AP Chemistry: 6.1 Endothermic and Exothermic Processes- Exam Style questions with Answer- FRQ

Question

 \(MgF_2(s) \Leftrightarrow Mg^{2+}(aq) + 2 F^{ -}(aq)\)

In a saturated solution of \(MgF_2\) at 18° C, the concentration of Mg2+ is \(1.21 x 10^ {-3}\) molar. The equilibrium is represented by the equation above.

(a) Write the expression for the solubility-product constant, \(K_{sp}\), and calculate its value at 18° C.

(b) Calculate the equilibrium concentration of \(Mg^{2+}\) in 1.000 liter of saturated \(MgF_2\) solution at 18°C to which 0.100 mole of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.

(c) Predict whether a precipitate of \(MgF_2\) will form when 100.0 milliliters of a \(3.00 x 10 ^{-3}\) molar \(Mg(NO_3)_2\) solution is mixed with 200.0 milliliters of a \(2.00 x 10 ^{-3}\) molar NaF solution at 18°C.
Calculations to support your prediction must be shown.

(d) At 27°C the concentration of \(Mg^{2+}\) in a saturated solution of \(MgF_2\) is \(1.17 x 10 ^{-3}\) molar. Is the dissolving of \(MgF_2\) in water an endothermic or an exothermic process? Give an explanation to support your conclusion.

▶️Answer/Explanation

Answer:

Question

The compound urea, H2NCONH2 , is widely used in chemical fertilizers. The complete Lewis electron-dot diagram for the urea molecule is shown above.
(a) Identify the hybridization of the valence orbitals of the carbon atom in the urea molecule.
(b) Urea has a high solubility in water, due in part to its ability to form hydrogen bonds. A urea molecule and four water molecules are represented in the box below. Draw ONE dashed line (—-) to indicate a possible location of a hydrogen bond between a water molecule and the urea molecule.

The dissolution of urea is represented by the equation above. A student determines that 5.39 grams of H2NCONH2 (molar mass 60.06 g/mol) can dissolve in water to make 5.00 mL of a saturated solution at 20.°C.

(c) Calculate the concentration of urea, in mol/L, in the saturated solution at 20.°C.
(d) The student also determines that the concentration of urea in a saturated solution at 25°C is 19.8 M. Based on this information, is the dissolution of urea endothermic or exothermic? Justify your answer in terms of Le Chatelier’s principle.

(e) The equipment shown above is provided so that the student can determine the value of the molar heat of solution for urea. Knowing that the specific heat of the solution is 4.18 J/(g⋅°C), list the specific measurements that are required to be made during the experiment.

(f) The entropy change for the dissolution of urea, ΔS0soln , is 70.1 J/(mol⋅K) at 25°C. Using the information in the table above, calculate the absolute molar entropy, S°, of aqueous urea.
(g) Using particle-level reasoning, explain why ΔS0soln is positive for the dissolution of urea in water.
(h) The student claims that ΔS° for the process contributes to the thermodynamic favorability of the dissolution of urea at 25°C. Use the thermodynamic information above to support the student’s claim.

▶️Answer/Explanation

Ans:

(a)

sp2

(b)

A dashed line should connect a hydrogen atom in water to a nitrogen or oxygen atom in urea or an oxygen atom in water to a hydrogen atom in urea. One possible correct response is shown above.

(c)

5.39 g H2NCONH2 × \(\frac{1 mol}{60.06 g}=0.0897 mol\)

\(\frac{0.0897 mol}{0.00500 L}=17.9 M\)

(d)

The increased solubility at the higher temperature implies that the dissolution of urea is endothermic. If a saturated solution of urea is heated, then the equilibrium system is stressed. The stress is counteracted by the endothermic dissolution of more urea.

(e)

mass of urea, mass of water, initial temperature of water, final temperature of solution

(f)

ΔS0soln  = S0  (H2NCONH2(aq)) – S0  (H2NCONH2(s))

70.1 J/(mol.K) = S0  (H2NCONH2(aq)) – 104.6 J/(mol.K)

S0  (H2NCONH2(aq)) = 174.7 J/(mol.K)

(g)

Urea molecules in solution have a greater number of possible arrangements than in solid urea. This increased number of arrangements corresponds to a positive ΔS0soln.  

(h)

Thermodynamic favorability for a process at standard conditions is determined by the sign of ΔG0  , with ΔG0  =ΔH0  -TΔS0 . Since ΔS0   is positive, the TΔS0  term makes the value of ΔG0  smaller and thus makes the dissolution more thermodynamically favorable.

Question

MgF2(s) ⇔ Mg2+(aq) + 2 F(aq)
In a saturated solution of MgF2 at 180 C, the concentration of Mg2+ is 1.21 × 10-3 molar. The equilibrium is represented by the equation above.
(a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 180 C.
(b) Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at 180 C to which 0.100 mole of solid KF has been added. The \(K_F\) dissolves completely. Assume the volume change is negligible.
(c) Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.00 × 10-3-molar Mg(N03)2 solution is mixed with 200.0 milliliters of a 2.00 × 10-3-molar NaF solution at 180 C. Calculations to support your prediction must be shown.
(d) At 270 C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 × 10-3 molar. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion.

▶️Answer/Explanation

Ans: 

(a) \(K_{sp}=\left [ Mg^{2+} \right ][F^{-}]^{2}\)
                       = (1.21 × 10-3) (2 × 1.21 × 10-3)2
                       = 7.09 × 10-9

Note: if number of significant figures in final answer differs by more than one from the appropriate number, 1 point is deducted ONCE PER PROBLEM.

(b) \(K_{sp}=\left [ Mg^{2+} \right ](2x+0.100)^{2}\)
\(7.09 \times 10^{-9}=\left [ Mg^{2+} \right ](0.010)^{2}\)
\(\left [ Mg^{2+} \right ]=(7.09 \times 10^{-9})/(10^{-2})\)
                                               = 7.09 × 10-7 M

Note: OK if 0.102 is used for [F], then Ksp = 6.76 x 10-7

(c) \(\left [ Mg^{2+} \right ]:100.0 \times 3.00 \times 10^{-3}=300.0\times \left [ Mg^{2+} \right ]\)

\(\left [ Mg^{2+} \right ]=1.00 \times 10^{-3}M\)

\(\left [ F^{-} \right ]=200.0 \times 2.00\times 10^{-3}=300.0\times \left [ F^{-} \right ]\)

\(\left [ F^{-} \right ]=1.33 \times 10^{-3}M\)

Q = Ion Product = \(=[Mg^{2+}][F^{-}]^{2}\)

                              = (1.00 × 10-3) (1.33 × 10-3)2  **

                              = 1.77 × 10-9

Since Q < Ksp, no precipitate will form

Note: conclusion must be consistent with Q value.
** Correct substitution and calculation of the wrong concentration values earns the second point, but not the first.

(d) Solubility of MgF2 decreases with increasing temperature, thus dissolution process is exothermic

Reason:
EITHER
i) Increased temperature puts a stress on the system (Le Chatelier). The system will reduce the stress by shifting the equilibrium in the endothermic (left) direction

OR,

ii) a data supported argument such as comparing ion concentrations, calculating second Ksp and giving proper interpretations.

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