbitals of chlorine atoms. On the spectrum shown, carefully draw the peak that would correspond to the electrons in the same orbitals of sulfur atoms.

\(3S_{8}+8OH^{-}\rightarrow 8S_{3}^{-}+4HOOH\)

In an experiment, a student studies the kinetics of the reaction represented above and obtains the data shown in the following table.

(e) Use the data in the table to do the following.
(i) Determine the order of the reaction with respect to \(S_8\) . Justify your answer.
(ii) Determine the value of\( [OH^−]\) that was used in trial 3, considering that the reaction is first order with respect to \(OH^−\). Justify your answer. The next day the student conducts trial 4 using the same concentrations of  \(S_8\) and \(OH^−\) as in trial 1, but the reaction occurs at a much slower rate than the reaction in trial 1. The student observes that the temperature in the lab is lower than it was the day before.
(f) Using particle-level reasoning, provide TWO explanations that help to account for the fact that the reaction rate is slower in trial 4.

▶️Answer/Explanation

(a) Two

(b)\(109.5^{\circ}\)

       Acceptable range: \(104^{\circ}\leq angle \leq 110^{\circ}\).
     (The experimentally determined angle is \(107^{\circ}8\)

(c) \(1s^{2} 2s^{2}2p^63s^2 3p^4\)

      OR [Ne]\(3s^23p^4\)

(i) Peak X represents electrons in a 1s orbital. A Cl atom has one more proton in its nucleus than does a S atom; therefore, the electrons in Cl are more strongly attracted to the nucleus, and the binding energy of the 1s electrons in the Cl atom is greater than that of the 1s electrons in the S atom.

(ii) See example of a correct response (dashed peak) above.

(e)(i)

e(ii) Comparing trials 2 and 3, [S8] is kept constant and the initial reaction rate doubles. Since the reaction is first order with respect to\( OH^{-}\), the concentration of \(OH^{-}\) intrial 3 must be\( 2 \times 0.0100 M \)= 0.0200 M.

(f) The temperature was lower on the second day so the average kinetic energy of the reactant particles was lower. Therefore, there were fewer collisions between particles with sufficient
energy to react. Since the temperature was lower, the kinetic energy was lower and the average speed of the particles was lower. At the lower speeds, the reactant particles collided less frequently.

 Question

A student investigates gas behavior using a rigid cylinder with a movable piston of negligible mass, as shown in the diagram above. The cylinder contains 0.325 mol of O2(g).
(a) The cylinder has a volume of 7.95 L at 25°C and 1.00 atm. Calculate the density of the O2(g), in g/L, under these conditions.

(b) Attempting to change the density of the O2(g), the student opens the valve on the side of the cylinder, pushes down on the piston to release some of the gas, and closes the valve again. The temperature of the gas remains constant at 25°C. Will this action change the density of the gas remaining in the cylinder? Justify your answer.

(c) The student tries to change the density of the O2(g) by cooling the cylinder to −55°C, which causes the volume of the gas to decrease. Using principles of kinetic molecular theory, explain why the volume of the O2(g) decreases when the temperature decreases to −55°C.

 

(d) The student further cools the cylinder to −180°C and observes that the measured volume of the O2(g) is substantially smaller than the volume that is calculated using the ideal gas law. Assume all equipment is functioning properly. Explain why the measured volume of the O2(g) is smaller than the calculated volume. (The boiling point of O2(l) is −183°C.)

▶️Answer/Explanation

Ans:

(a) For the correct calculated value:
Accept one of the following:

• \(0.325 mol O_{2}\times \frac{32.00 gO_{2}}{1 mol O_{2}}=10.4 gO_{2}\)

\(D = \frac{m}{V}=\frac{10.4 g}{7.95 L}=1.31 g/L\)

• \(D = \frac{m}{V}=\frac{P(MM)}{RT}=\frac{(1.0 ati)(32.00 g/mol)}{(0.08206\frac{L\cdot atm}{mol\cdot K})(298K)}=1.31 g/L\)

(b) For the correct answer and a valid justification:
Accept one of the following:
• No, the density of the gas remains constant because P, R, and T remain constant AND the mass and volume of O2 decrease proportionately.
• A mathematical justification is shown below

\(D = \frac{m}{V}=\frac{n ~moles~ of ~O_{2}\times molar ~mass ~of ~O_{2}}{\frac{nRT}{P}}=\frac{P\times (molar ~mass ~of~ O_{2})}{RT}\)

(c) For a valid explanation:
Accept one of the following:
• As the gas cools, the average kinetic energy (speed) of the O2 molecules decreases. The molecules rebound with less energy when they collide with each other and the walls of
the container. The spacing between particles decreases, causing the volume occupied by the gas to decrease.
• As the gas cools, the average kinetic energy (speed) of the O2 molecules decreases. The molecules rebound with less energy when they collide with each other and the walls of the container. The only way for the molecules to maintain a constant rate of collisions with the walls of the container (maintaining a pressure of 1.00 atm) is for the volume of the gas to decrease.

(d) For a valid explanation:
The ideal gas law assumes that gas particles do not experience interparticle attractions. As a real gas cools further, the intermolecular forces have greater effect as the average speed of the molecules decreases, resulting in inelastic collisions. To maintain a gas pressure of 1.00 atm, the volume must decrease to accommodate more collisions with less energy.

Question

A student is doing experiments with CO2(g). Originally, a sample of the gas is in a rigid container at 299 K and 0.70 atm. The student increases the temperature of the CO2(g) in the container to 425 K.
(a) Describe the effect of raising the temperature on the motion of the CO2(g) molecules.
(b) Calculate the pressure of the CO2(g) in the container at 425 K.
(c) In terms of kinetic molecular theory, briefly explain why the pressure of the CO2(g) in the container changes as it is heated to 425 K.
(d) The student measures the actual pressure of the CO2(g) in the container at 425 K and observes that it is less than the pressure predicted by the ideal gas law. Explain this observation.

▶️Answer/Explanation

Ans:

(a)

The average speed of the molecules increases as temperature increases.

(b)

Both volume and the number of molecules are constant, therefore

\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\Rightarrow \frac{0.70 atm}{299 K}=\frac{P_{2}}{425 K}\Rightarrow P_{2}=0.99 atm\)

(c)

Faster-moving gas particles collide more frequently with the walls of the container, thus increasing the pressure.
OR
Faster-moving gas particles collide more forcefully with the walls of the container, thus increasing the pressure.

(d)

The attractive forces between CO2 molecules result in a pressure that is lower than that predicted by the ideal gas law.
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