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Chemistry 30 Diploma study materials
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Chemistry 30 Diploma practice questions

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Question 1Thermochemical Changes

Which statement best defines the enthalpy of a system?

  1. The energy change when one mole of a compound forms from its elements
  2. The average kinetic energy of the particles in the sample
  3. The energy required to break one mole of a bond in the gas phase
  4. The internal energy of the system plus the product of its pressure and volume Correct answer.

Explanation

Enthalpy is defined as H = U + PV, so it combines internal energy with the pressure–volume term. At constant pressure, the change in enthalpy equals the heat exchanged with the surroundings.

Question 2Thermochemical Changes

Which pair of statements correctly describes an exothermic reaction at constant pressure?

  1. ΔH is negative and the surroundings gain energy Correct answer.
  2. ΔH is positive and the surroundings lose energy
  3. ΔH is positive and the surroundings gain energy
  4. ΔH is negative and the surroundings lose energy

Explanation

In an exothermic reaction the system releases energy, so the surroundings absorb it and their temperature rises. Because the system loses enthalpy, ΔH has a negative sign.

Question 3Thermochemical Changes

Which unit is used to express a molar enthalpy of reaction?

  1. J/°C
  2. kJ/mol Correct answer.
  3. J
  4. J/(g·°C)

Explanation

Molar enthalpy is an energy change per mole of a specified substance, so its unit is energy divided by amount, such as kJ/mol. The joule alone expresses energy with no reference to amount of substance.

Question 4Thermochemical Changes

The quantity that expresses the energy needed to raise the temperature of 1 g of a substance by 1 °C is called the

  1. specific heat capacity Correct answer.
  2. molar enthalpy
  3. heat of vaporization
  4. calorimeter constant

Explanation

Specific heat capacity is defined per unit mass and per degree of temperature change, usually in J/(g·°C). It appears in the relation q = mcΔT.

Question 5Thermochemical Changes

Which statement is the basis of Hess's law?

  1. The rate of a reaction doubles for every 10 °C rise in temperature
  2. The total energy of the universe is constant in any process
  3. The enthalpy change of an overall reaction depends only on the initial and final states, not on the pathway Correct answer.
  4. The enthalpy change of a reaction equals the activation energy of the forward reaction minus that of the reverse reaction

Explanation

Because enthalpy is a state function, the net enthalpy change depends only on the reactants and products. This allows multi-step reactions to be added together to find an overall ΔH.

Question 6Thermochemical Changes

The standard molar enthalpy of formation of a compound is the enthalpy change when

  1. one mole of the compound burns completely in oxygen
  2. one mole of bonds in the compound is broken in the gas phase
  3. one mole of the compound dissolves in water
  4. one mole of the compound forms from its elements in their standard states Correct answer.

Explanation

Formation reactions always start from the elements in their standard states and produce exactly one mole of the compound. By convention, the enthalpy of formation of an element in its standard state is zero.

Question 7Thermochemical Changes

Activation energy is best described as

  1. the difference in enthalpy between reactants and products
  2. the energy released as products form
  3. the total energy stored in the bonds of the reactants
  4. the minimum energy colliding particles need in order to react Correct answer.

Explanation

Only collisions with at least the activation energy can lead to product formation. On a potential energy diagram, it is the height from the reactants to the transition state.

Question 8Thermochemical Changes

How does adding a catalyst affect the energy profile of a reaction?

  1. It increases ΔH for the forward reaction
  2. It lowers the activation energy and also lowers ΔH
  3. It lowers the activation energy without changing ΔH Correct answer.
  4. It raises the activation energy of the forward reaction only

Explanation

A catalyst provides an alternative pathway with a lower energy barrier. The energies of the reactants and products are unchanged, so ΔH stays the same.

Question 9Thermochemical Changes

A 50.0 g sample of water is heated from 21.0 °C to 33.5 °C. Using c = 4.19 J/(g·°C), how much energy is absorbed?

