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constant-pressure conditions, as commonly applies to reactions performed in open containers."},{"text":"Heat energy change at constant volume.","correct":false,"explanation":"Heat energy measured at constant volume corresponds more closely to internal energy change. Enthalpy change measures the heat energy change at constant pressure."},{"text":"Temperature change at constant pressure.","correct":false,"explanation":"Temperature change may be measured during calorimetry, but it is used to calculate transferred heat. Enthalpy change measures heat energy change at constant pressure."},{"text":"Activation energy change at constant pressure.","correct":false,"explanation":"Activation energy is the minimum energy needed for a reaction to occur and is not an overall energy change. Enthalpy change measures heat energy change at constant pressure."}]},{"id":"b7499fde-be87-4203-8691-cea947a7104f","cardType":"cloze","points":null,"front":"A standard enthalpy change ($$ \\Delta H^\\circ $$) requires substances in their standard states at [ ] and 298 K.","back":"A standard enthalpy change ($$ \\Delta H^\\circ $$) requires substances in their standard states at 100 kPa and 298 K.","mcqOptions":[{"text":"100 kPa","correct":true,"explanation":"Standard enthalpy changes are defined with all substances in their standard states under a standard pressure of 100 kPa, with the temperature usually specified as 298 K."},{"text":"101 kPa","correct":false,"explanation":"This confuses the current standard pressure with the older convention of one atmosphere. For these definitions, standard enthalpy changes use 100 kPa at 298 K."},{"text":"1000 kPa","correct":false,"explanation":"This is ten times the required pressure, likely from a place-value error. The standard pressure used for standard enthalpy changes is 100 kPa."},{"text":"1.00 kPa","correct":false,"explanation":"This is atmospheric pressure expressed with the wrong magnitude. Standard enthalpy changes require substances in their standard states at 100 kPa."}]},{"id":"0a98cd3b-388a-4368-a7ac-a5461b9b85e2","cardType":"cloze","points":null,"front":"A standard enthalpy change ($$ \\Delta H^\\circ $$) requires substances in their standard states at 100 kPa and [ ].","back":"A standard enthalpy change ($$ \\Delta H^\\circ $$) requires substances in their standard states at 100 kPa and 298 K.","mcqOptions":[{"text":"298 K","correct":true,"explanation":"Standard enthalpy data are normally quoted at 298 K, equivalent to approximately 25 °C, with substances in their standard states at 100 kPa."},{"text":"25 K","correct":false,"explanation":"This is the standard temperature in Celsius written incorrectly as kelvin. Approximately 25 °C corresponds to 298 K, which is the required standard temperature."},{"text":"273 K","correct":false,"explanation":"This is the freezing point of water and is not the usual temperature for standard enthalpy data. The specified standard temperature is 298 K."},{"text":"373 K","correct":false,"explanation":"This is the boiling point of water at atmospheric pressure, not the usual reference temperature for standard enthalpy changes. The correct temperature is 298 K."}]},{"id":"82537440-d170-4ada-970d-9e0cd276d438","cardType":"basic","points":2,"front":"If the temperature of the surroundings increases, what is the reaction type and sign of $$ \\Delta H $$?","back":"

Exothermic
Negative sign

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

Exothermic
Negative sign

","correct":true,"explanation":"A rise in the surroundings’ temperature shows that the reaction releases heat, making it exothermic. The reacting system loses enthalpy, so its enthalpy change is negative."},{"text":"

Endothermic
Positive sign

","correct":false,"explanation":"An endothermic reaction absorbs heat and would tend to cool the surroundings. A temperature increase indicates an exothermic reaction with a negative enthalpy change."},{"text":"

Exothermic
Positive sign

","correct":false,"explanation":"The reaction type is identified correctly, but the sign is reversed. Heat released by an exothermic system means its enthalpy decreases, giving a negative enthalpy change."},{"text":"

