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Equilibrium

Question 1
2022ChemistryAPConcept Practice
12 marks

CH3OH(g)⇌CO(g)+2H2(g)ΔH∘=90.0,kJ/molrxnCH_3OH(g) \rightleftharpoons CO(g) + 2H_2(g) \quad \Delta H^\circ = 90.0 , kJ/mol_{rxn}CH3​OH(g)⇌CO(g)+2H2​(g)ΔH∘=90.0,kJ/molrxn​

2.a. Methanol vapor decomposes to form carbon monoxide gas and hydrogen gas at high temperatures in the presence of a platinum catalyst, as represented by the balanced chemical equation given.

2.a.i. Are the hydrogen atoms oxidized or are they reduced in the forward reaction? Justify your answer in terms of oxidation numbers.

The hydrogen atoms are oxidized. In CH3OHCH_3OHCH3​OH, the oxidation number of H is +1. In H2H_2H2​, the oxidation number of H is 0. An increase in oxidation number indicates oxidation.

2.a.ii. In the following box, draw the complete Lewis electron-dot diagram for the carbon monoxide molecule in which every atom obeys the octet rule. Show all bonding and nonbonding valence electrons.

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3 marks

2.b. The values of the standard molar entropies of the compounds involved in the reaction are given in the following table.

SubstanceS∘S^\circS∘ (J/(K·mol))
CH3OH(g)CH_3OH(g)CH3​OH(g)240.
CO(g)CO(g)CO(g)198.
H2(g)H_2(g)H2​(g)131.

2.b.i. Use the data in the table to calculate the value of the standard entropy change, ΔS∘\Delta S^\circΔS∘, in J/(K·molrxn_{rxn}rxn​), for the reaction.

ΔS∘=∑Sproducts∘−∑Sreactants∘\Delta S^\circ = \sum S^\circ_{products} - \sum S^\circ_{reactants}ΔS∘=∑Sproducts∘​−∑Sreactants∘​ ΔS∘=(198+(2×131))−240=220,J/(K⋅molrxn)\Delta S^\circ = (198 + (2 \times 131)) - 240 = 220 , J/(K \cdot mol_{rxn})ΔS∘=(198+(2×131))−240=220,J/(K⋅molrxn​)

2 marks

2.c. Calculate the value of ΔG∘\Delta G^\circΔG∘, in kJ/molrxn_{rxn}rxn​, for the reaction at 375 K. Assume that ΔH∘\Delta H^\circΔH∘ and ΔS∘\Delta S^\circΔS∘ are independent of temperature.

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circΔG∘=ΔH∘−TΔS∘ ΔG∘=90.0,kJ/mol−(375,K)(0.220,kJ/(K⋅mol))=7.5,kJ/mol\Delta G^\circ = 90.0 , kJ/mol - (375 , K)(0.220 , kJ/(K \cdot mol)) = 7.5 , kJ/molΔG∘=90.0,kJ/mol−(375,K)(0.220,kJ/(K⋅mol))=7.5,kJ/mol

2 marks

2.d. The following particle-level diagram shows a representative sample of the equilibrium mixture represented by the equation given.

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2.d.i. Use information from the particle diagram to calculate the partial pressure of CO at equilibrium when the total pressure of the equilibrium mixture is 12.0 atm.

There are a total of 8 particles in the diagram. There is 1 particle of CO. So the partial pressure of CO is (1/8)(12.0 atm) = 1.5 atm

2.d.ii. Write the expression for the equilibrium constant, KpK_pKp​, for the reaction.

Kp=PCOPH22PCH3OHK_p = \frac{P_{CO}P_{H_2}^2}{P_{CH_3OH}}Kp​=PCH3​OH​PCO​PH2​2​​

2 marks

2.e. CH3OH(g)⇌CO(g)+2H2(g)CH_3OH(g) \rightleftharpoons CO(g) + 2H_2(g)CH3​OH(g)⇌CO(g)+2H2​(g)

The reaction system represented by the equation is allowed to achieve equilibrium at a different temperature. The following table gives the partial pressure of each species in the equilibrium mixture.

SubstancePartial Pressure at Different Temperature
CH3OH(g)CH_3OH(g)CH3​OH(g)2.7 atm
CO(g)CO(g)CO(g)4.2 atm
H2(g)H_2(g)H2​(g)8.4 atm

2.e.i. Use the information in the table to calculate the value of the equilibrium constant, KpK_pKp​, at the new temperature.

Kp=PCOPH22PCH3OH=(4.2)(8.4)22.7=110K_p = \frac{P_{CO}P_{H_2}^2}{P_{CH_3OH}} = \frac{(4.2)(8.4)^2}{2.7} = 110Kp​=PCH3​OH​PCO​PH2​2​​=2.7(4.2)(8.4)2​=110

2.e.ii. The volume of the container is rapidly doubled with no change in temperature. As equilibrium is re-established, does the number of moles of CH3OH(g)CH_3OH(g)CH3​OH(g) increase, decrease, or remain the same? Justify your answer by comparing the value of the reaction quotient, Q, with the value of the equilibrium constant, KpK_pKp​.

The number of moles of CH3OH(g)CH_3OH(g)CH3​OH(g) will increase. When the volume is doubled, the partial pressures of all species are halved. The new value of Q will be: Q=(2.1)(4.2)21.35=27.6Q = \frac{(2.1)(4.2)^2}{1.35} = 27.6Q=1.35(2.1)(4.2)2​=27.6. Since Q < KpK_pKp​, the reaction will shift to the right, increasing the amount of products and decreasing the amount of reactants, so the number of moles of CH3OH(g)CH_3OH(g)CH3​OH(g) will increase.

3 marks
Question 2
2024ChemistryAPConcept Practice
4 marks

Hydrogen gas and iodine gas react to form hydrogen iodide at an elevated temperature, as represented by the following equation.

H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)H2​(g)+I2​(g)⇌2HI(g) DeltaHrxnDelta H_{rxn}DeltaHrxn​ = −12.19 kJ / molrxn_{rxn}rxn​

5.a) Write the expression for the equilibrium constant, KcK_cKc​, for this reaction.

1 mark

5.b) H2(g)H_2(g)H2​(g) and I2(g)I_2(g)I2​(g) are added to a previously evacuated container and allowed to react.

5.b.i) At a certain time, the value of the reaction quotient, QQQ, is 0.67. The following particle diagram is an incomplete representation of the system at this time. The diagram shows the relative number of H2(g)H_2(g)H2​(g) and I2(g)I_2(g)I2​(g) molecules, but the HI(g)HI(g)HI(g) molecules are not included. Draw the number of HI(g)HI(g)HI(g) molecules needed to complete the diagram so that it accurately represents the system.

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5.b.ii) A student monitors the number of moles of HI(g)HI(g)HI(g) over time. Hypothesize an experimental change that could have been applied to the system in the rigid container at time ttt to result in the change in the number of moles of HI(g)HI(g)HI(g) shown in the graph. Assume that the student did not add more HI(g)HI(g)HI(g) to the system.

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5.b.iii) After equilibrium is established, the mixture is transferred to a larger container at constant temperature. As a result, would the number of moles of HI(g)HI(g)HI(g) increase, decrease, or remain the same? Justify your answer.

3 marks
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