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23:[["$","$L32",null,{"feature":"flashcards"}],"$L33",["$","$L34",null,{"href":"/as-level/chemistry/edexcel/as-hess-cycles/flashcards","label":"Hess cycles","subjectCode":"as_level_chemistry","surface":"flashcards"}],["$","$L35",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L36",null,{"decks":[{"id":"e9c734c3-8378-4e1d-8097-45b266745c70","title":"Hess cycles Flashcards","topicId":"df89202e-4d77-5c30-b626-9cc39249b76c","topicSlug":"as-hess-cycles","cards":[{"id":"27ba3218-1efb-44cd-bc0c-00580a03de31","cardType":"basic","points":1,"front":"What is an enthalpy change ($ΔH$)?","back":"The heat energy change for a reaction at constant pressure.","mcqOptions":[{"text":"The heat energy change for a reaction at constant pressure.","correct":true,"explanation":"At constant pressure, the enthalpy change equals the heat transferred between the reacting system and its surroundings, indicating whether heat is released or absorbed."},{"text":"The activation energy required for a reaction to begin at constant pressure.","correct":false,"explanation":"Activation energy is the minimum energy needed to initiate a reaction, not its overall heat change. Enthalpy change is the heat energy change at constant pressure."},{"text":"The temperature change for a reaction occurring at constant pressure.","correct":false,"explanation":"Temperature change may be measured to calculate transferred heat, but it is not enthalpy change itself. Enthalpy change is the heat energy change at constant pressure."},{"text":"The heat energy change for a reaction at constant volume.","correct":false,"explanation":"Constant volume relates heat transfer directly to internal energy rather than enthalpy. Enthalpy change is defined here as the heat energy change at constant pressure."}]},{"id":"70b65dd9-25e4-4a45-a78f-96e432a7c9cf","cardType":"basic","points":2,"front":"What is the significance of the sign of $ΔH$?","back":"

Negative: exothermic (heat released)
Positive: endothermic (heat absorbed)

","mcqOptions":[{"text":"

Negative: exothermic (heat released)
Positive: endothermic (heat absorbed)

","correct":true,"explanation":"An exothermic reaction transfers heat from the system, so its enthalpy decreases and ΔH is negative. An endothermic reaction absorbs heat, making ΔH positive."},{"text":"

Negative: endothermic (heat absorbed)
Positive: exothermic (heat released)

","correct":false,"explanation":"This reverses the sign convention. Heat release lowers the system’s enthalpy, giving a negative ΔH, while heat absorption raises it, giving a positive ΔH."},{"text":"

Negative: exothermic (heat absorbed)
Positive: endothermic (heat released)

","correct":false,"explanation":"The reaction labels match the signs, but the directions of heat transfer are reversed. Exothermic reactions release heat, whereas endothermic reactions absorb heat."},{"text":"

Negative: endothermic (heat released)
Positive: exothermic (heat absorbed)

","correct":false,"explanation":"Both reaction labels are mismatched with their heat transfers and signs. A negative ΔH means exothermic heat release, while a positive ΔH means endothermic heat absorption."}]},{"id":"a44ef767-3999-4599-99a0-92a238f23c49","cardType":"cloze","points":null,"front":"Hess’s law states that the total enthalpy change for a reaction is [ ], provided the initial and final conditions are the same.","back":"Hess’s law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.","mcqOptions":[{"text":"independent of the route taken","correct":true,"explanation":"Enthalpy is a state function, so the overall enthalpy change depends only on the initial and final states, regardless of the intermediate steps followed."},{"text":"dependent on the route taken","correct":false,"explanation":"This contradicts Hess’s law by treating enthalpy as a pathway-dependent quantity. The total enthalpy change is independent of the route when initial and final conditions match."},{"text":"equal to the activation energy","correct":false,"explanation":"Activation energy describes the energy barrier for a particular reaction pathway, not the overall enthalpy change. Hess’s law says ΔH is independent of the route taken."},{"text":"independent of the initial conditions","correct":false,"explanation":"Enthalpy change depends on both the initial and final states, so changing the initial conditions can change ΔH. It is the route taken that does not matter."}]},{"id":"4c8ab093-d51e-413a-8a8b-105715b452da","cardType":"basic","points":3,"front":"What are the standard conditions used in energetics?","back":"

