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The reaction system absorbs this energy, so breaking bonds is an endothermic process."},{"text":"Breaking bonds is exothermic because it requires energy.","correct":false,"explanation":"Calling the process exothermic confuses energy absorption with energy release. Breaking bonds requires energy from the surroundings, so it is endothermic rather than exothermic."},{"text":"Breaking bonds is endothermic because it releases energy.","correct":false,"explanation":"The classification is right, but the reason is reversed. Breaking bonds requires an input of energy to overcome attractions; energy is released when new bonds form."},{"text":"Breaking bonds is exothermic because it releases energy.","correct":false,"explanation":"This describes bond formation, not bond breaking. Breaking bonds absorbs energy to overcome attractions between atoms, making the process endothermic rather than exothermic."}]},{"id":"19807450-eadf-484c-8e5a-22b538716bef","cardType":"cloze","points":null,"front":"Making bonds is [ ] because it [ ].","back":"Making bonds is exothermic because it releases energy.","mcqOptions":[{"text":"Making bonds is exothermic because it releases energy.","correct":true,"explanation":"When a bond forms, atoms move to a more stable, lower-energy arrangement. The excess energy is transferred to the surroundings, so bond formation is exothermic."},{"text":"Making bonds is endothermic because it releases energy.","correct":false,"explanation":"Releasing energy makes a process exothermic, not endothermic. Bond formation transfers energy from the system to the surroundings as atoms enter a lower-energy arrangement."},{"text":"Making bonds is exothermic because it requires energy.","correct":false,"explanation":"The classification is right, but the reason describes bond breaking. Making bonds releases energy as atoms enter a more stable arrangement; it does not require a net energy input."},{"text":"Making bonds is endothermic because it requires energy.","correct":false,"explanation":"This confuses bond formation with bond breaking. Energy is required to break existing bonds, whereas making new bonds releases energy and is therefore exothermic."}]},{"id":"3408b2af-ae10-4584-9bd3-2d73ef1bcdb4","cardType":"cloze","points":null,"front":"For the reaction system, exothermic reactions have [ ] ΔH and endothermic reactions have [ ] ΔH.","back":"For the reaction system, exothermic reactions have negative ΔH and endothermic reactions have positive ΔH.","mcqOptions":[{"text":"For the reaction system, exothermic reactions have negative ΔH and endothermic reactions have positive ΔH.","correct":true,"explanation":"An exothermic system loses enthalpy to the surroundings, giving a negative ΔH. An endothermic system gains enthalpy from the surroundings, giving a positive ΔH."},{"text":"For the reaction system, exothermic reactions have positive ΔH and endothermic reactions have negative ΔH.","correct":false,"explanation":"The signs have been reversed. Exothermic reactions release energy and therefore have negative ΔH, while endothermic reactions absorb energy and have positive ΔH."},{"text":"For the reaction system, exothermic reactions have negative ΔH and endothermic reactions have negative ΔH.","correct":false,"explanation":"Only the exothermic sign is correct. Endothermic reactions increase the system’s enthalpy by absorbing energy, so their ΔH is positive rather than negative."},{"text":"For the reaction system, exothermic reactions have positive ΔH and endothermic reactions have positive ΔH.","correct":false,"explanation":"Only the endothermic sign is correct. Exothermic reactions transfer energy out of the system, decreasing its enthalpy and producing a negative ΔH."}]},{"id":"dce47eaf-bb16-44be-a051-517809417368","cardType":"basic","points":1,"front":"What temperature change do the surroundings usually show in an exothermic reaction?","back":"The surroundings show a temperature rise.","mcqOptions":[{"text":"The surroundings show a temperature rise.","correct":true,"explanation":"An exothermic reaction transfers heat from the reaction system to the surroundings. This added thermal energy usually causes the measured temperature of the surroundings to rise."},{"text":"The surroundings show a temperature fall.","correct":false,"explanation":"A temperature fall is normally associated with an endothermic reaction, which absorbs heat. An exothermic reaction releases heat to the surroundings, so their temperature rises."},{"text":"The surroundings show no temperature change.","correct":false,"explanation":"Ignoring heat transfer leads to this misconception. Exothermic reactions release heat into the surroundings, so the surroundings usually show a temperature rise rather than remaining unchanged."},{"text":"The surroundings show a temperature fluctuation.","correct":false,"explanation":"A random fluctuation is not the characteristic thermal change. Because heat is transferred from an exothermic system to its surroundings, the surroundings usually show a temperature rise."}]},{"id":"63eeecf8-d003-4878-b847-21b86defb016","cardType":"basic","points":1,"front":"In an endothermic reaction, which way does heat transfer?","back":"Heat transfers from the surroundings to the system.","mcqOptions":[{"text":"Heat transfers from the surroundings to the system.","correct":true,"explanation":"An endothermic reaction absorbs heat, increasing the energy of the reaction system. The heat therefore flows into the system from its surroundings."},{"text":"Heat transfers from the system to the surroundings.","correct":false,"explanation":"This direction describes an exothermic reaction, where the system releases heat. In an endothermic reaction, heat is absorbed and transfers from the surroundings to the system."},{"text":"Heat transfers from the products to the reactants.","correct":false,"explanation":"Heat transfer is defined across the boundary between the system and surroundings, not simply between products and reactants. In an endothermic reaction, heat enters the system from the surroundings."},{"text":"Heat transfers from the system to the reactants.","correct":false,"explanation":"The reactants are part of the reaction system, so this does not describe heat crossing the system boundary. Endothermic reactions absorb heat from the surroundings into the system."}]},{"id":"8ec6559d-dab0-4396-9856-0f0875f17dea","cardType":"basic","points":2,"front":"How do product enthalpies compare with reactants in reaction profiles?","back":"