Endothermic
Negative sign

","correct":false,"explanation":"The negative sign indicates heat release, not heat absorption. If the surroundings warm, the reaction is exothermic and has a negative enthalpy change."}]},{"id":"223dae9b-142f-402f-b36d-9dc5fb7424dc","cardType":"basic","points":1,"front":"Define specific heat capacity ($c$).","back":"Energy required to raise the temperature of 1 g of a substance by 1 K.","mcqOptions":[{"text":"Energy required to raise the temperature of 1 g of a substance by 1 K.","correct":true,"explanation":"Specific heat capacity is the energy needed per unit mass for a one-kelvin temperature rise. With units of J g⁻¹ K⁻¹, the specified mass is 1 g."},{"text":"Energy required to raise the temperature of 1 mol of a substance by 1 K.","correct":false,"explanation":"Using one mole defines molar heat capacity rather than specific heat capacity by mass. Here, specific heat capacity is the energy needed to heat 1 g by 1 K."},{"text":"Energy required to raise the temperature of 1 g of a substance by 10 K.","correct":false,"explanation":"A ten-kelvin increase would give ten times the energy associated with the definition. Specific heat capacity uses the energy required to heat 1 g by exactly 1 K."},{"text":"Energy released to raise the temperature of 1 g of a substance by 1 K.","correct":false,"explanation":"Specific heat capacity concerns energy absorbed to raise temperature, not energy released as temperature rises. It is the energy required to heat 1 g by 1 K."}]},{"id":"91e0eab5-139a-4be5-afd9-c26bb6e44cbe","cardType":"basic","points":2,"front":"In the calorimetry equation $$ q = mc\\Delta T $$, what are the units for $q$ and $m$?","back":"

$q$ is in Joules (J)
$m$ is in grams (g)

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

$q$ is in Joules (J)
$m$ is in grams (g)

","correct":true,"explanation":"When specific heat capacity is expressed in J g⁻¹ K⁻¹ and temperature change in kelvin, mass must be in grams, making the calculated heat transfer measured in joules."},{"text":"

$q$ is in Joules (J)
$m$ is in kilograms (kg)

","correct":false,"explanation":"This mixes incompatible prefixes: using kilograms with a heat capacity in J g⁻¹ K⁻¹ creates a factor-of-1000 error. The required units are joules and grams."},{"text":"

$q$ is in kilojoules (kJ)
$m$ is in grams (g)

","correct":false,"explanation":"Using kilojoules instead of joules makes the reported heat value 1000 times too large unless a conversion is applied. The stated equation convention uses joules and grams."},{"text":"

$q$ is in kilojoules (kJ)
$m$ is in kilograms (kg)