Temperature: 298 K
Pressure: 100 kPa
Concentration: 1 mol dm⁻³

","mcqOptions":[{"text":"

Temperature: 298 K
Pressure: 100 kPa
Concentration: 1 mol dm⁻³

","correct":true,"explanation":"Standard energetics data use 298 K and 100 kPa, with solutions at a concentration of 1 mol dm⁻³, allowing enthalpy values to be compared consistently."},{"text":"

Temperature: 273 K
Pressure: 100 kPa
Concentration: 1 mol dm⁻³

","correct":false,"explanation":"273 K is the freezing point of water and is often confused with standard temperature. Energetics standard conditions use 298 K, 100 kPa and 1 mol dm⁻³."},{"text":"

Temperature: 298 K
Pressure: 101 kPa
Concentration: 1 mol dm⁻³

","correct":false,"explanation":"101 kPa approximates one atmosphere and appears in older conventions, but the stated standard pressure for energetics is 100 kPa, alongside 298 K and 1 mol dm⁻³."},{"text":"

Temperature: 298 K
Pressure: 100 kPa
Concentration: 1 g dm⁻³

","correct":false,"explanation":"Standard concentration is an amount concentration, not a mass concentration. It must be 1 mol dm⁻³, together with a temperature of 298 K and pressure of 100 kPa."}]},{"id":"2d42a49d-9c44-4056-acf9-de2c93f7aa00","cardType":"basic","points":1,"front":"What defines the standard state of a substance?","back":"Its physical state under standard conditions (298 K and 100 kPa).","mcqOptions":[{"text":"Its physical state under standard conditions (298 K and 100 kPa).","correct":true,"explanation":"A substance’s standard state is its physical form under the specified standard temperature and pressure, so states such as solid, liquid and gas must be identified correctly."},{"text":"Its lowest energy level under standard conditions (298 K and 100 kPa).","correct":false,"explanation":"Energy level refers to an electronic or quantum state, not the thermodynamic standard state. Standard state identifies the substance’s physical state at 298 K and 100 kPa."},{"text":"Its reaction rate under standard conditions (298 K and 100 kPa).","correct":false,"explanation":"Reaction rate is a kinetic quantity and does not define standard state. Standard state refers to the substance’s physical state under conditions of 298 K and 100 kPa."},{"text":"Its chemical reactivity under standard conditions (298 K and 100 kPa).","correct":false,"explanation":"Chemical reactivity describes how readily a substance reacts, not its standard state. The standard state is its physical state at 298 K and 100 kPa."}]},{"id":"18f1f180-8280-48b3-80a5-3b74ffed25ed","cardType":"cloze","points":null,"front":"Hess’s law works because enthalpy is a [ ], meaning it only depends on the current state of the system.","back":"Hess’s law works because enthalpy is a state function, meaning it only depends on the current state of the system.","mcqOptions":[{"text":"state function","correct":true,"explanation":"A state function depends only on the system’s current state, not on how it reached that state. Therefore, enthalpy changes can be combined using Hess’s law."},{"text":"path function","correct":false,"explanation":"A path function depends on the route followed, which would prevent Hess cycles from working. Enthalpy is instead a state function determined by initial and final states."},{"text":"rate constant","correct":false,"explanation":"A rate constant is a kinetic quantity describing reaction speed at a particular temperature. Enthalpy is a state function, allowing route-independent enthalpy changes."},{"text":"transition state","correct":false,"explanation":"A transition state is a short-lived high-energy arrangement along a reaction pathway. Enthalpy is a state function, meaning only the initial and final states determine ΔH."