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Enthalpy Quiz

12 questions chemistry Grades 9-12

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  1. According to the first law of thermodynamics, how is the change in internal energy expressed?

    • ΔU = q − w
    • ΔU = q + w
    • ΔU = w − q
    • ΔU = q × w
    Reveal answer

    Answer: ΔU = q + w

    Source evidence

    PDF page 265: Thermochemistry is a branch of chemical thermodynamics, the science that deals with the relationships between heat, work, and other forms of energy in the context of chemical and physical processes. As we concentrate on thermochemistry in this chapter, we need to consider some widely used concepts of thermodynamics. Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic energy of its atoms or molecules is raised. The greater kinetic energy may be in the form of increased translations (travel or straight-line motions), vibrations, or rotations of the atoms or molecules. When thermal energy is lost, the intensities of these motions decrease and the kinetic energy falls. The total of all possible kinds of energy present in a substance is called the internal energy (U), sometimes symbolized as E. As a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system. Energy is transferred into a system when it absorbs heat (q) from the surroundings or when the surroundings do work (w) on the system. For example, energy is transferred into room-temperature metal wire if it is immersed in hot water (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire, which is reflected in an increase in the wire’s temperature. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings. The relationship between internal energy, heat, and work can be represented by the equation: ΔU = q + w as shown in Figure 5.19. This is one version of the first law of thermodynamics, and it shows that the internal energy of a system changes through heat flow into or out of the system (positive q is heat flow in; negative q is heat flow out) or work done on or by the system. The work, w, is positive if it is done on the system and negative if it is done by the system.

    PDF page 280: first law of thermodynamics internal energy of a system changes due to heat flow in or out of the system or work done on or by the system

  2. At constant pressure with only expansion work, how are heat flow (qp) and enthalpy change (ΔH) related?

    • They are opposite in sign
    • They are equal
    • qp is always zero
    • ΔH is twice qp
    Reveal answer

    Answer: They are equal

    Source evidence

    PDF page 267: pressure (P) and volume (V): H = U + PV Enthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the enthalpy change (ΔH) is: ΔH = ΔU + PΔV The mathematical product PΔV represents work (w), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of ΔV and w will always be opposite: PΔV = −w Substituting this equation and the definition of internal energy into the enthalpy-change equation yields: ΔH = ΔU + PΔV = qp + w − w = qp where qp is the heat of reaction under conditions of constant pressure. And so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow (qp) and enthalpy change (ΔH) for the process are equal. The heat given off when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since it occurs at the essentially constant pressure of the atmosphere. On the other hand, the heat produced by a reaction measured in a bomb calorimeter (Figure 5.17) is not equal to ΔH because the closed, constant-volume metal container prevents the pressure from remaining constant (it may increase or decrease if the reaction yields increased or decreased amounts of gaseous species). Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with q = ΔH, which makes enthalpy the most convenient choice for determining heat changes for chemical reactions. The following conventions apply when using ΔH:

  3. A negative value of ΔH indicates what type of reaction?

    • Endothermic
    • Neutral
    • Exothermic
    • Reversible
    Reveal answer

    Answer: Exothermic

    Source evidence

    PDF page 267: • A negative value of an enthalpy change, ΔH < 0, indicates an exothermic reaction; a positive value, ΔH > 0,

  4. What happens to the sign of ΔH if a chemical equation's direction is reversed?

    • It becomes zero
    • It doubles
    • It stays the same
    • It changes sign
    Reveal answer

    Answer: It changes sign

    Source evidence

    PDF page 267: indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its ΔH is changed (a process that is endothermic in one direction is exothermic in the opposite direction).

    PDF page 276: 2. ΔH for a reaction in one direction is equal in magnitude and opposite in sign to ΔH for the reaction in the

  5. When 1 mole of H2(g) and half a mole of O2(g) form liquid water, how much heat is released?

    • 242 kJ
    • 286 kJ
    • 572 kJ
    • 143 kJ
    Reveal answer

    Answer: 286 kJ

    Source evidence

    PDF page 267: thermochemical equation, the enthalpy change of a reaction is shown as a ΔH value following the equation for the reaction. This ΔH value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products as shown in the chemical equation. For example, consider this equation: 1 H (g) + O (g) ⟶ H O(l) ΔH = −286 kJ 2 2 2 2 1 This equation indicates that when 1 mole of hydrogen gas and mole of oxygen gas at some temperature and 2 pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor (ΔH is an extensive property): (two-fold increase in amounts) 2H (g) + O (g) ⟶ 2H O(l)

  6. If gaseous water forms instead of liquid water from the same reactants, how much heat is released?

    • 286 kJ
    • 393 kJ
    • 242 kJ
    • 572 kJ
    Reveal answer

    Answer: 242 kJ

    Source evidence

    PDF page 268: water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released. 1 H (g) + O (g) ⟶ H O(g) ΔH = −242 kJ 2 2 2 2

  7. Enthalpy is classified as which kind of function?

    • A path function
    • A state function
    • A wave function
    • A rate function
    Reveal answer

