AP Chemistry: 7.10 Reaction Quotient and Le Châtelier’s Principle – Exam Style questions with Answer- FRQ

Question

\(CaO(s)+H_{2}O(l)\rightarrow Ca(OH)_{2}(s) \Delta H^{\circ}_{rxn}=-63.7kj/mol_{rxn}\)

Calcium oxide, CaO(s), has been proposed as a substance that can be used to heat water quickly for portable heating packs or for cooking. When placed in water, CaO(s) reacts as shown by the equation above.
(a) A student wants to design a heating pad that could heat a 150.0 g sample of water from \(25.0^{\circ}C to 60.0^{\circ}C. 4.18J/(g.^{\circ}C).) \)
(i) Calculate the amount of heat, in joules, that the water must absorb for its temperature to change by this amount. (Assume that the specific heat capacity of the water is \(4.18 J/(g.^{\circ}C)\)
(ii) Calculate the minimum mass of CaO(s) that the student would need to use in order to cause this temperature change.
(b) The student hypothesizes that the design of the heating pad could be changed to enable it to heat 150.0 g of water from \(25.0^{\circ}C to 90.0^{\circ}C\) by using a greater mass of CaO(s).
(i) Use the data in the table below to determine the standard entropy change, \(\Delta S^{\circ}_{rxn}, in J/(K.mol_{rxn}) \Delta S^{\circ}_{rxn} and \Delta H^{\circ}_{rxn} \)for the reaction.

(ii) Is the reaction thermodynamically favorable at 90.0°C? Justify your answer with a calculation. (Assume that both \(\Delta S^{\circ}_{rxn}\) and \( \Delta H^{\circ}_{rxn}\)  are constant between 25.0°C and 90.0°C.) The student learns that the \(Ca(OH)_2\) produced from the reaction is relatively insoluble and that it dissolves in water according to the equation below.

\(Ca(OH)_{2}(s)\rightarrow Ca^{2+}(aq)+2OH^{-}(aq)\)

(c) The student prepares a saturated solution of Ca(OH)2 and determines that the \([Ca^{2+}] \)is 0.011 M. Calculate
the value of\( K_{sp}\) for \(Ca(OH)_{2}\)

The student wishes to significantly increase the molar solubility of the. \(Ca(OH)_{2}(s)\) and has access to the following substances.

                                              15 mL of distilled water 15 mL of 1.0 M KOH(aq)
                                              15 mL of 1.0 M\( CaCl_2\)(aq) 15 mL of 1.0 M HCl(aq)

(d) Which substance, when added to the \(Ca(OH)_2\) mixture, would increase the molar solubility most significantly? Justify your choice based on concepts of chemical equilibrium, such as Le Châtelier’s principle.

▶️Answer/Explanation

a(i)\(q=mc\Delta T = 150.0 g x 4.18 J/(g °C) x (60.0°C -25.0°C) = 2.19 x 104 J\)

a(ii) 

b(i) \(\Delta S°_rxn \) = ∑S° products – ∑S° reactants
= \(40. J/(K.mol) – [83 J/(K.mol) + 70. J/(K.mol)]\)
= \(-113 J/(K.mol_{rxn})\)

b(ii) Yes, because \(\Delta G° is negative. \Delta G° = \Delta H° – T \Delta S° =-63.7 kJ/mol_{rxn} K (-113 J/(K.mol_{rxn})) ×\frac{1kj}{1000j} = -22.7 kJ/mol_{rxn}\)

(c) \( [OH^-] = 2 × [Ca^2+] = 2 (0.011 M) = 0.022 \)\(M K_sp=[Ca^2+] [OH^-]^2  = (0.011)(0.022)^2 = 5.3 × 10^-6\)

(d) HCl
The \(H^+\) from the HCl reacts with \(OH^−\), decreasing \([OH^−]\). The loss of\( OH^−\) results in non-equilibrium conditions in the\( Ca(OH)_2\) dissolution. More\( Ca(OH)_2\) must dissolve to increase\( [OH^{−}]\), increasing the molar solubility as Q approaches K and the system returns to equilibrium.

Question

A student is studying the properties of CaSO4 and PbSO4. The student has samples of both compounds, which are white powders.
(a) The student tests the electrical conductivity of each solid and observes that neither solid conducts electricity. Describe the structures of the solids that account for their inability to conduct electricity.

The student places excess CaSO4 (s) in a beaker containing 100 mL of water and places excess PbSO4 (s) in another beaker containing 100 mL of water. The student stirs the contents of the beakers and then measures the electrical conductivity of the solution in each beaker. The student observes that the conductivity of the solution in the beaker containing the CaSO4 (s) is higher than the conductivity of the solution in the beaker containing the PbSO4 (s).

(b) Which compound is more soluble in water, CaSO4(s) or PbSO4(s) ? Justify your answer based on the results of the conductivity test.

The left side of the diagram below shows a particulate representation of the contents of the beaker containing the CaSO4(s) from the solution conductivity experiment.

(c) Draw a particulate representation of PbSO4(s) and the ions dissolved in the solution in the beaker on the right in the diagram. Draw the particles to look like those shown to the right of the beaker. Draw an appropriate number of dissolved ions relative to the number of dissolved ions in the beaker on the left.
(d) The student attempts to increase the solubility of CaSO4(s) by adding 10.0 mL of 2 M H2SO4 (aq) to the beaker, and observes that additional precipitate forms in the beaker. Explain this observation.

▶️Answer/Explanation

Ans:

(a) For a correct description:
Ionic solids do not have free-moving ions that are required to carry an electric current. Therefore, there is no conduction of electricity.

(b) For the correct answer and a valid justification:
CaSO4. The greater electrical conductivity of the CaSO4 solution relative to the PbSO4 solution implies a higher concentration of ions, which comes from the dissolution (dissociation) of CaSO4 to a greater extent.

(c) For a correct drawing that shows an equal number of cations and anions:
The drawing shows solid PbSO4 at the bottom of the beaker (similar to the solid shown for CaSO4) and fewer dissociated Pb2+ and SO42− ions in the solution.

(d) For a correct explanation:
The additional precipitate is CaSO4 that forms in response to the increased [SO42−] in solution. According to Le Chatelier’s principle (Q > Ksp), the introduction of SO42− as a
common ion shifts the equilibrium towards the formation of more CaSO4(s).

Scroll to Top