  1. 1.31 kJ
  2. 21.0 kJ
  3. 2.62 kJ Correct answer.
  4. 262 kJ

Explanation

Use q = mcΔT with ΔT = 33.5 − 21.0 = 12.5 °C. Then q = (50.0 g)(4.19 J/(g·°C))(12.5 °C) = 2.62 × 10³ J, or 2.62 kJ.

Question 10Thermochemical Changes

When 2.00 g of a fuel with molar mass 46.0 g/mol burns completely, 88.0 kJ of energy is released. What is the molar enthalpy of combustion of the fuel?

  1. -2.02 × 10³ kJ/mol Correct answer.
  2. -44.0 kJ/mol
  3. -1.01 × 10³ kJ/mol
  4. -4.05 × 10³ kJ/mol

Explanation

The amount burned is 2.00 g ÷ 46.0 g/mol = 0.0435 mol. Dividing the released energy per mole gives ΔH = -88.0 kJ ÷ 0.0435 mol ≈ -2.02 × 10³ kJ/mol.

Question 11Thermochemical Changes

Given: (1) C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ; (2) 2CO(g) + O₂(g) → 2CO₂(g), ΔH = -566.0 kJ. What is ΔH for 2C(s) + O₂(g) → 2CO(g)?

  1. -110.5 kJ
  2. +221.0 kJ
  3. -221.0 kJ Correct answer.
  4. -1259.0 kJ

Explanation

Double equation (1) to get 2C + 2O₂ → 2CO₂ with ΔH = -787.0 kJ, then reverse equation (2), which changes its sign to +566.0 kJ. Adding gives 2C + O₂ → 2CO with ΔH = -787.0 + 566.0 = -221.0 kJ.

Question 12Thermochemical Changes

Using ΔHf° values of -74.6 kJ/mol for CH₄(g), -393.5 kJ/mol for CO₂(g), and -285.8 kJ/mol for H₂O(l), calculate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

  1. -965.1 kJ
  2. -604.7 kJ
  3. -890.5 kJ Correct answer.
  4. +890.5 kJ

Explanation

ΔH = ΣΔHf°(products) − ΣΔHf°(reactants). The products total -393.5 + 2(-285.8) = -965.1 kJ, and the reactants total -74.6 kJ, so ΔH = -965.1 − (-74.6) = -890.5 kJ.

Question 13Thermochemical Changes

Using average bond energies H-H = 436 kJ/mol, Cl-Cl = 243 kJ/mol, and H-Cl = 431 kJ/mol, calculate ΔH for H₂(g) + Cl₂(g) → 2HCl(g).

  1. -862 kJ
  2. +183 kJ
  3. -679 kJ
  4. -183 kJ Correct answer.

Explanation

Bonds broken total 436 + 243 = 679 kJ, and bonds formed total 2(431) = 862 kJ. ΔH = 679 − 862 = -183 kJ.

Question 14Thermochemical Changes

A potential energy diagram shows the products at a higher energy level than the reactants. Which description of the reaction is correct?

  1. Exothermic, with a negative ΔH
  2. Endothermic, with a positive ΔH Correct answer.
  3. Endothermic, with a negative ΔH
  4. Exothermic, with a positive ΔH

Explanation

When the products lie above the reactants on the diagram, the system has absorbed energy, which is endothermic. Absorbing energy corresponds to a positive ΔH.

Question 15Thermochemical Changes

Which of the following phase changes is exothermic?

  1. H₂O(l) → H₂O(g)
  2. H₂O(s) → H₂O(l)
  3. H₂O(g) → H₂O(l) Correct answer.
  4. H₂O(s) → H₂O(g)

Explanation

Condensation releases the energy that was absorbed during vaporization, so the process gives off heat and ΔH is negative. Melting and boiling both require energy input.

Question 16Thermochemical Changes

A 2.50 g sample is burned in a bomb calorimeter containing 250 g of water (c = 4.18 J/(g·°C)). The water temperature rises by 4.20 °C. Assuming the calorimeter itself absorbs negligible energy, what is the energy released per gram of sample?