}]},{"id":"7750c850-54a1-4677-b394-8c3d499074e5","cardType":"basic","points":1,"front":"Define standard enthalpy of formation ($ΔH_f^∘$).","back":"Enthalpy change when one mole of a compound is formed from its elements in their standard states.","mcqOptions":[{"text":"Enthalpy change when one mole of a compound is formed from its elements in their standard states.","correct":true,"explanation":"Standard enthalpy of formation refers to forming exactly one mole of a compound directly from its constituent elements, with all substances in their standard states."},{"text":"Enthalpy change when one mole of a compound is burned in excess oxygen in standard states.","correct":false,"explanation":"Burning one mole completely in oxygen defines standard enthalpy of combustion, not formation. Formation requires one mole of compound to form from its elements in standard states."},{"text":"Enthalpy change when one mole of an element is converted into gaseous atoms in standard states.","correct":false,"explanation":"Producing gaseous atoms from an element describes enthalpy of atomisation. Standard enthalpy of formation instead forms one mole of a compound from its elements."},{"text":"Enthalpy change when two moles of a compound are formed from their elements in standard states.","correct":false,"explanation":"The definition is normalised to exactly one mole of compound, not two. It is the enthalpy change when one mole forms from its elements in standard states."}]},{"id":"a029fee3-6891-4cae-9a80-f2e51e1efd0b","cardType":"basic","points":1,"front":"What is the $ΔH_f^∘$ value for elements in their standard states?","back":"0 kJ mol⁻³","mcqOptions":[{"text":"0 kJ mol⁻³","correct":true,"explanation":"The standard enthalpy of formation of an element in its standard state is defined as zero because no formation change is required to produce it from itself."},{"text":"1 kJ mol⁻³","correct":false,"explanation":"Assigning a small non-zero value is a common misconception, but the value is fixed by definition rather than measurement. Elements in their standard states have ΔH_f^∘ = 0 kJ mol⁻³."},{"text":"-1 kJ mol⁻³","correct":false,"explanation":"A negative value would imply exothermic formation, but an element in its standard state is the reference from which formation enthalpies are measured, so its value is zero."},{"text":"298 kJ mol⁻³","correct":false,"explanation":"This confuses the standard temperature of 298 K with an enthalpy value. By definition, the formation enthalpy of an element in its standard state is 0 kJ mol⁻³."}]},{"id":"527cd917-fb65-40e9-8a22-49812266273d","cardType":"basic","points":1,"front":"What is the formula to calculate $ΔH_r^∘$ using enthalpies of formation?","back":"$$$ΔH_r^∘ = ∑ ΔH_f^∘ \\text{ (products)} - ∑ ΔH_f^∘ \\text{ (reactants)}$$","mcqOptions":[{"text":"$$$ΔH_r^∘ = ∑ ΔH_f^∘ \\text{ (products)} - ∑ ΔH_f^∘ \\text{ (reactants)}$$","correct":true,"explanation":"This subtracts the sum of the standard enthalpies of formation of the reactants from the sum for the products, giving the reaction enthalpy change."},{"text":"$$$ΔH_r^∘ = ∑ ΔH_f^∘ \\text{ (reactants)} - ∑ ΔH_f^∘ \\text{ (products)}$$","correct":false,"explanation":"The order has been reversed, which changes the sign of the reaction enthalpy. Using formation enthalpies requires products minus reactants, because the reaction forms products from reactants."},{"text":"$$$ΔH_r^∘ = ∑ ΔH_c^∘ \\text{ (products)} - ∑ ΔH_c^∘ \\text{ (reactants)}$$","correct":false,"explanation":"This uses enthalpies of combustion instead of enthalpies of formation. For formation data, calculate products minus reactants; combustion data uses the opposite arrangement."},{"text":"$$$ΔH_r^∘ = ∑ ΔH_f^∘ \\text{ (products)} + ∑ ΔH_f^∘ \\text{ (reactants)}$$","correct":false,"explanation":"Adding both sets of formation enthalpies ignores that reactants must be subtracted when applying Hess’s law. The correct expression is products minus reactants."