","correct":false,"explanation":"Both prefixes conflict with the usual calorimetry convention using specific heat capacity in J g⁻¹ K⁻¹. In that convention, heat is in joules and mass is in grams."}]},{"id":"bd0369cf-bdcb-4e30-bb35-fe28ea848c1d","cardType":"cloze","points":null,"front":"When calculating the mass of a dilute aqueous solution from its volume, the density is assumed to be [ ].","back":"When calculating the mass of a dilute aqueous solution from its volume, the density is assumed to be 1.00 g cm⁻³.","mcqOptions":[{"text":"1.00 g cm⁻³","correct":true,"explanation":"Dilute aqueous solutions are approximated as having water’s density, 1.00 g cm⁻³. Therefore, each cubic centimetre of solution is treated as having a mass of one gram."},{"text":"10.0 g cm⁻³","correct":false,"explanation":"This value is ten times water’s approximate density and would make the calculated solution mass ten times too large. Dilute aqueous solutions use 1.00 g cm⁻³."},{"text":"0.001 g cm⁻³","correct":false,"explanation":"This is a factor-of-1000 place-value error and would give an unrealistically small mass for the solution. The assumed density is 1.00 g cm⁻³."},{"text":"0.80 g cm⁻³","correct":false,"explanation":"Although this may resemble the density of a concentrated or different liquid, dilute aqueous solutions are approximated as water and assigned a density of 1.00 g cm⁻³."}]},{"id":"5823271e-3579-4c09-a1bc-2a27e0f27f01","cardType":"basic","points":2,"front":"Why is a cooling-curve extrapolation used in solution calorimetry?","back":"To estimate the temperature change at the instant of mixing, reducing the effect of heat loss.","mcqOptions":[{"text":"To estimate the temperature change at the instant of mixing, reducing the effect of heat loss.","correct":true,"explanation":"Heat transfer begins immediately after mixing, so the measured peak may be lower than the true value. Extrapolation estimates the instantaneous temperature change and compensates for heat loss."},{"text":"To estimate the mass of solution at the instant of mixing, reducing the effect of density changes.","correct":false,"explanation":"Extrapolation does not determine the solution’s mass; mass is obtained from volume and assumed density. It estimates the temperature change at mixing to reduce heat-loss error."},{"text":"To estimate the concentration at the instant of mixing, reducing the effect of dilution.","correct":false,"explanation":"A cooling curve cannot determine reactant concentration, which must be measured when solutions are prepared. It estimates the initial temperature change while correcting for heat loss."},{"text":"To estimate the reaction time at the instant of mixing, reducing the effect of incomplete reaction.","correct":false,"explanation":"Extrapolation corrects for unwanted heat transfer rather than incomplete reaction. Its purpose is to estimate the temperature change at the instant of mixing, reducing heat-loss effects."}]},{"id":"c488358d-f28f-4776-a78c-54b503ad198a","cardType":"cloze","points":null,"front":"To convert heat energy into molar enthalpy, use the formula $$ \\Delta H = $$ [ ].","back":"To convert heat energy into molar enthalpy, use the formula $$ \\Delta H = $$ -q / n.","mcqOptions":[{"text":"-q / n","correct":true,"explanation":"Dividing the heat energy by the amount of substance gives energy per mole, while the negative sign converts heat released by the reaction into a negative enthalpy change."},{"text":"q / n","correct":false,"explanation":"This omits the negative sign, so heat released would incorrectly produce a positive enthalpy change. The correct conversion is $\\Delta H = -q / n$."},{"text":"-q × n","correct":false,"explanation":"Multiplying by the amount of substance gives a total energy rather than an energy per mole. Enthalpy change per mole requires division: $\\Delta H = -q / n$."},{"text":"q × n","correct":false,"explanation":"This both omits the sign change and multiplies instead of dividing by the amount of substance. The correct expression is $\\Delta H = -q / n$."}]},{"id":"d5630b53-209c-4726-9935-3c03547a1345","cardType":"basic","points":1,"front":"Define standard enthalpy change of combustion ($$ \\Delta H_c $$).","back":"Enthalpy change when 1 mol of a substance burns completely in oxygen under standard conditions.","mcqOptions":[{"text":"Enthalpy change when 1 mol of a substance burns completely in oxygen under standard conditions.","correct":true,"explanation":"Standard enthalpy change of combustion is defined for exactly 1 mol of substance undergoing complete combustion in oxygen under standard conditions."},{"text":"Enthalpy change when 1 mol of a substance reacts incompletely in oxygen under standard conditions.","correct":false,"explanation":"Incomplete combustion does not meet the definition because combustion must be complete. The correct definition specifies 1 mol burning completely in oxygen under standard conditions."},{"text":"Enthalpy change when 1 mol of a substance burns completely in air under standard conditions.","correct":false,"explanation":"The standard definition refers specifically to combustion in oxygen, not simply air. It also requires 1 mol to burn completely under standard conditions."},{"text":"Enthalpy change when 1 mol of a substance burns completely in oxygen at any temperature and pressure.","correct":false,"explanation":"Standard enthalpy changes require standard conditions, so arbitrary temperature and pressure are not acceptable. The definition includes complete combustion of 1 mol in oxygen under standard conditions."}]},{"id":"c9ff8c66-a0fd-45af-9bb5-340fecff96dc","cardType":"basic","points":2,"front":"Give two reasons why experimental combustion calorimetry values are less exothermic than data book values.","back":"