  1. 8.78 kJ/g Correct answer.
  2. 0.878 kJ/g
  3. 17.6 kJ/g
  4. 4.39 kJ/g

Explanation

First find the heat absorbed by the water: q = (250 g)(4.18 J/(g·°C))(4.20 °C) = 4.39 × 10³ J. Dividing by the 2.50 g of sample gives 8.78 kJ/g.

Question 17Thermochemical Changes

For the reaction 2H₂(g) + O₂(g) → 2H₂O(g), ΔH = -484 kJ. How much energy is released when 6.0 mol of H₂ reacts completely?

  1. 726 kJ
  2. 484 kJ
  3. 1452 kJ Correct answer.
  4. 2904 kJ

Explanation

The thermochemical equation releases 484 kJ per 2 mol of H₂. For 6.0 mol, the ratio is 3, so the energy released is 3 × 484 kJ = 1452 kJ.

Question 18Thermochemical Changes

When solid ammonium nitrate dissolves in water, the temperature of the solution decreases. Which statement describes the process?

  1. It is endothermic and ΔH is negative
  2. It is exothermic and ΔH is positive
  3. It is exothermic and ΔH is negative
  4. It is endothermic and ΔH is positive Correct answer.

Explanation

A temperature drop means the system absorbed energy from the water, so the dissolving is endothermic. Endothermic processes have a positive ΔH.

Question 19Thermochemical Changes

Methane has molar mass 16.0 g/mol and ΔHcomb = -890 kJ/mol. Octane has molar mass 114.0 g/mol and ΔHcomb = -5470 kJ/mol. Which fuel releases more energy per gram, and by how much?

  1. Methane, by about 55.6 kJ/g
  2. Octane, by about 48 kJ/g
  3. Octane, by about 7.6 kJ/g
  4. Methane, by about 7.6 kJ/g Correct answer.

Explanation

Energy per gram is the molar enthalpy of combustion divided by the molar mass: methane gives 890 ÷ 16.0 = 55.6 kJ/g and octane gives 5470 ÷ 114.0 = 48.0 kJ/g. Methane therefore releases about 7.6 kJ more per gram.

Question 20Thermochemical Changes

Given N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ, what is ΔH for 2NO(g) → N₂(g) + O₂(g)?

  1. -180.6 kJ Correct answer.
  2. +90.3 kJ
  3. +180.6 kJ
  4. -90.3 kJ

Explanation

Reversing a reaction reverses the roles of reactants and products, so the magnitude of ΔH stays the same but the sign flips. The reverse reaction therefore has ΔH = -180.6 kJ.

Question 21Thermochemical Changes

Given: (1) 2C(s) + 3H₂(g) → C₂H₆(g), ΔH = -84.7 kJ; (2) 2C(s) + 2H₂(g) → C₂H₄(g), ΔH = +52.3 kJ. What is ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g)?

  1. -137.0 kJ Correct answer.
  2. -32.4 kJ
  3. +137.0 kJ
  4. -189.3 kJ

Explanation

Reverse equation (2) to obtain C₂H₄ → 2C + 2H₂ with ΔH = -52.3 kJ. Adding this to equation (1) cancels the carbon and leaves C₂H₄ + H₂ → C₂H₆ with ΔH = -84.7 + (-52.3) = -137.0 kJ.

Question 22Thermochemical Changes

A 2.50 g sample of fuel is burned in a bomb calorimeter that has a heat capacity of 8.50 kJ/°C and contains 1.00 kg of water (c = 4.18 J/(g·°C)). The temperature rises from 20.00 °C to 24.50 °C. What is the energy released per gram of fuel?

  1. 22.8 kJ/g
  2. 7.52 kJ/g
  3. 57.1 kJ/g
  4. 15.3 kJ/g Correct answer.