}]},{"id":"46b5ebde-4973-421f-9edb-064596870cf3","cardType":"cloze","points":null,"front":"In a Hess cycle, going [ ] an arrow means you must reverse the sign of the enthalpy value.","back":"In a Hess cycle, going against an arrow means you must reverse the sign of the enthalpy value.","mcqOptions":[{"text":"against","correct":true,"explanation":"Following a Hess cycle against an arrow means travelling in the reverse direction to the equation shown, so the enthalpy change must have its sign reversed."},{"text":"with","correct":false,"explanation":"Travelling with an arrow follows the equation as written, so its enthalpy sign stays unchanged. The sign is reversed only when travelling against an arrow."},{"text":"alongside","correct":false,"explanation":"‘Alongside’ does not describe the direction of travel relative to a Hess cycle arrow. Moving against the arrow reverses the equation and therefore reverses its enthalpy sign."},{"text":"between","correct":false,"explanation":"The word ‘between’ does not indicate reverse direction in a Hess cycle. The enthalpy sign changes when the cycle is followed against an arrow."}]},{"id":"b5cb07f0-16f2-49d0-9d37-536022babd65","cardType":"cloze","points":null,"front":"In a Hess cycle, going against an arrow means you must [ ] of the enthalpy value.","back":"In a Hess cycle, going against an arrow means you must reverse the sign of the enthalpy value.","mcqOptions":[{"text":"reverse the sign","correct":true,"explanation":"Going against an arrow means using the thermochemical equation in reverse. Reversing a reaction changes an exothermic enthalpy to positive or an endothermic one to negative."},{"text":"keep the sign","correct":false,"explanation":"Keeping the sign applies when travelling with the arrow in the direction written. Going against it reverses the equation, so the enthalpy sign must also reverse."},{"text":"double the value","correct":false,"explanation":"Doubling an enthalpy is needed only when the equation is multiplied by two. Travelling against an arrow instead reverses the sign without necessarily changing its magnitude."},{"text":"swap the units","correct":false,"explanation":"Changing direction in a Hess cycle does not alter the units, which remain typically kJ mol⁻¹. It reverses the sign of the enthalpy value."}]},{"id":"5eba4b25-c998-4017-8811-d14993ba64af","cardType":"basic","points":1,"front":"Define standard enthalpy of combustion ($ΔH_c^∘$).","back":"Enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.","mcqOptions":[{"text":"Enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.","correct":true,"explanation":"Standard enthalpy of combustion is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions, with substances in their standard states."},{"text":"Enthalpy change when one mole of a substance is partially burned in oxygen under standard conditions.","correct":false,"explanation":"Incomplete or partial combustion does not meet the definition. Standard enthalpy of combustion requires one mole of substance to burn completely in oxygen under standard conditions."},{"text":"Enthalpy change when one mole of oxygen is completely reacted with a substance under standard conditions.","correct":false,"explanation":"The defined amount is one mole of the substance being combusted, not one mole of oxygen. It must also be completely burned in oxygen under standard conditions."},{"text":"Enthalpy change when one mole of a substance is completely burned in air at any temperature.","correct":false,"explanation":"The process must occur in oxygen under standard conditions, not simply in air at any temperature. Standard conditions and complete combustion are essential parts of the definition."}]},{"id":"4e16183c-ac30-4a28-bdee-80c6a671e5f4","cardType":"basic","points":1,"front":"What is the formula to calculate $ΔH_r^∘$ using enthalpies of combustion?","back":"$$$ΔH_r^∘ = ∑ ΔH_c^∘ \\text{ (reactants)} - ∑ ΔH_c^∘ \\text{ (products)}$$","mcqOptions":[{"text":"$$$ΔH_r^∘ = ∑ ΔH_c^∘ \\text{ (reactants)} - ∑ ΔH_c^∘ \\text{ (products)}$$","correct":true,"explanation":"In a combustion cycle, both sides burn to the same products, usually carbon dioxide and water. Therefore, the reaction enthalpy is reactant combustion values minus product values."