Exothermic: products lower.
Endothermic: products higher.

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

Exothermic: products lower.
Endothermic: products higher.

","correct":true,"explanation":"Exothermic reactions release energy, leaving products at lower enthalpy than reactants. Endothermic reactions absorb energy, leaving products at higher enthalpy than reactants."},{"text":"

Exothermic: products higher.
Endothermic: products lower.

","correct":false,"explanation":"Both relationships have been reversed. Exothermic products are lower because energy is released, while endothermic products are higher because energy is absorbed."},{"text":"

Exothermic: products lower.
Endothermic: products lower.

","correct":false,"explanation":"The exothermic relationship is correct, but endothermic products are higher than reactants. Their increased enthalpy reflects the energy absorbed by the reaction system."},{"text":"

Exothermic: products higher.
Endothermic: products higher.

","correct":false,"explanation":"The endothermic relationship is correct, but exothermic products are lower than reactants. Their decreased enthalpy corresponds to energy released to the surroundings."}]},{"id":"82c21ecb-892f-4c0c-b343-b48927ab39c3","cardType":"basic","points":1,"front":"On a reaction profile, what does the vertical difference between reactants and products represent?","back":"The enthalpy change, ΔH.","mcqOptions":[{"text":"The enthalpy change, ΔH.","correct":true,"explanation":"ΔH is the difference between the enthalpy levels of the products and reactants. Its direction and sign show whether the reaction is exothermic or endothermic."},{"text":"The activation energy, Ea.","correct":false,"explanation":"Activation energy is measured from the reactant level to the peak of the profile, not between reactants and products. Their vertical difference is the enthalpy change, ΔH."},{"text":"The mean bond enthalpy.","correct":false,"explanation":"Mean bond enthalpy is the average energy needed to break a particular bond in gaseous molecules. The profile’s product-reactant difference represents the reaction enthalpy change, ΔH."},{"text":"The reaction rate, k.","correct":false,"explanation":"A reaction profile displays energy rather than reaction rate. The vertical separation between reactants and products is the enthalpy change, ΔH, while rate depends on kinetic factors."}]},{"id":"4972ba6d-dc5b-43d3-8877-ab37a001b20c","cardType":"basic","points":1,"front":"On a reaction profile, what does the hump represent?","back":"The activation energy barrier.","mcqOptions":[{"text":"The activation energy barrier.","correct":true,"explanation":"The hump shows the high-energy transition state that must be reached before products can form. The energy needed to rise from the reactants to this peak is the activation energy."},{"text":"The enthalpy change barrier.","correct":false,"explanation":"Enthalpy change is the vertical difference between reactants and products, not the hump. The hump represents the activation energy barrier that must be overcome for reaction."},{"text":"The bond enthalpy barrier.","correct":false,"explanation":"Bond enthalpy refers to the energy required to break a specified bond, usually as a mean value. The profile’s hump represents the overall activation energy barrier."},{"text":"The heat transfer barrier.","correct":false,"explanation":"The hump does not measure heat exchanged with the surroundings. It represents the activation energy needed for reacting particles to reach the transition state."}]},{"id":"e1449e51-df28-48c5-a1f4-1a8849c0d3e8","cardType":"cloze","points":null,"front":"Activation energy, E_a, is measured from the [ ] to the [ ].","back":"Activation energy, E_a, is measured from the reactants to the top of the energy barrier.","mcqOptions":[{"text":"Activation energy, E_a, is measured from the reactants to the top of the energy barrier.","correct":true,"explanation":"The forward activation energy is the energy difference between the reactants and the transition state, represented by the highest point of the energy barrier."