Heat loss to the surroundings/apparatus
Incomplete combustion of the fuel

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

Heat loss to the surroundings/apparatus
Incomplete combustion of the fuel

","correct":true,"explanation":"Heat escaping to the surroundings or apparatus means less measured energy, while incomplete combustion releases less energy than complete combustion, making the result less exothermic."},{"text":"

Heat gain from the surroundings/apparatus
Complete combustion of the fuel

","correct":false,"explanation":"These effects would tend to increase the measured temperature change, not reduce it. The correct causes are heat loss and incomplete combustion of the fuel."},{"text":"

Heat loss to the surroundings/apparatus
Excess oxygen around the fuel

","correct":false,"explanation":"Heat loss is a valid cause, but excess oxygen generally promotes complete combustion rather than making values less exothermic. The second cause is incomplete combustion."},{"text":"

Heat gain from the surroundings/apparatus
Incomplete combustion of the fuel

","correct":false,"explanation":"Incomplete combustion lowers the energy released, but heat gained from outside would increase the measured temperature change. Heat loss is the required first cause."}]},{"id":"e992c451-c6ed-448f-84b6-72cb931cdbdc","cardType":"cloze","points":null,"front":"Breaking chemical bonds is an [ ] process, while forming bonds is exothermic.","back":"Breaking chemical bonds is an endothermic process, while forming bonds is exothermic.","mcqOptions":[{"text":"endothermic","correct":true,"explanation":"Breaking a bond requires energy to overcome the electrostatic attraction between bonded atoms, so bond breaking is endothermic."},{"text":"exothermic","correct":false,"explanation":"Exothermic describes bond formation, when energy is released. Breaking bonds requires energy input, so the correct term is endothermic."},{"text":"oxidising","correct":false,"explanation":"Oxidising describes electron loss or an increase in oxidation state, not the energy change associated with breaking a chemical bond. Bond breaking is endothermic."},{"text":"neutral","correct":false,"explanation":"Breaking a bond cannot be treated as energetically neutral because energy must be supplied to separate the bonded atoms. The process is endothermic."}]},{"id":"451f617e-bdcc-4cba-9bbb-f19355ef5c73","cardType":"cloze","points":null,"front":"Breaking chemical bonds is an endothermic process, while forming bonds is [ ].","back":"Breaking chemical bonds is an endothermic process, while forming bonds is exothermic.","mcqOptions":[{"text":"exothermic","correct":true,"explanation":"When a new bond forms, the bonded system moves to a lower energy state and releases energy, so bond formation is exothermic."},{"text":"endothermic","correct":false,"explanation":"Endothermic describes bond breaking, which requires energy input. Bond formation releases energy, so the correct term is exothermic."},{"text":"oxidising","correct":false,"explanation":"Oxidising refers to electron transfer or oxidation state changes, not whether energy is released during bond formation. Forming bonds is exothermic."},{"text":"neutral","correct":false,"explanation":"Bond formation releases energy as the system becomes more stable, so it is not energetically neutral. The correct description is exothermic."}]},{"id":"9042192e-b653-4e73-b16c-1d42a13bcad4","cardType":"basic","points":1,"front":"Define mean bond enthalpy.","back":"The average enthalpy change to break 1 mol of a covalent bond in gaseous molecules.","mcqOptions":[{"text":"The average enthalpy change to break 1 mol of a covalent bond in gaseous molecules.","correct":true,"explanation":"Mean bond enthalpy is the average energy required to break 1 mol of a specified covalent bond in gaseous molecules, allowing for different molecular environments."},{"text":"The enthalpy change to break 1 mol of an ionic bond in solid compounds.","correct":false,"explanation":"This describes neither the bond type nor the physical state used for mean bond enthalpies. The definition concerns average enthalpy changes for covalent bonds in gaseous molecules."