},{"text":"$$$ΔH_r^∘ = ∑ ΔH_c^∘ \\text{ (products)} - ∑ ΔH_c^∘ \\text{ (reactants)}$$","correct":false,"explanation":"This reverses the required order and gives the negative reaction enthalpy. With combustion data, use the sum for reactants minus the sum for products."},{"text":"$$$ΔH_r^∘ = ∑ ΔH_f^∘ \\text{ (reactants)} - ∑ ΔH_f^∘ \\text{ (products)}$$","correct":false,"explanation":"The formula uses enthalpies of formation rather than combustion. Formation data requires products minus reactants, whereas combustion data requires reactants minus products."},{"text":"$$$ΔH_r^∘ = ∑ ΔH_c^∘ \\text{ (reactants)} + ∑ ΔH_c^∘ \\text{ (products)}$$","correct":false,"explanation":"Adding the two combustion sums fails to account for the reversed path through the combustion products. The correct calculation subtracts product combustion enthalpies from reactant values."}]},{"id":"5fac8d87-c8c1-419b-bf6a-b8443b4a61dc","cardType":"basic","points":2,"front":"Why must enthalpy values be multiplied by coefficients in Hess cycle calculations?","back":"Enthalpy data is given per mole, but equations may involve multiple moles of a substance.","mcqOptions":[{"text":"Enthalpy data is given per mole, but equations may involve multiple moles of a substance.","correct":true,"explanation":"Because tabulated enthalpy values usually refer to one mole, multiplying by each equation coefficient scales the value to the actual amount reacting or forming."},{"text":"Enthalpy data is given per molecule, but equations may involve multiple molecules of a substance.","correct":false,"explanation":"Enthalpy data is normally quoted per mole, not per molecule. Coefficients are used to scale the per-mole value to the number of moles in the equation."},{"text":"Enthalpy data is given per gram, but equations may involve different masses of a substance.","correct":false,"explanation":"Hess cycle enthalpies are generally supplied per mole rather than per gram. Multiplication by coefficients accounts for stoichiometric mole amounts, not differing sample masses."},{"text":"Enthalpy data is given as a total, but equations may involve fractions of a substance.","correct":false,"explanation":"The tabulated value is not automatically the total enthalpy for every equation; it is usually per mole. Coefficients scale that value to the equation’s stoichiometric amounts."}]},{"id":"ebdbabeb-c7b0-466e-b724-e692cdc27ea3","cardType":"basic","points":1,"front":"In a combustion cycle, what are the reactants and products usually burned into at the bottom of the cycle?","back":"CO₂(g) and H₂O(l)","mcqOptions":[{"text":"CO₂(g) and H₂O(l)","correct":true,"explanation":"Combustion cycles usually burn every substance completely to carbon dioxide gas and liquid water. These common products provide the shared bottom point of the Hess cycle."},{"text":"CO(g) and H₂O(l)","correct":false,"explanation":"Carbon monoxide indicates incomplete combustion, not the usual complete-combustion endpoint. The standard bottom products are carbon dioxide gas and liquid water."},{"text":"CO₂(g) and H₂(g)","correct":false,"explanation":"Hydrogen gas is not the usual product when hydrogen-containing substances burn completely. Complete combustion produces liquid water alongside carbon dioxide gas."},{"text":"C(s) and H₂O(l)","correct":false,"explanation":"Solid carbon represents unburned or incompletely oxidised carbon. In a combustion cycle, carbon is normally fully oxidised to carbon dioxide gas, with water also formed."}]},{"id":"b36e3ea2-9acb-4796-98f0-f478fc66ebe1","cardType":"basic","points":1,"front":"If you reverse a thermochemical equation, what happens to its enthalpy change?","back":"The sign is changed (e.g., positive becomes negative).","mcqOptions":[{"text":"The sign is changed (e.g., positive becomes negative).","correct":true,"explanation":"Reversing a thermochemical equation reverses the direction of the process. The enthalpy change therefore changes sign, while its magnitude and units remain unchanged."