AP Chemistry: 6.8 Enthalpy of Formation – Exam Style questions with Answer- FRQ

Question

             \(3Ag(s)+4HNO_{3}(aq)\rightarrow 3AgNO_{3}(aq)+NO(g)+2H_{2}O(l)\)

A student investigates the reaction between Ag(s) and \(HNO_3\)(aq) represented by the equation above.
(a) Predict the sign of the entropy change, ΔS°, for the reaction. Justify your answer.
(b) Use the information in the table below to calculate the value of \(\Delta H^{\circ}_{rxn}\), the standard enthalpy change for the reaction, in\( kJ/mol_{rxn}\) .

(c) Based on your answers to parts (a) and (b), is the reaction more likely to be thermodynamically favorable at 25°C, or at 95°C? Justify your answer.
(d) The student runs the reaction using a 3 to 4 mole ratio of Ag(s) to\( HNO_3\)(aq). Suggest a method the student can use to isolate solid \(AgNO_3 \)from the other products of the reaction.

▶️Answer/Explanation

(a) The entropy change is positive because the reaction has one mole of gas in the products and none in the reactants.

(b)\( \Delta H^{\circ}_{rxn} = 3(-101) + 90. +2(-286) -4(-207)
= 43 kJ/molrxn\)

(c) \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\) The reaction is more likely to be favorable at \(95G^{\circ}\). At the higher temperature, the term\( T\Delta S^{\circ}\) is larger and positive; thus, when subtracted from \(\Delta H^{\circ}\), the value of  \(\Delta G^{\circ} \) is more likely to be negative.

(d) The student can evaporate the water, leaving behind solid silver nitrate.

Question

Na 2S2O3(aq) + 4 NaOCl(aq) + 2 NaOH(aq) → 2 Na2SO4(aq) + 4 NaCl(aq) + H2O(l)

A student performs an experiment to determine the value of the enthalpy change, ΔH0rxn , for the oxidation-reduction reaction represented by the balanced equation above.
(a) Determine the oxidation number of Cl in NaOCl.
(b) Calculate the number of grams of Na2S2O3 needed to prepare 100.00 mL of 0.500 M Na2S2O3 (aq).
In the experiment, the student uses the solutions shown in the table below.

(c) Using the balanced equation for the oxidation-reduction reaction and the information in the table above,
determine which reactant is the limiting reactant. Justify your answer.

The solutions, all originally at 20.0 0C, are combined in an insulated calorimeter. The temperature of the reaction mixture is monitored, as shown in the graph below.

(d) According to the graph, what is the temperature change of the reaction mixture?
(e) The mass of the reaction mixture inside the calorimeter is 15.21 g.
(i) Calculate the magnitude of the heat energy, in joules, that is released during the reaction. Assume that the specific heat of the reaction mixture is 3.94 J/(g·°C) and that the heat absorbed by the calorimeter is negligible.
(ii) Using the balanced equation for the oxidation-reduction reaction and your answer to part (c), calculate D the value of the enthalpy change of the reaction, ΔH0rxn , in kJ/molrxn . Include the appropriate algebraic sign with your answer.
The student repeats the experiment, but this time doubling the volume of each of the reactants, as shown in the table below.

 

 

(f) The magnitude of the enthalpy change, ΔH0rxn, in kJ/molrxn , calculated from the results of the second
experiment is the same as the result calculated in part (e)(ii). Explain this result.

(g) Write the balanced net ionic equation for the given reaction.

▶️Answer/Explanation

Ans:

(a)

+1

(b)

\(100.00mL\times \frac{0.500 mol Na_{2}S_{2}O_{3}}{1000 mL}\times \frac{158.10 g Na_{2}S_{2}O_{3}}{1 mol Na_{2}S_{2}O_{3}}\)

= 7.90 g Na2S2O3

(c)

NaOCl is the limiting reactant.

Given that equal numbers of moles of each reactant were present initially, it follows from the coefficients of the reactants in the balanced equation that NaOCl will be depleted first.

(d)

From the graph the final temperature is 32.50 C.
ΔT = Tf – Ti = 32.50 C – 20.00 C = 12.50 C.

(e) (i)

q = mcΔT
   = (15.21 g) (3.94 J/(g.0C))(12.50 C) = 749 J

(ii)

(f)

By doubling the volumes, the number of moles of the reactants are
doubled, which doubles the amount of energy produced. Therefore the amount of heat per mole will remain the same.
OR
In the second experiment, ΔH0rxn = \(\frac{2mc\Delta T}{2n}=\frac{mc\Delta T}{n}\) = ΔH0rxn
Thus the magnitude is the same as calculated in the first experiment.

(g)

S2O32-(aq) + 4OCl (aq) + 2 OH(aq) → 2SO42-(aq) + 4 Cl(aq) + H2O(l)

Question

 The reaction below is used to produce methanol:
\(CO(g) + 2H_2 (g) \rightarrow CH_3OH (l) \)           \(\Delta H_{rxn}=-128 kj\)
(a) Calculate the C-H bond energy given the following data:

(b) The tabulated value of the (C-H) bond energy is 413 kJ/mol. Explain why there is a difference between the
number you have calculated in (a) and the tabulated value.

▶️Answer/Explanation

Ans:

(a)

(b) Tabulated values, like those in Table 8.4, are averaged from many bond energies measured for C-H bonds in many
different molecules.

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