Explanation

Total heat absorbed equals (C_calorimeter + C_water)ΔT. The water contributes 4.18 kJ/°C, so the combined capacity is 12.68 kJ/°C and q = (12.68)(4.50) = 57.1 kJ; dividing by 2.50 g gives about 22.8 kJ/g.

Question 23Thermochemical Changes

For 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g), ΔH = +129.2 kJ. Given ΔHf° values of -1130.7 kJ/mol for Na₂CO₃(s), -241.8 kJ/mol for H₂O(g), and -393.5 kJ/mol for CO₂(g), what is ΔHf° of NaHCO₃(s)?

  1. -1766.0 kJ/mol
  2. -473.8 kJ/mol
  3. -947.6 kJ/mol Correct answer.
  4. -1895.2 kJ/mol

Explanation

The products sum to -1130.7 − 241.8 − 393.5 = -1766.0 kJ. Setting +129.2 = -1766.0 − 2x and solving gives 2x = -1895.2, so x = -947.6 kJ/mol.

Question 24Thermochemical Changes

Using average bond energies C-H = 413 kJ/mol, O=O = 498 kJ/mol, C=O = 745 kJ/mol, and O-H = 463 kJ/mol, estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).

  1. -2648 kJ
  2. -3342 kJ
  3. +694 kJ
  4. -694 kJ Correct answer.

Explanation

Bonds broken: 4(413) + 2(498) = 2648 kJ. Bonds formed: 2(745) + 4(463) = 3342 kJ. ΔH = 2648 − 3342 = -694 kJ.

Question 25Thermochemical Changes

The molar enthalpy of combustion of methane is -890 kJ/mol. What mass of methane must burn to raise the temperature of 2.00 kg of water from 15.0 °C to 85.0 °C? Use c = 4.18 J/(g·°C) and assume no heat loss.

  1. 10.5 g Correct answer.
  2. 5.27 g
  3. 0.658 g
  4. 20.9 g

Explanation

The water needs q = (2000 g)(4.18 J/(g·°C))(70.0 °C) = 5.85 × 10⁵ J = 585 kJ. That requires 585 ÷ 890 = 0.658 mol of methane, which has a mass of 0.658 mol × 16.04 g/mol ≈ 10.5 g.

Question 26Thermochemical Changes

For an exothermic reaction at constant pressure, which statement describes the enthalpy change of the system?

  1. ΔH is negative; heat flows to the surroundings Correct answer.
  2. ΔH is negative; heat flows into the system
  3. ΔH is zero; no heat is exchanged
  4. ΔH is positive; heat flows to the surroundings

Explanation

At constant pressure, the enthalpy change equals the heat exchanged. In an exothermic reaction the system loses enthalpy, so ΔH is negative and energy leaves as heat.

Question 27Thermochemical Changes

A 100.0 g sample of water absorbs 836 J of energy and its temperature rises by 2.00 °C. What is the specific heat capacity of water?

  1. 2.09 J/(g·°C)
  2. 0.418 J/(g·°C)
  3. 8.36 J/(g·°C)
  4. 4.18 J/(g·°C) Correct answer.

Explanation

Use q = mcΔT and rearrange to c = q/(mΔT). Substituting gives c = 836 J / (100.0 g × 2.00 °C) = 4.18 J/(g·°C).

Question 28Thermochemical Changes

A potential energy diagram shows reactants at 35 kJ and products at 80 kJ. What is ΔH for the forward reaction?

  1. -45 kJ
  2. +115 kJ
  3. +45 kJ Correct answer.
  4. +35 kJ

Explanation

ΔH is the enthalpy of the products minus the enthalpy of the reactants. Here, 80 kJ − 35 kJ = +45 kJ, so the forward reaction is endothermic.

Question 29Thermochemical Changes

Which statement about chemical bonds and enthalpy is correct?

