American Chemical Society General Chemistry Thermochemistry
Heat, enthalpy, Hess's law, entropy, and Gibbs free energy.
These questions are original. This is not an official American Chemical Society exam, and FreeExam is not affiliated with the American Chemical Society.
1.A student heats 25.0 g of liquid water in an open beaker from 22.0 °C to 30.0 °C. The specific heat of water is 4.18 J/(g·°C). What is q for the water?
2.Ethanol, 80.0 g, is warmed in a capped flask from 15.0 °C to 25.0 °C. Its specific heat is 2.46 J/(g·°C). How much heat does the ethanol gain?
3.An aluminum bar of mass 150 g sits on a hot plate and climbs from 20.0 °C to 35.0 °C. Take c = 0.900 J/(g·°C). Calculate q for the bar.
4.A 40.0 g iron nail left in the sun goes from 25.0 °C to 65.0 °C. The specific heat of iron is 0.450 J/(g·°C). What is q for the nail?
5.A copper pipe weighing 200 g warms from 18.0 °C to 28.0 °C as hot water passes through it. Use c = 0.385 J/(g·°C) for copper. Find q for the pipe.
6.A 60.0 g window pane heats from 16.0 °C to 41.0 °C in the afternoon. Glass has c = 0.840 J/(g·°C). What heat does the pane absorb?
7.A 30.0 g silver spoon dipped into hot soup rises from 20.0 °C to 80.0 °C. Silver's specific heat is 0.235 J/(g·°C). Determine q for the spoon.
8.Dry playground sand, 500 g, warms from 24.0 °C to 28.0 °C. Use the given specific heat 0.800 J/(g·°C). How much heat does the sand take in?
9.Liquid mercury in a thermometer, 12.0 g, climbs from 19.0 °C to 44.0 °C. Its specific heat is 0.140 J/(g·°C). What is q for the mercury?
10.In a coffee-cup calorimeter, dissolving NaOH raises 100.0 g of solution by 2.00 °C. Use c = 4.18 J/(g·°C). What is q for the dissolving reaction?
11.Ammonium nitrate dissolves in a coffee cup and 50.0 g of solution cools by 3.00 °C. The solution specific heat is 4.18 J/(g·°C). Find q for the dissolution.
12.Neutralizing HCl with NaOH in a coffee cup warms 200.0 g of solution by 1.50 °C. Take c = 4.18 J/(g·°C). What is q of the neutralization?
13.Potassium chloride dissolves and cools 250.0 g of solution by 0.800 °C in a coffee-cup calorimeter. c = 4.18 J/(g·°C). Calculate q for dissolving.
14.Magnesium ribbon reacts with acid in a coffee cup. The 120.0 g solution rises 5.00 °C. Use c = 4.18 J/(g·°C). What is q for the metal-acid reaction?
15.Urea dissolves in a foam cup and the 75.0 g solution falls 2.00 °C. Specific heat of the solution: 4.18 J/(g·°C). Determine q for dissolving the urea.
16.Calcium chloride dissolves and warms 400.0 g of solution by 0.500 °C in a coffee cup. c = 4.18 J/(g·°C). Find the reaction heat q.
17.Citric acid and sodium bicarbonate react in a coffee cup, and 25.0 g of solution cools by 4.00 °C. Use c = 4.18 J/(g·°C). What is q for that reaction?
18.Solid sodium carbonate dissolves in a coffee-cup calorimeter and 150.0 g of solution warms by 3.00 °C. c = 4.18 J/(g·°C). Calculate q for the dissolution.
19.Dissolving a salt sample releases 30.0 kJ, and a follow-up reaction in the same mixture absorbs 74.0 kJ. What is the net ΔH, and is the overall process endothermic or exothermic?
20.A two-step synthesis first absorbs 18.5 kJ and then releases 63.5 kJ. What is the net enthalpy change, and which label matches its sign?
21.Both stages of a combustion cleanup release heat: 15.2 kJ and then 9.80 kJ. What is the net ΔH, and is it endothermic or exothermic?
22.An ice-salt bath absorbs 40.0 kJ in the first change and 12.5 kJ in the second. Give the net ΔH and whether it is endothermic or exothermic.
23.A metal-oxide reduction releases 88.0 kJ, after which a drying step absorbs 20.0 kJ. What net ΔH follows, and which sign label fits?
24.One beaker step absorbs 6.40 kJ and the next step releases 6.40 kJ. What is the net ΔH, and how should its sign be classified?
25.Roasting a sulfide sample releases 102 kJ, and the gas-scrubbing step that follows absorbs 140 kJ. Find the net ΔH and its endothermic or exothermic label.
26.Dissolving a pellet absorbs 11.0 kJ. A follow-up precipitation then releases 47.0 kJ. What is the combined ΔH, and is it endothermic or exothermic?