},{"text":"The average enthalpy change to form 1 mol of a covalent bond in gaseous molecules.","correct":false,"explanation":"Mean bond enthalpy is defined for breaking bonds, which requires energy, rather than forming them. It is the average enthalpy change to break covalent bonds in gaseous molecules."},{"text":"The average enthalpy change to break 1 mol of a covalent bond in liquid molecules.","correct":false,"explanation":"The physical state is wrong: bond enthalpies are defined for gaseous molecules so intermolecular forces do not affect the value. The correct definition specifies gaseous molecules."}]},{"id":"04785c2e-0f21-401a-87a1-9fc6b4f071b5","cardType":"cloze","points":null,"front":"Mean bond enthalpies are averaged over [ ].","back":"Mean bond enthalpies are averaged over a range of compounds.","mcqOptions":[{"text":"a range of compounds","correct":true,"explanation":"The same type of covalent bond can have slightly different strengths in different molecular environments, so mean bond enthalpies are averaged over a range of compounds."},{"text":"one specific compound","correct":false,"explanation":"A single compound provides one bond environment rather than a mean value. Mean bond enthalpies are averaged over a range of compounds."},{"text":"a range of elements","correct":false,"explanation":"The averaging concerns bond strengths in different compounds, not isolated elements. Mean bond enthalpies are averaged over a range of compounds."},{"text":"one specific bond length","correct":false,"explanation":"Bond length is a structural measurement, not the set of molecular environments used to calculate a mean bond enthalpy. Values are averaged over a range of compounds."}]},{"id":"ab39bea1-24e2-47d0-a82e-08174c780298","cardType":"basic","points":1,"front":"What is the formula for estimating enthalpy change using bond enthalpies?","back":"$$$ \\Delta H \\approx \\sum \\text{bonds broken} - \\sum \\text{bonds formed} $$","mcqOptions":[{"text":"$$$ \\Delta H \\approx \\sum \\text{bonds broken} - \\sum \\text{bonds formed} $$","correct":true,"explanation":"Breaking bonds requires energy and contributes positively, while forming bonds releases energy and contributes negatively, giving bonds broken minus bonds formed."},{"text":"$$$ \\Delta H \\approx \\sum \\text{bonds formed} - \\sum \\text{bonds broken} $$","correct":false,"explanation":"This reverses the energy signs: bond breaking requires energy and bond formation releases energy. The correct estimate is bonds broken minus bonds formed."},{"text":"$$$ \\Delta H \\approx \\sum \\text{bonds broken} + \\sum \\text{bonds formed} $$","correct":false,"explanation":"Adding both terms treats bond formation as energy absorbed, but forming bonds releases energy. The correct formula subtracts bonds formed from bonds broken."},{"text":"$$$ \\Delta H \\approx \\sum \\text{bonds formed} + \\sum \\text{bonds broken} $$","correct":false,"explanation":"This both places the terms in the wrong order and adds them, so it cannot represent energy released during bond formation. Use bonds broken minus bonds formed."}]},{"id":"be588b2b-058f-4486-b872-c831b1f42518","cardType":"basic","points":2,"front":"Why are bond enthalpy calculations only approximations?","back":"They use average values and only apply to gaseous substances.","mcqOptions":[{"text":"They use average values and only apply to gaseous substances.","correct":true,"explanation":"Average bond enthalpies combine measurements from many different compounds, although bond strengths depend on molecular environment. They also describe bonds between gaseous species, so other state changes are omitted."},{"text":"They use exact values and only apply to liquid substances.","correct":false,"explanation":"Bond enthalpies are not exact values that apply only to liquids. The calculations use average bond enthalpies for gaseous substances, so molecular environments and energy changes between physical states are not fully represented."},{"text":"They use measured values and only apply to solid substances.","correct":false,"explanation":"Measured values for solids are not the basis of bond enthalpy calculations. The data are average values for gaseous substances, meaning intermolecular forces and state-change enthalpies are excluded."