  1. Breaking bonds releases energy and is exothermic
  2. Breaking bonds requires energy and is endothermic Correct answer.
  3. Forming bonds absorbs energy and is endothermic
  4. Forming bonds requires energy and is endothermic

Explanation

Breaking a bond always requires an input of energy, so bond breaking is endothermic. Forming a bond releases energy, so bond formation is exothermic.

Question 30Thermochemical Changes

Standard enthalpy changes are measured under which conditions?

  1. 298 K and 101.3 kPa
  2. 298 K and 100 kPa Correct answer.
  3. 273 K and 100 kPa
  4. 273 K and 101.3 kPa

Explanation

Standard thermodynamic conditions are 298 K (25 °C) and 100 kPa (1 bar). These conditions define standard enthalpies such as ΔHf°.

Question 31Thermochemical Changes

The molar enthalpy of combustion is the enthalpy change when

  1. one mole of a compound forms from its elements
  2. one mole of a substance reacts with oxygen to form oxides
  3. one mole of a substance burns completely in excess oxygen Correct answer.
  4. one mole of a fuel is vaporized at constant pressure

Explanation

Molar enthalpy of combustion is defined per mole of fuel burned completely in excess oxygen. The products are the stable oxides formed under the stated conditions.

Question 32Thermochemical Changes

For the reaction 2A → B, ΔH = -150 kJ. What is ΔH for B → 2A?

  1. -150 kJ
  2. -300 kJ
  3. +300 kJ
  4. +150 kJ Correct answer.

Explanation

Reversing a reaction changes only the sign of ΔH, not its magnitude. The reverse reaction therefore has ΔH = +150 kJ.

Question 33Thermochemical Changes

A reaction absorbs 50 kJ of heat from its surroundings. What is ΔH for the reaction?

  1. -50 kJ
  2. 0 kJ
  3. +25 kJ
  4. +50 kJ Correct answer.

Explanation

When a system absorbs heat at constant pressure, its enthalpy increases. The enthalpy change has the same magnitude as the heat absorbed, so ΔH = +50 kJ.

Question 34Thermochemical Changes

A 25.0 g piece of aluminum absorbs 450 J and its temperature rises by 20.0 °C. What is its specific heat capacity?

  1. 1.80 J/(g·°C)
  2. 4.18 J/(g·°C)
  3. 0.450 J/(g·°C)
  4. 0.900 J/(g·°C) Correct answer.

Explanation

Rearrange q = mcΔT to c = q/(mΔT). Substituting gives c = 450 J / (25.0 g × 20.0 °C) = 0.900 J/(g·°C).

Question 35Thermochemical Changes

A 0.500 g fuel sample burns and raises the temperature of a calorimeter with heat capacity 10.0 kJ/°C by 3.00 °C. How much energy is released?

  1. 15.0 kJ
  2. 3.00 kJ
  3. 33.3 kJ
  4. 30.0 kJ Correct answer.

Explanation

For a calorimeter with a known heat capacity, q = CΔT. Substituting gives q = 10.0 kJ/°C × 3.00 °C = 30.0 kJ.

Question 36Thermochemical Changes

A 1.20 g sample (molar mass 60.0 g/mol) burns in a bomb calorimeter containing 500 g of water (c = 4.18 J/(g·°C)) and a calorimeter with heat capacity 2.00 kJ/°C. The temperature rises by 4.00 °C. What is the molar enthalpy of combustion, ΔHcomb?

  1. -818 kJ/mol Correct answer.
  2. -836 kJ/mol
  3. -418 kJ/mol
  4. -409 kJ/mol

Explanation

Heat absorbed by water is 500 g × 4.18 J/(g·°C) × 4.00 °C = 8.36 kJ, and the calorimeter absorbs 2.00 kJ/°C × 4.00 °C = 8.00 kJ. The total is 16.36 kJ for 0.0200 mol, so ΔHcomb = -16.36/0.0200 = -818 kJ/mol.

Question 37Thermochemical Changes

Given A → B, ΔH = -50 kJ and B → C, ΔH = -20 kJ, what is ΔH for A → C?