    Answer: A state function

    Source evidence

    PDF page 267: pressure (P) and volume (V): H = U + PV Enthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the enthalpy change (ΔH) is: ΔH = ΔU + PΔV The mathematical product PΔV represents work (w), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of ΔV and w will always be opposite: PΔV = −w Substituting this equation and the definition of internal energy into the enthalpy-change equation yields: ΔH = ΔU + PΔV = qp + w − w = qp where qp is the heat of reaction under conditions of constant pressure. And so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow (qp) and enthalpy change (ΔH) for the process are equal. The heat given off when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since it occurs at the essentially constant pressure of the atmosphere. On the other hand, the heat produced by a reaction measured in a bomb calorimeter (Figure 5.17) is not equal to ΔH because the closed, constant-volume metal container prevents the pressure from remaining constant (it may increase or decrease if the reaction yields increased or decreased amounts of gaseous species). Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with q = ΔH, which makes enthalpy the most convenient choice for determining heat changes for chemical reactions. The following conventions apply when using ΔH:

    PDF page 281: state function property depending only on the state of a system, and not the path taken to reach that state

  8. Which quantity is given as an example of a state function in the source?

    • Altitude or elevation
    • Distance traveled
    • Path length
    • Speed of climbing
    Reveal answer

    Answer: Altitude or elevation

    Source evidence

    PDF page 266: As discussed, the relationship between internal energy, heat, and work can be represented as ΔU = q + w. Internal energy is an example of a state function (or state variable), whereas heat and work are not state functions. The value of a state function depends only on the state that a system is in, and not on how that state is reached. If a quantity is not a state function, then its value does depend on how the state is reached. An example of a state function is altitude or elevation. If you stand on the summit of Mt. Kilimanjaro, you are at an altitude of 5895 m, and it does not matter whether you hiked there or parachuted there. The distance you traveled to the top of Kilimanjaro, however, is not a state function. You could climb to the summit by a direct route or by a more roundabout, circuitous path (Figure 5.20). The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function).

  9. Why is Hess's law valid?

    • Work is conserved
    • Energy is created in steps
    • Enthalpy is a state function
    • Heat is a state function
    Reveal answer

    Answer: Enthalpy is a state function

    Source evidence

    PDF page 274: There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment. This type of calculation usually involves the use of Hess’s law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written: ΔH° = −394 kJ C(s) + O (g) ⟶ CO (g) 2 2 In the two-step process, first carbon monoxide is formed:

  10. What is the standard enthalpy of combustion of ethanol?

    • −393.5 kJ/mol
    • −5461 kJ/mol
    • −890.8 kJ/mol
    • −1366.8 kJ/mol
    Reveal answer

    Answer: −1366.8 kJ/mol

    Source evidence

    PDF page 269: ° Standard enthalpy of combustion (ΔH ) is the enthalpy change when 1 mole of a substance burns (combines C vigorously with oxygen) under standard state conditions; it is sometimes called “heat of combustion.” For example, the enthalpy of combustion of ethanol, −1366.8 kJ/mol, is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 °C and 1 atmosphere pressure, yielding products also at 25 °C and 1 atm. C H OH(l) + 3O (g) ⟶ 2CO + 3H O(l) ΔH° = −1366.8 kJ 2 2 2 2 5 Enthalpies of combustion for many substances have been measured; a few of these are listed in Table 5.2. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline.

    PDF page 270: Standard Molar Enthalpies of Combustion Substance Combustion Reaction Enthalpy of Combustion, ΔH°c kJ ( at25 °C) mol C(s) + O (g) ⟶ CO (g) carbon −393.5 2 2 1 H (g) + O (g) ⟶ H O(l) −285.8 hydrogen 2 2 2 2 1 Mg(s) + O (g) ⟶ MgO(s) magnesium −601.6 2 2 S(s) + O (g) ⟶ SO (g) −296.8 sulfur 2 2 carbon 1 CO(g) + O (g) ⟶ CO (g) −283.0 2 2 2 monoxide CH (g) + 2O (g) ⟶ CO (g) + 2H O(l) methane −890.8 4 2 2 2 5 C H (g) + O (g) ⟶ 2CO (g) + H O(l) acetylene −1301.1 2 2 2 2 2 2 C H OH(l) + 3O (g) ⟶ 2CO (g) + 3H O(l) −1366.8 ethanol 2 2 2 2 5 3 CH OH(l) + O (g) ⟶ CO (g) + 2H O(l) methanol −726.1 3 2 2 2 2 25 C H (l) + O (g) ⟶ 8CO (g) + 9H O(l) isooctane −5461 8 18 2 2 2 2

  11. A standard enthalpy of formation involves forming how many moles of a pure substance from its elements?

    • 0.5 mole
    • Any amount
    • Exactly 1 mole
    • 2 moles
    Reveal answer

    Answer: Exactly 1 mole

    Source evidence

    PDF page 273: A standard enthalpy of formation is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess’s law. The standard enthalpy of formation of CO2(g) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

    PDF page 281: standard enthalpy of formation (ΔH° ) enthalpy change of a chemical reaction in which 1 mole of a pure f substance is formed from its elements in their most stable states under standard state conditions

  12. By definition, the standard enthalpy of formation of an element in its most stable form equals:

    • −393.5 kJ/mol
    • +286 kJ/mol
    • +33.2 kJ/mol
    • Zero
    Reveal answer

    Answer: Zero

    Source evidence

    PDF page 273: N (g) + O (g) ⟶ NO (g) 2 2 2 f 2 1 A reaction equation with mole of N2 and 1 mole of O2 is correct in this case because the standard enthalpy of 2 formation always refers to 1 mole of product, NO2(g). You will find a table of standard enthalpies of formation of many common substances in Appendix G. These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of P4O10) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C2H2). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.

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