27.Three heats in one run are a release of 8.50 kJ, an absorption of 8.50 kJ, and an absorption of 3.00 kJ. What is the net ΔH, and how is it classified?
28.For C(graphite) + O2(g) → CO2(g), ΔHf is 0 kJ/mol for C(graphite), 0 for O2, and −394 kJ/mol for CO2. What is ΔH for the reaction?
29.Compute ΔH for 2 H2(g) + O2(g) → 2 H2O(l). ΔHf is 0 for H2, 0 for O2, and −286 kJ/mol for H2O(l).
30.Find ΔH for CaCO3(s) → CaO(s) + CO2(g). ΔHf values: CaCO3 −1207 kJ/mol, CaO −635 kJ/mol, CO2 −394 kJ/mol.
31.What is ΔH for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)? ΔHf: CH4 −75 kJ/mol, O2 0, CO2 −394 kJ/mol, H2O(l) −286 kJ/mol.
32.Calculate ΔH for N2(g) + 3 H2(g) → 2 NH3(g). ΔHf is 0 for N2, 0 for H2, and −46 kJ/mol for NH3.
33.The reaction is 2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(s). ΔHf: Al 0, Fe 0, Fe2O3 −824 kJ/mol, Al2O3 −1676 kJ/mol. Find ΔH.
34.Diamond converts to graphite: C(diamond) → C(graphite). ΔHf is +2 kJ/mol for diamond and 0 for graphite, the standard state of carbon. What is ΔH?
35.S(rhombic) + O2(g) → SO2(g). ΔHf is 0 for rhombic sulfur, 0 for O2, and −297 kJ/mol for SO2. What is ΔH of this formation reaction?
36.Find ΔH for H2(g) + Cl2(g) → 2 HCl(g). ΔHf: H2 0, Cl2 0, HCl −92 kJ/mol.
37.Use Hess's law. (1) C(s) + O2(g) → CO2(g), ΔH = −394 kJ. (2) CO(g) + 1/2 O2(g) → CO2(g), ΔH = −283 kJ. Find ΔH for C(s) + 1/2 O2(g) → CO(g).
38.Equation (1) is 2 H2(g) + O2(g) → 2 H2O(l), ΔH = −572 kJ. What ΔH does Hess's law give for H2O(l) → H2(g) + 1/2 O2(g)?
39.Find ΔH for C2H4(g) + H2(g) → C2H6(g). (1) C2H4(g) + 3 O2 → 2 CO2 + 2 H2O(g), ΔH = −1410 kJ. (2) C2H6(g) + 7/2 O2 → 2 CO2 + 3 H2O(g), ΔH = −1560 kJ. (3) H2 + 1/2 O2 → H2O(g), ΔH = −286 kJ.
40.Target: P4 + 10 Cl2 → 4 PCl5. (1) P4 + 6 Cl2 → 4 PCl3, ΔH = −1200 kJ. (2) PCl3 + Cl2 → PCl5, ΔH = −90 kJ. What does Hess's law give?
41.Find ΔH for SO2(g) + 1/2 O2(g) → SO3(g). (1) S(s) + O2 → SO2, ΔH = −297 kJ. (2) S(s) + 3/2 O2 → SO3, ΔH = −396 kJ.
42.Target: 2 CO2(g) → 2 CO(g) + O2(g). (1) C(s) + O2 → CO2, ΔH = −394 kJ. (2) 2 C(s) + O2 → 2 CO, ΔH = −222 kJ. Apply Hess's law.
43.Target: CaO(s) + CO2(g) → CaCO3(s). (1) Ca + 1/2 O2 → CaO, ΔH = −635 kJ. (2) C + O2 → CO2, ΔH = −394 kJ. (3) Ca + C + 3/2 O2 → CaCO3, ΔH = −1207 kJ. What is ΔH?
44.Target: C(s) + 2 H2(g) → CH4(g). (1) CH4 + 2 O2 → CO2 + 2 H2O(l), ΔH = −890 kJ. (2) C + O2 → CO2, ΔH = −394 kJ. (3) H2 + 1/2 O2 → H2O(l), ΔH = −286 kJ. Find ΔH.
45.Target: 2 Al(s) + 3 CuO(s) → Al2O3(s) + 3 Cu(s). (1) 2 Al + 3/2 O2 → Al2O3, ΔH = −1680 kJ. (2) Cu + 1/2 O2 → CuO, ΔH = −150 kJ. What is ΔH?
46.Methane's combustion enthalpy is −800 kJ/mol. The molar mass of CH4 is 16.0 g/mol. What is ΔH per gram of methane?
47.Ethanol burns with ΔH = −1380 kJ/mol. Its molar mass is 46.0 g/mol. What is the enthalpy change per gram of ethanol?