},{"text":"They use specific values and only apply to aqueous substances.","correct":false,"explanation":"Bond enthalpies do not use compound-specific values for aqueous substances. They are averages for gaseous species, so they neglect variations between molecular environments and effects associated with other physical states."}]},{"id":"68dc70b4-00ab-45bd-946f-32d3a72de2a2","cardType":"cloze","points":null,"front":"If a reaction involves [ ] or solids, bond enthalpy calculations are less accurate because they ignore energy changes like vaporisation.","back":"If a reaction involves liquids or solids, bond enthalpy calculations are less accurate because they ignore energy changes like vaporisation.","mcqOptions":[{"text":"liquids","correct":true,"explanation":"Liquids must be converted into gaseous species before gaseous bond enthalpies can be applied. Ignoring the enthalpy of vaporisation therefore reduces the accuracy of the calculation."},{"text":"gases","correct":false,"explanation":"Bond enthalpies are defined for gaseous species, so gases themselves do not require vaporisation. The missing state-change energy becomes important when liquids are involved, making the correct term liquids."},{"text":"atoms","correct":false,"explanation":"Atoms are already individual gaseous particles in bond enthalpy cycles and do not undergo vaporisation as molecular liquids do. The relevant physical state omitted from the calculation is liquids."},{"text":"catalysts","correct":false,"explanation":"Catalysts change a reaction’s pathway and activation energy but do not create the omitted vaporisation term. Bond enthalpy estimates become less accurate when liquids are involved because gaseous data are used."}]},{"id":"174dc3d5-1253-4b28-8853-56d8b4eeff95","cardType":"basic","points":1,"front":"What specific quantity does $m$ represent when using $$ q = mc\\Delta T $$ for combustion calorimetry?","back":"The mass of water being heated (not the mass of fuel).","mcqOptions":[{"text":"The mass of water being heated (not the mass of fuel).","correct":true,"explanation":"The heat calculation refers to the substance whose temperature change is measured. In combustion calorimetry, that substance is the water, so its mass is used with water’s specific heat capacity."},{"text":"The mass of fuel being burned (not the mass of water).","correct":false,"explanation":"The fuel’s mass loss is used to calculate moles burned, not the heat absorbed using $q=mc\\Delta T$. Here, $m$ is the mass of water being heated."},{"text":"The mass of burner holding the fuel (not the mass of water).","correct":false,"explanation":"The burner does not undergo the measured temperature change and its mass includes material that is not being combusted. In $q=mc\\Delta T$, $m$ is the mass of heated water."},{"text":"The mass of apparatus being used (not the mass of water).","correct":false,"explanation":"The apparatus is not represented by water’s specific heat capacity in this simplified calculation. The required mass is that of the water whose temperature rise is measured."}]},{"id":"b9bdf6c3-8894-4fcd-b4dc-b1e29796f8bc","cardType":"basic","points":1,"front":"How do you calculate the moles of fuel burned in a spirit burner experiment?","back":"$$$ n = \\frac{\\text{mass of burner lost}}{\\text{molar mass of fuel}} $$","mcqOptions":[{"text":"$$$ n = \\frac{\\text{mass of burner lost}}{\\text{molar mass of fuel}} $$","correct":true,"explanation":"The burner’s decrease in mass equals the mass of fuel combusted. Dividing that mass by the fuel’s molar mass converts it into the number of moles burned."},{"text":"$$$ n = \\frac{\\text{initial mass of burner}}{\\text{molar mass of fuel}} $$","correct":false,"explanation":"The initial burner mass includes the burner itself and all fuel present, rather than only the fuel consumed. Moles burned use the burner’s mass loss divided by fuel molar mass."},{"text":"$$$ n = \\frac{\\text{mass of fuel remaining}}{\\text{molar mass of fuel}} $$","correct":false,"explanation":"The final fuel mass measures what remains rather than what reacted. The mass burned is found from the burner’s decrease in mass, then divided by the fuel’s molar mass."},{"text":"$$$ n = \\text{mass of burner lost} \\times \\text{molar mass of fuel} $$","correct":false,"explanation":"Multiplying mass by molar mass gives incorrect units and does not calculate amount of substance. The mass lost must be divided by the fuel’s molar mass."