},{"text":"The sign is unchanged (e.g., positive remains positive).","correct":false,"explanation":"The sign cannot remain unchanged when the reaction direction is reversed. Reversing a thermochemical equation changes a positive enthalpy to negative, or vice versa."},{"text":"The magnitude is doubled (e.g., positive becomes twice as positive).","correct":false,"explanation":"Reversing an equation does not multiply the reaction by two. It changes only the sign; the magnitude remains the same unless the equation is also scaled."},{"text":"The units are changed (e.g., kJ mol⁻¹ becomes J mol⁻¹).","correct":false,"explanation":"Changing reaction direction does not convert units. Reversal changes the sign of the enthalpy, whereas unit conversion would require multiplying by a numerical conversion factor."}]},{"id":"55c109e9-fb5f-4d89-a9c0-c6a07c8c7630","cardType":"cloze","points":null,"front":"If you multiply a thermochemical equation by a factor of 2, you must [ ].","back":"If you multiply a thermochemical equation by a factor of 2, you must multiply ΔH by 2.","mcqOptions":[{"text":"multiply ΔH by 2","correct":true,"explanation":"Enthalpy change is an extensive quantity, so doubling every stoichiometric coefficient doubles the amount reacting and therefore doubles the equation’s ΔH value."},{"text":"divide ΔH by 2","correct":false,"explanation":"Dividing confuses scaling an equation with reversing the scaling process. Doubling all coefficients doubles the quantities reacting, so the correct operation is to multiply ΔH by 2."},{"text":"leave ΔH unchanged","correct":false,"explanation":"Leaving ΔH unchanged treats enthalpy change as independent of the amount reacting. It is an extensive quantity, so doubling the equation requires multiplying ΔH by 2."},{"text":"multiply ΔH by 4","correct":false,"explanation":"Squaring the factor is an area-style scaling error. Enthalpy change scales directly with the stoichiometric amounts, so a factor of 2 requires multiplying ΔH by 2, not 4."}]},{"id":"e48c1cb1-ab0b-4094-9ebe-9732e42a2a3d","cardType":"basic","points":2,"front":"Why is Hess’s law useful for reactions that cannot be measured directly?","back":"The reaction might be too slow, incomplete, or occur alongside side reactions.","mcqOptions":[{"text":"The reaction might be too slow, incomplete, or occur alongside side reactions.","correct":true,"explanation":"These difficulties prevent the desired reaction’s heat change from being isolated accurately. Hess’s law instead combines measurable enthalpy changes to calculate the inaccessible value indirectly."},{"text":"The reaction might be very fast, complete, or occur without side reactions.","correct":false,"explanation":"Fast, complete reactions without competing pathways are generally suitable for direct measurement. Hess’s law is especially useful when reactions are too slow, incomplete, or accompanied by side reactions."},{"text":"The enthalpy change might depend on the route, vary with catalysts, or change with the number of steps.","correct":false,"explanation":"This incorrectly assumes enthalpy change depends on the pathway taken. Hess’s law works because ΔH is route-independent and helps with reactions that are too slow, incomplete, or affected by side reactions."},{"text":"The reaction might have zero activation energy, 100% yield, or produce only one product.","correct":false,"explanation":"These features would normally make a reaction easier, not impossible, to measure directly. The genuine obstacles are a reaction being too slow, incomplete, or accompanied by side reactions."}]},{"id":"7437e1dc-32ae-4358-a7ce-69ab42289d45","cardType":"basic","points":3,"front":"Give three reasons why experimental combustion enthalpies are less exothermic than data book values.","back":"