48.A 2.00 g sample (molar mass 40.0 g/mol) releases 50.0 kJ when it reacts completely. What is ΔH in kJ per mole?
49.Glucose has a combustion enthalpy of −2700 kJ/mol and a molar mass of 180.0 g/mol. Express that enthalpy change per gram of glucose.
50.A reaction consumes 0.250 mol of a reactant, which is also 10.0 g of that reactant, and releases 40.0 kJ. What is ΔH per mole?
51.Dissolving 5.00 g of a solid (molar mass 100.0 g/mol) absorbs 15.0 kJ. What is the enthalpy change per gram?
52.Propane's combustion ΔH is −2200 kJ/mol. The molar mass of C3H8 is 44.0 g/mol. Find ΔH per gram of propane.
53.Magnesium, 6.00 g, reacts and releases 90.0 kJ. The molar mass of Mg is 24.0 g/mol. What is ΔH per mole of Mg?
54.A 0.500 mol portion of a reactant has a mass of 20.0 g and absorbs 25.0 kJ as it reacts. Report the enthalpy change per gram.
55.A gas absorbs 200 J of heat while it expands and does 80 J of work. Use ΔU = q + w, with w negative when the system expands. What is ΔU?
56.The system releases 150 J of heat and expands, performing 50 J of work. With ΔU = q + w and expansion work negative, find ΔU.
57.Surroundings push a gas inward, doing 100 J of work on the system, while the system absorbs 400 J of heat. Use ΔU = q + w, and take work done on the system as positive. What is ΔU?
58.A cylinder loses 60 kJ of heat and the gas inside expands, doing 25 kJ of work. Use ΔU = q + w, with w negative for expansion. Find ΔU.
59.Heat absorbed by a gas is 90 kJ, and the gas expands and does 90 kJ of work. Apply ΔU = q + w with w negative on expansion. What is ΔU?
60.A gas absorbs 1500 J and expands against a constant external pressure of 2.0 atm. The volume increases by 5.0 L. Use 101 J = 1 L·atm, w = −P_ext ΔV, and ΔU = q + w. What is ΔU?
61.A gas is compressed by the surroundings, which do 200 J of work on it, while the gas gives off 500 J of heat. ΔU = q + w, and work done on the system is positive. Find ΔU.
62.A gas absorbs 200 J and expands by 1.00 L against 1.00 atm. Use 101 J = 1 L·atm, w = −P_ext ΔV, and ΔU = q + w, with expansion work negative. What is ΔU?
63.A gas expands into vacuum. q = 0, and the external pressure is 0, so w = −P_ext ΔV = 0. A classmate uses the gas pressure and gets w = −200 J. Using ΔU = q + w, what is ΔU?
64.For CaCO3(s) → CaO(s) + CO2(g), how does the entropy of the system change, based on the moles of gas?
65.Predict the sign of ΔS for 2 H2(g) + O2(g) → 2 H2O(l) from the change in moles of gas.
66.What is the entropy direction for N2(g) + 3 H2(g) → 2 NH3(g), judged by moles of gas?
67.Heating decomposes chlorate as 2 KClO3(s) → 2 KCl(s) + 3 O2(g). Which statement about ΔS is right?
68.N2O4(g) → 2 NO2(g) is the dissociation of dinitrogen tetroxide. What happens to the entropy?
69.Iron rusting can be written 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s). How does the system entropy change?
70.Liquid water boils: H2O(l) → H2O(g). Which entropy conclusion follows from that phase change?
71.In 2 SO2(g) + O2(g) → 2 SO3(g), what is the sign of ΔS from the gas-mole change?
72.Calcium in water: Ca(s) + 2 H2O(l) → Ca(OH)2(s) + H2(g). Which entropy statement is correct?
73.At 298 K a precipitation has ΔG = −42.0 kJ. What does that sign mean for the forward reaction?
74.A synthesis has ΔG = +18.5 kJ at the stated temperature. Which conclusion follows from that sign?
75.A combustion has ΔH = −200 kJ and ΔS = +50 J/K. Over what temperatures is the forward reaction spontaneous? Use the signs in ΔG = ΔH − TΔS.
76.For a reaction, ΔH = +80 kJ and ΔS = −40 J/K. When is the forward reaction spontaneous?
77.Decomposition of a carbonate has ΔH = +120 kJ and ΔS = +200 J/K. The two signs compete. When is the forward reaction spontaneous?
78.Freezing a pure liquid sample has ΔH = −90 kJ and ΔS = −150 J/K for the amounts given. Temperature decides the winner. When is freezing spontaneous?
79.A reaction is held at a temperature where ΔG = 0 exactly. What does that value say about spontaneity?
80.Vaporizing a liquid sample has ΔH = +15 kJ and ΔS = +80 J/K. Which description of spontaneity is right?