  1. -20 kJ
  2. -70 kJ Correct answer.
  3. -30 kJ
  4. +70 kJ

Explanation

Hess's law allows enthalpy changes of successive steps to be added. The overall change is (-50 kJ) + (-20 kJ) = -70 kJ.

Question 38Thermochemical Changes

Given X → Y, ΔH = +100 kJ and Z → Y, ΔH = +40 kJ, what is ΔH for X → Z?

  1. +140 kJ
  2. -60 kJ
  3. -140 kJ
  4. +60 kJ Correct answer.

Explanation

Use X → Y as given and reverse Z → Y to get Y → Z with ΔH = -40 kJ. Adding gives (+100 kJ) + (-40 kJ) = +60 kJ.

Question 39Thermochemical Changes

Using ΔHf° values of -277.6 kJ/mol for C2H5OH(l), -393.5 kJ/mol for CO2(g), and -285.8 kJ/mol for H2O(l), calculate ΔH for C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l).

  1. -956.7 kJ
  2. -1025.1 kJ
  3. +1366.8 kJ
  4. -1366.8 kJ Correct answer.

Explanation

For products, 2(-393.5) + 3(-285.8) = -1644.4 kJ. Subtracting the reactant enthalpy, -1644.4 - (-277.6) = -1366.8 kJ.

Question 40Thermochemical Changes

Using average bond energies N≡N = 945 kJ/mol, H-H = 436 kJ/mol, and N-H = 391 kJ/mol, estimate ΔH for N2(g) + 3H2(g) → 2NH3(g).

  1. -1401 kJ
  2. -93 kJ Correct answer.
  3. +1401 kJ
  4. +93 kJ

Explanation

Energy absorbed to break bonds is 945 + 3(436) = 2253 kJ. Energy released forming six N-H bonds is 6(391) = 2346 kJ. Thus ΔH = 2253 - 2346 = -93 kJ.

Question 41Thermochemical Changes

A potential energy diagram has reactants at 25 kJ, an activated complex at 85 kJ, and products at 40 kJ. What is the activation energy for the reverse reaction?

  1. 45 kJ Correct answer.
  2. 60 kJ
  3. 85 kJ
  4. 15 kJ

Explanation

The reverse activation energy is the height from the products to the activated complex. That is 85 kJ − 40 kJ = 45 kJ.

Question 42Thermochemical Changes

For C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g), ΔH = -2220 kJ. What mass of propane (M = 44.0 g/mol) must burn to release 444 kJ?

  1. 4.40 g
  2. 17.6 g
  3. 44.0 g
  4. 8.80 g Correct answer.

Explanation

The desired energy is one-fifth of 2220 kJ, so 0.200 mol of propane must burn. Mass is 0.200 mol × 44.0 g/mol = 8.80 g.

Question 43Thermochemical Changes

50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH are mixed, and the temperature rises by 6.80 °C. Assume the total mass is 100.0 g and c = 4.18 J/(g·°C). What is the molar enthalpy of neutralization?

  1. -113.6 kJ/mol
  2. -28.4 kJ/mol
  3. -56.8 kJ/mol Correct answer.
  4. +56.8 kJ/mol

Explanation

Heat absorbed is 100.0 g × 4.18 J/(g·°C) × 6.80 °C = 2.84 kJ. The limiting amounts are 0.0500 mol each, so ΔH = -2.84 kJ / 0.0500 mol = -56.8 kJ/mol.

Question 44Thermochemical Changes

Which fuel releases the most energy per gram? H2: -286 kJ/mol (2.0 g/mol); C3H8: -2220 kJ/mol (44.0 g/mol); CH3OH: -726 kJ/mol (32.0 g/mol); C2H5OH: -1367 kJ/mol (46.0 g/mol).

