Enthalpy Changes (College Board AP® Chemistry): Flashcards
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- FrontEnthalpy of Reaction
Define standard enthalpy change of reaction (ΔH°r).
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Cards in this collection (23)
Define standard enthalpy change of reaction (ΔH°r).
The enthalpy change when the reactants in the stoichiometric equation react to give products under standard conditions (100 kPa, 1 M concentration, standard states, usually 25°C).
True or False?
The standard enthalpy change of neutralisation (ΔH°neut) is always exothermic.
True.
ΔH°neut is defined as the enthalpy change when one mole of water is formed from the reaction of an acid and a base — this process always releases energy to the surroundings.
Why is temperature not part of the definition of standard state, yet 25°C (298 K) is almost always quoted alongside standard enthalpy values?
Standard state is defined solely by pressure (100 kPa) and the physical form of each substance — temperature is not fixed by the definition. However, because enthalpy changes vary with temperature, a reference temperature of 25°C is conventionally specified so that values from different sources can be compared.
True or False?
The standard enthalpy change of combustion (ΔH°c) can be either exothermic or endothermic.
False.
ΔH°c is always exothermic. Combustion is defined as burning one mole of a substance in excess oxygen — this always releases energy, giving a negative ΔH value.
If ΔH°f [Fe2O3 (s)] = −824 kJ mol−1, what is ΔHr for the reaction: 4Fe (s) + 3O2 (g) → 2Fe2O3 (s)?
ΔHr = 2 × (−824) = −1648 kJ
Two moles of Fe2O3 are formed, so the enthalpy of formation value is multiplied by the stoichiometric coefficient 2.
The symbol ° on a ΔH value indicates that the reaction was carried out under .......... conditions: a pressure of 100 kPa, a concentration of 1 M, and each substance in its .......... state.
The symbol ° on a ΔH value indicates that the reaction was carried out under standard conditions: a pressure of 100 kPa, a concentration of 1 M, and each substance in its standard state.
Define average bond energy.
The energy needed to break one mole of a particular type of bond in a gaseous molecule, averaged over similar compounds. It is always a positive value because bond breaking is endothermic.
True or False?
Breaking a chemical bond is an exothermic process.
False.
Breaking a bond requires energy input from the surroundings — it is endothermic. Bond formation is the exothermic process, releasing energy to the surroundings.
Why are bond energies described as average values rather than exact values?
The energy required to break a bond is influenced by the surrounding atoms in the molecule (its chemical environment). Because the same bond type exists in many different compounds with slightly different environments, an average across many similar molecules is calculated and used as a representative value.
True or False?
In the Haber process (N2 + 3H2 → 2NH3), using the bond energies N≡N = 945 kJ mol-1, H–H = 436 kJ mol-1, and N–H = 391 kJ mol-1, the reaction is exothermic.
True.
Bonds broken: 945 + (3 × 436) = +2253 kJ mol-1
Bonds formed: 6 × 391 = −2346 kJ mol-1
ΔH°r = +2253 + (−2346) = −93 kJ mol-1 (approximate, as bond energies are average values) — negative, so exothermic.
How does the relative stability of products compared to reactants determine whether a reaction is exothermic or endothermic?
If the products are more stable than the reactants, more energy is released forming new bonds than is required to break the original bonds — the reaction is exothermic (negative ΔH). If the products are less stable, more energy is needed to break bonds than is released on forming new ones — the reaction is endothermic (positive ΔH).
The formula for calculating enthalpy of reaction using bond energies is: ΔH°r = enthalpy change for bonds .......... + enthalpy change for bonds .........., where bond breaking values are .......... (endothermic) and bond forming values are .......... (exothermic).
The formula for calculating enthalpy of reaction using bond energies is: ΔH°r = enthalpy change for bonds broken + enthalpy change for bonds formed, where bond breaking values are positive (endothermic) and bond forming values are negative (exothermic).
Define standard enthalpy of formation (ΔH°f).
The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (100 kPa, 25°C).
True or False?
The standard enthalpy of formation of oxygen gas, O2 (g), is zero.
True.
By definition, the standard enthalpy of formation of any element in its standard state is zero — no enthalpy change occurs when an element forms from itself.
Using the formula ΔHr = ΣΔH°f (products) − ΣΔH°f (reactants), why do the arrows in a Hess's Law formation cycle always point upward from elements to compounds?
The definition of standard enthalpy of formation requires compounds to be formed from their elements. In the cycle diagram, elements sit at the bottom and compounds at the top, so the formation arrows necessarily point upward — their direction encodes the definition and ensures the arithmetic of the Hess's Law calculation is consistent.
True or False?
Given ΔH°f: B2H6 (g) = +31.4 kJ mol-1, B2O3 (s) = −1270 kJ mol-1, H2O (g) = −242 kJ mol-1, the enthalpy of combustion of B2H6 (g) is approximately −2027 kJ mol-1.
True.
ΔH°f (products) = −1270 + 3(−242) = −1996 kJ
ΔH°f (reactants) = +31.4 + 0 = +31.4 kJ
ΔH°c = −1996 − (+31.4) = −2027.4 kJ mol-1
The Hess's Law formula using standard enthalpies of formation is: ΔHr = ΣΔH°f (..........) − ΣΔH°f (.......... ).
The Hess's Law formula using standard enthalpies of formation is: ΔHr = ΣΔH°f (products) − ΣΔH°f (reactants).
Define Hess's Law.
The total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place, provided the initial and final conditions are the same. This is a consequence of the conservation of energy.
True or False?
If the enthalpy change for a forward reaction is −2044 kJ, then the enthalpy change for the reverse reaction is also −2044 kJ.
False.
Reversing a reaction reverses the sign of ΔH. The reverse reaction has ΔH = +2044 kJ — it is endothermic by the same magnitude.
Why can Hess's Law be used to calculate enthalpy changes that cannot be measured directly in the laboratory?
Because enthalpy is a state function, its change depends only on the initial and final states — not the pathway taken. Any indirect route using known ΔH values gives the same overall ΔH as the direct route.
True or False?
When applying Hess's Law, if the stoichiometric coefficients of an equation are multiplied by 2, the ΔH°r value is also multiplied by 2.
True.
Enthalpy change is an extensive property — it scales with the amount of substance reacting. Doubling all coefficients doubles the number of moles reacting and therefore doubles the total enthalpy change.
Using Hess's Law:
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O (l) ΔH = −1411 kJ
C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (l) ΔH = −1560 kJ
H2 (g) + 0.5O2 (g) → H2O (l) ΔH = −286 kJ
What is ΔHr for C2H4 (g) + H2 (g) → C2H6 (g)?
ΔHr = (−1411) + (+1560) + (−286) = −137 kJ
The second equation is reversed (giving +1560 kJ), then all three are added. Common intermediates (CO2, H2O, O2) cancel, leaving the target equation.
When applying Hess's Law, if an equation is .........., the sign of ΔH must also be .........., and if all coefficients are multiplied by a constant, the ΔH value is .......... by that same constant.
When applying Hess's Law, if an equation is reversed, the sign of ΔH must also be reversed, and if all coefficients are multiplied by a constant, the ΔH value is multiplied by that same constant.
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