81.A reaction has ΔH = −10 kJ and ΔS = +5 J/K. Which statement matches ΔG = ΔH − TΔS?
82.A reaction is studied at 300 K. ΔH = −80.0 kJ and ΔS = −100 J/K. Compute ΔG = ΔH − TΔS. Convert ΔS to kJ/K before multiplying by T.
83.Inside a bath held at 250 K, a dissolution has ΔH = +40.0 kJ and ΔS = +200 J/K. What is ΔG? Use kelvin and convert joules to kilojoules.
84.A fuel-cell reaction has ΔH = −20.0 kJ and ΔS = +50.0 J/K at T = 400 K. Find ΔG = ΔH − TΔS.
85.At 200 K, a laboratory run has ΔH = +100 kJ and ΔS = +250 J/K. Calculate ΔG = ΔH − TΔS, with ΔS converted to kJ/K.
86.A closed flask is held at 500 K, where ΔH = −60.0 kJ and ΔS = −150 J/K. What is ΔG?
87.Use T = 300 K, ΔH = +25.0 kJ, and ΔS = +50.0 J/K for a salt dissolving in water. Find ΔG. Convert ΔS from J/K to kJ/K.
88.Combustion data at 350 K are ΔH = −150 kJ and ΔS = +100 J/K. Compute ΔG = ΔH − TΔS.
89.A dimerization has ΔH = +10.0 kJ and ΔS = −20.0 J/K. The temperature is 400 K. What is ΔG?
90.Calculate ΔG at 200 K for a reaction with ΔH = −45.0 kJ and ΔS = +150 J/K. Convert the entropy unit.
91.Limestone decomposition has ΔH = +50.0 kJ and ΔS = +100 J/K, so a crossover exists. Above what temperature is it spontaneous? Use T = ΔH/ΔS with matching units.
92.For the reaction with ΔH = +100 kJ and ΔS = +200 J/K, find the temperature above which the forward direction becomes spontaneous.
93.A reaction has ΔH = +30.0 kJ and ΔS = +100 J/K. Both signs allow a high-temperature crossover. What is that temperature, and on which side of it is the reaction spontaneous?
94.ΔH = +80.0 kJ and ΔS = +200 J/K for a decomposition. Calculate the temperature above which ΔG becomes negative.
95.Given ΔH = +120 kJ and ΔS = +400 J/K for vapor escaping a solution, above which temperature is the process spontaneous?
96.ΔH = +25.0 kJ and ΔS = +50.0 J/K. Find T = ΔH/ΔS and say whether the reaction is spontaneous above or below that temperature.
97.A dissolution has ΔH = +60.0 kJ and ΔS = +150 J/K. At what temperatures is it spontaneous?
98.ΔH = +200 kJ and ΔS = +500 J/K for a reaction that can cross over. Above what temperature is ΔG negative?
99.With ΔH = +45.0 kJ and ΔS = +150 J/K, determine the temperature above which the reaction is spontaneous.
100.Estimate ΔH for H2 + Cl2 → 2 HCl from bond enthalpies: H–H 436 kJ/mol, Cl–Cl 244 kJ/mol, H–Cl 430 kJ/mol. Use bonds broken minus bonds formed.
101.Using bond enthalpies H–H 436, Br–Br 194, and H–Br 365 kJ/mol, estimate ΔH for H2 + Br2 → 2 HBr. Subtract bonds formed from bonds broken.
102.Treat the bond-enthalpy estimate as the answer for H2 + F2 → 2 HF. Bond enthalpies: H–H 436, F–F 164, and H–F 570, all in kJ/mol. Use broken minus formed.
103.Estimate ΔH for 2 H2 + O2 → 2 H2O from these bond enthalpies in kJ/mol: H–H 440, O=O 500, O–H 465. Use broken minus formed.
104.For N2 + 3 H2 → 2 NH3, bond enthalpies are N≡N 945, H–H 435, and N–H 390 kJ/mol. Estimate ΔH as bonds broken minus bonds formed.
105.For hydrogenation C2H4 + H2 → C2H6, break one C=C (610 kJ/mol) and one H–H (435). Form one C–C (345) and two C–H (415 each). Estimate ΔH.
106.Estimate ΔH for H2 + I2 → 2 HI. Bond enthalpies: H–H 440, I–I 150, and H–I 300 kJ/mol. Use bonds broken minus bonds formed.
107.Estimate ΔH for N2 + O2 → 2 NO using bond enthalpies N≡N 950, O=O 500, and NO 630 kJ/mol. Report bonds broken minus bonds formed.
108.Treat the bond-enthalpy estimate as the answer for CH4 + Cl2 → CH3Cl + HCl. Break one C–H (415 kJ/mol) and one Cl–Cl (240). Form one C–Cl (330) and one H–Cl (430).