}]},{"id":"333f4f08-4b95-468e-8c01-d51753f62095","cardType":"basic","points":1,"front":"Which species is used to calculate $n$ in the molar enthalpy formula if the reactants are not in stoichiometric amounts?","back":"The limiting reagent.","mcqOptions":[{"text":"The limiting reagent.","correct":true,"explanation":"The limiting reagent is consumed first and therefore determines the maximum reaction extent. Its amount is used to relate the measured heat transfer to one mole of reaction."},{"text":"The excess reagent.","correct":false,"explanation":"The excess reagent is not completely consumed, so its initial amount does not represent the amount that reacted. The limiting reagent determines the reaction extent and supplies the required value of $n$."},{"text":"The catalyst.","correct":false,"explanation":"A catalyst provides an alternative reaction pathway but is regenerated rather than consumed stoichiometrically. The amount reacting is determined from the limiting reagent, which is consumed first."},{"text":"The solvent.","correct":false,"explanation":"The solvent provides the reaction medium but does not normally determine the stoichiometric extent of reaction. The moles used are those of the limiting reagent after considering the equation’s ratios."}]},{"id":"4eecd280-06d1-44d7-ac58-00ab6676f292","cardType":"cloze","points":null,"front":"In calorimetry, if the solution cools down, the reaction is [ ] and the value of $$ \\Delta H $$ must be positive.","back":"In calorimetry, if the solution cools down, the reaction is endothermic and the value of $$ \\Delta H $$ must be positive.","mcqOptions":[{"text":"endothermic","correct":true,"explanation":"Cooling of the solution means the reaction absorbs thermal energy from its surroundings. Heat absorption gives the reacting system a positive enthalpy change, so the reaction is endothermic."},{"text":"exothermic","correct":false,"explanation":"An exothermic reaction releases heat and would normally raise the solution’s temperature. Cooling indicates that heat is absorbed, so the reaction is endothermic and $\\Delta H$ is positive."},{"text":"thermoneutral","correct":false,"explanation":"A thermoneutral reaction has essentially no enthalpy change and would not cause cooling through reaction heat. A falling temperature indicates heat absorption, making the reaction endothermic."},{"text":"spontaneous","correct":false,"explanation":"Spontaneity depends on Gibbs free energy and cannot be inferred solely from whether the solution cools. Cooling identifies heat absorption, so the required energetic description is endothermic."}]},{"id":"a9e63c6c-9fe5-402d-b123-c67a1e78d959","cardType":"cloze","points":null,"front":"In calorimetry, if the solution cools down, the reaction is endothermic and the value of $$ \\Delta H $$ must be [ ].","back":"In calorimetry, if the solution cools down, the reaction is endothermic and the value of $$ \\Delta H $$ must be positive.","mcqOptions":[{"text":"positive","correct":true,"explanation":"An endothermic reaction absorbs heat from the surroundings, increasing the reacting system’s enthalpy. Products therefore have higher enthalpy than reactants, making $\\Delta H$ positive."},{"text":"negative","correct":false,"explanation":"A negative enthalpy change corresponds to heat release by an exothermic reaction. Because the cooling solution supplies heat to an endothermic reaction, $\\Delta H$ must be positive."},{"text":"zero","correct":false,"explanation":"A zero enthalpy change would describe a thermoneutral process with no net reaction heat. The observed cooling shows that energy is absorbed, so $\\Delta H$ is positive."},{"text":"constant","correct":false,"explanation":"Enthalpy change is a signed energy difference, not a quantity described as constant here. Heat absorption by the endothermic reaction means its calculated $\\Delta H$ must be positive."}]}]}],"examCountdown":{"examDate":"2027-05-11","paperLabel":"Paper 1"},"heading":"Calorimetry and bond enthalpies","title":"Calorimetry and bond enthalpies Flashcards","topbar":{"backHref":"/as-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"ac495f8c-d79d-5f5b-8c75-f6eabeb45401","walled":"$L39"}],"$L3a","$L3b"]