  1. C3H8
  2. H2 Correct answer.
  3. CH3OH
  4. C2H5OH

Explanation

Divide each molar enthalpy by its molar mass to get energy per gram. Hydrogen gives about 143 kJ/g, propane about 50.5 kJ/g, methanol about 22.7 kJ/g, and ethanol about 29.7 kJ/g.

Question 45Thermochemical Changes

During melting at constant pressure, the energy added to a solid is used mainly to

  1. increase potential energy as intermolecular forces are overcome Correct answer.
  2. decrease the enthalpy of the system
  3. break chemical bonds within each molecule
  4. increase average kinetic energy and raise temperature

Explanation

During a phase change, temperature remains constant, so average kinetic energy does not increase. The added energy increases potential energy by overcoming intermolecular attractions.

Question 46Thermochemical Changes

When solid NaOH dissolves in water, the solution temperature rises. Which statement is correct?

  1. ΔH_solution is negative; dissolving is exothermic Correct answer.
  2. ΔH_solution is positive; dissolving is exothermic
  3. ΔH_solution is negative; dissolving is endothermic
  4. ΔH_solution is zero; heat is not exchanged

Explanation

A temperature rise means the solution gains heat from the dissolving process, so the process releases heat. Releasing heat corresponds to a negative enthalpy change and an exothermic process.

Question 47Thermochemical Changes

Which equation represents the standard molar enthalpy of formation of H2O(l)?

  1. H2(g) + ½O2(g) → H2O(l) Correct answer.
  2. H2(g) + O2(g) → H2O2(l)
  3. H2O(l) → H2(g) + ½O2(g)
  4. 2H2(g) + O2(g) → 2H2O(l)

Explanation

Formation enthalpy is defined for one mole of product formed from elements in their standard states. The correct equation forms exactly one mole of H2O(l) from H2(g) and O2(g).

Question 48Thermochemical Changes

Given (1) 2S(s) + 3O2(g) → 2SO3(g), ΔH = -792 kJ and (2) S(s) + O2(g) → SO2(g), ΔH = -297 kJ, what is ΔH for 2SO2(g) + O2(g) → 2SO3(g)?

  1. -495 kJ
  2. +198 kJ
  3. -990 kJ
  4. -198 kJ Correct answer.

Explanation

Reverse and double equation (2) to get 2SO2 → 2S + 2O2 with ΔH = +594 kJ. Adding equation (1) cancels S and gives 2SO2 + O2 → 2SO3 with ΔH = +594 - 792 = -198 kJ.

Question 49Thermochemical Changes

A 5.00 g sample of ethanol (M = 46.0 g/mol, theoretical ΔHcomb = -1367 kJ/mol) burns and raises 500.0 g of water by 40.0 °C. Using c = 4.18 J/(g·°C), what is the percent efficiency?

  1. 44.2%
  2. 28.1%
  3. 56.3% Correct answer.
  4. 76.5%

Explanation

Heat absorbed by water is 500.0 g × 4.18 J/(g·°C) × 40.0 °C = 83.6 kJ. The ethanol amount is 5.00/46.0 = 0.1087 mol, so theoretical heat is 0.1087 × 1367 = 148.6 kJ. Efficiency is 83.6/148.6 × 100% = 56.3%.

Question 50Thermochemical Changes

Given (1) C(s) + O2(g) → CO2(g), ΔH = -393.5 kJ; (2) H2(g) + ½O2(g) → H2O(l), ΔH = -285.8 kJ; (3) CH3OH(l) + 1½O2(g) → CO2(g) + 2H2O(l), ΔH = -726.4 kJ. What is ΔHf° for CH3OH(l)?

  1. -238.7 kJ/mol Correct answer.
  2. +238.7 kJ/mol
  3. -1305.7 kJ/mol
  4. -204.9 kJ/mol

Explanation

Formation of CH3OH(l) is C(s) + 2H2(g) + ½O2(g) → CH3OH(l). Add equation (1) and twice equation (2), then subtract equation (3): -393.5 + 2(-285.8) - (-726.4) = -238.7 kJ/mol.