9701 · 5.2
Hess's law flashcards
Revision flashcards for Cambridge 9701 Hess's law (syllabus 5.2). Flip, recall, then mark a real past-paper question.
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What is Hess's Law?
Hess's Law states that the total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same.
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Why is Hess's Law useful?
It allows us to calculate enthalpy changes for reactions that are difficult or impossible to measure directly, such as those that are too slow, too fast, or produce unwanted side products.
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What is the standard enthalpy change of formation, $\Delta H_f^\ominus$?
The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K and 100 kPa).
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What is the standard enthalpy change of formation of an element in its standard state?
Zero. For example, $\Delta H_f^\ominus[\text{O}_2(\text{g})]$ = 0 kJ mol⁻¹.
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What is the standard enthalpy change of combustion, $\Delta H_c^\ominus$?
The enthalpy change when one mole of a substance is completely combusted in excess oxygen under standard conditions (298 K and 100 kPa).
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When constructing an enthalpy cycle using $\Delta H_f^\ominus$ data, where do the arrows point from?
The arrows point UP from the elements in their standard states to the reactants and products. The indirect route goes 'down then up'.
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When constructing an enthalpy cycle using $\Delta H_c^\ominus$ data, where do the arrows point to?
The arrows point DOWN from the reactants and products to the combustion products (usually CO₂ and H₂O). The indirect route goes 'down then up'.
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What is the general formula for calculating $\Delta H_r^\ominus$ using formation data?
$\Delta H_r^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})$
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What is the general formula for calculating $\Delta H_r^\ominus$ using combustion data?
$\Delta H_r^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})$ (Note: Reactants minus Products)
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What happens to the value of $\Delta H$ if you reverse the direction of a reaction?
The sign of $\Delta H$ is reversed. If A → B has $\Delta H = +50$ kJ mol⁻¹, then B → A has $\Delta H = -50$ kJ mol⁻¹.
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What common mistake must be avoided when using Hess's Law calculations?
Forgetting to multiply the standard enthalpy values by the stoichiometric coefficients from the balanced chemical equation.