Heat loss to surroundings
Incomplete combustion
Evaporation of the fuel

","mcqOptions":[{"text":"

Heat loss to surroundings
Incomplete combustion
Evaporation of the fuel

","correct":true,"explanation":"Heat escaping lowers the measured temperature rise, incomplete combustion releases less energy, and evaporated fuel is counted as burned. Each makes the calculated combustion enthalpy less negative."},{"text":"

Heat gain from surroundings
Complete combustion
Condensation of the fuel

","correct":false,"explanation":"These changes would increase the measured heat or favour ideal combustion, rather than explain a less exothermic result. The main causes are heat loss, incomplete combustion, and fuel evaporation."},{"text":"

Better insulation of apparatus
Excess oxygen supplied
No evaporation of the fuel

","correct":false,"explanation":"All three measures improve experimental accuracy and should bring the result closer to the data book value. Less exothermic values instead arise from heat loss, incomplete combustion, and fuel evaporation."},{"text":"

Higher flame temperature
Complete oxidation to CO2
Accurate fuel mass measurement

","correct":false,"explanation":"Complete oxidation and accurate mass measurement reduce error, while a higher flame temperature is not itself a source of the discrepancy. Heat loss, incomplete combustion, and fuel evaporation cause lower values."}]},{"id":"3c224323-1ea0-47cb-b1f7-4639e22ba044","cardType":"cloze","points":null,"front":"In calorimetry, the formula $ΔH = -q/n$ includes a negative sign because the heat change of the [ ] is opposite to the heat change of the surroundings.","back":"In calorimetry, the formula $ΔH = -q/n$ includes a negative sign because the heat change of the reaction is opposite to the heat change of the surroundings.","mcqOptions":[{"text":"reaction","correct":true,"explanation":"The measured heat, q, is gained or lost by the surroundings. Energy conservation means the reaction undergoes an equal heat change with the opposite sign, giving $ΔH = -q/n$."},{"text":"calorimeter","correct":false,"explanation":"The calorimeter forms part of the surroundings whose heat change is represented by q; it is not the opposing process. The reaction has the equal and opposite heat change."},{"text":"solution","correct":false,"explanation":"The solution is normally part of the surroundings used to calculate q from its temperature change. The reaction, not the solution, has the opposing heat change represented by the negative sign."},{"text":"fuel","correct":false,"explanation":"The fuel may be a reactant, but the sign convention applies to the entire chemical reaction rather than one substance. The reaction’s heat change is opposite to that of the surroundings."}]},{"id":"81f03039-196e-4b22-823d-c600c29013dc","cardType":"cloze","points":null,"front":"In calorimetry, the formula $ΔH = -q/n$ includes a negative sign because the heat change of the reaction is opposite to the heat change of the [ ].","back":"In calorimetry, the formula $ΔH = -q/n$ includes a negative sign because the heat change of the reaction is opposite to the heat change of the surroundings.","mcqOptions":[{"text":"surroundings","correct":true,"explanation":"Calorimetry calculates q from the temperature change of the solution or calorimeter, which belongs to the surroundings. The reaction’s heat change is equal in magnitude but opposite in sign."},{"text":"reaction","correct":false,"explanation":"A heat change cannot be described as opposite to itself. The reaction exchanges energy with the surroundings, whose equal and opposite heat change gives the negative sign."},{"text":"products","correct":false,"explanation":"Products belong to the reacting chemical system, not the external body absorbing or releasing measured heat. The reaction’s heat change is opposite to that of the surroundings."},{"text":"catalyst","correct":false,"explanation":"A catalyst provides an alternative pathway but is not the body whose heat change is measured. The negative sign relates the reaction’s heat change to the opposite change of the surroundings."}]},{"id":"1f249b0d-02fc-4349-8d30-ad5661a11a10","cardType":"basic","points":2,"front":"Contrast the direction of arrows in formation vs combustion Hess cycles.","back":"

Formation: arrows point up from elements to the reaction.
Combustion: arrows point down from the reaction to combustion products.

","mcqOptions":[{"text":"

Formation: arrows point up from elements to the reaction.
Combustion: arrows point down from the reaction to combustion products.

","correct":true,"explanation":"Formation enthalpies connect elements in their standard states upward to the substances in the reaction. Combustion enthalpies connect those substances downward to their common combustion products."},{"text":"

Formation: arrows point down from the reaction to elements.
Combustion: arrows point up from combustion products to the reaction.

","correct":false,"explanation":"Both directions have been reversed. Formation arrows run from elements up to the reaction substances, whereas combustion arrows run down from those substances to common combustion products."},{"text":"

Formation: arrows point down from the reaction to elements.
Combustion: arrows point down from the reaction to combustion products.

","correct":false,"explanation":"The formation direction has been confused with decomposition. Formation arrows point up from elements to the reaction substances, while combustion arrows correctly point down to combustion products."},{"text":"

Formation: arrows point up from elements to the reaction.
Combustion: arrows point up from combustion products to the reaction.

","correct":false,"explanation":"The combustion direction has been reversed. Formation arrows point up from elements to reaction substances, but combustion arrows point down from those substances to their common combustion products."}]}]}],"examCountdown":{"examDate":"2027-05-11","paperLabel":"Paper 1"},"heading":"Hess cycles","title":"Hess cycles Flashcards","topbar":{"backHref":"/as-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"df89202e-4d77-5c30-b626-9cc39249b76c","walled":"$L38"}],"$L39","$L3a"]