},{"text":"Activation energy, E_a, is measured from the products to the top of the energy barrier.","correct":false,"explanation":"Measuring from the products gives the activation energy for the reverse reaction. The forward activation energy is measured from the reactants to the top of the energy barrier."},{"text":"Activation energy, E_a, is measured from the reactants to the products.","correct":false,"explanation":"The difference between reactants and products is the reaction enthalpy change, not the activation energy. Activation energy extends from the reactants to the barrier’s highest point."},{"text":"Activation energy, E_a, is measured from the top of the energy barrier to the reactants.","correct":false,"explanation":"This reverses the required measurement and would give a negative energy difference. Forward activation energy is measured upward from the reactants to the top of the energy barrier."}]},{"id":"f44332bb-9d05-45b2-a25b-2fa7b4b6dbf2","cardType":"basic","points":1,"front":"If a thermochemical equation is doubled, what happens to ΔH?","back":"ΔH is doubled.","mcqOptions":[{"text":"ΔH is doubled.","correct":true,"explanation":"Enthalpy change is proportional to the quantities in a thermochemical equation. Doubling every stoichiometric amount therefore doubles the associated value of ΔH."},{"text":"ΔH is unchanged.","correct":false,"explanation":"Leaving ΔH unchanged ignores its dependence on the quantities reacting. When all coefficients in the thermochemical equation are doubled, ΔH must also be doubled."},{"text":"ΔH is halved.","correct":false,"explanation":"Halving ΔH would correspond to halving the equation’s coefficients, not doubling them. Doubling the stated molar quantities causes ΔH to be doubled."},{"text":"ΔH is sign-reversed.","correct":false,"explanation":"A sign change occurs when a thermochemical equation is reversed, not when its coefficients are scaled. Doubling the equation doubles ΔH without changing its sign."}]},{"id":"e153e7cf-a335-411e-982f-3aa690d526a4","cardType":"basic","points":1,"front":"If a thermochemical equation is reversed, what happens to ΔH?","back":"The sign changes.","mcqOptions":[{"text":"The sign changes.","correct":true,"explanation":"Reversing a reaction interchanges the initial and final states, so the energy transferred has the opposite direction. The magnitude of ΔH remains unchanged, but its sign changes."},{"text":"The sign stays the same.","correct":false,"explanation":"The sign cannot remain unchanged because the direction of energy transfer is reversed. An exothermic forward reaction becomes an endothermic reverse reaction, so the sign changes."},{"text":"The magnitude doubles.","correct":false,"explanation":"Doubling the magnitude applies when all equation coefficients are doubled. Reversing the equation leaves the magnitude unchanged and changes only the sign of ΔH."},{"text":"The value becomes zero.","correct":false,"explanation":"Reversing a reaction does not remove its enthalpy difference. The reverse reaction has the same magnitude of ΔH as the forward reaction, but the opposite sign."}]},{"id":"ed1e1cd6-e851-46c1-a041-88585631805c","cardType":"basic","points":2,"front":"N₂ + 3H₂ → 2NH₃, ΔH = −92.0 kJ mol⁻¹. What is ΔH for making 1 mol NH₃?","back":"−46.0 kJ mol⁻¹.","mcqOptions":[{"text":"−46.0 kJ mol⁻¹.","correct":true,"explanation":"The stated equation produces 2 mol of NH₃ with ΔH = −92.0 kJ mol⁻¹. Halving the equation to produce 1 mol halves ΔH to −46.0 kJ mol⁻¹."},{"text":"+46.0 kJ mol⁻¹.","correct":false,"explanation":"Halving the equation halves the magnitude but does not reverse the reaction, so the sign must remain negative. The correct value for 1 mol of NH₃ is −46.0 kJ mol⁻¹."},{"text":"−92.0 kJ mol⁻¹.","correct":false,"explanation":"This retains the enthalpy change for producing 2 mol of NH₃. Producing only 1 mol requires every coefficient and ΔH to be halved, giving −46.0 kJ mol⁻¹."},{"text":"+92.0 kJ mol⁻¹.","correct":false,"explanation":"This keeps the magnitude for 2 mol and also changes the sign without reversing the reaction. For 1 mol of NH₃, ΔH is −46.0 kJ mol⁻¹."}]},{"id":"2c43b864-4d94-4487-8588-5f63594e12fa","cardType":"cloze","points":null,"front":"Standard conditions are usually [ ] and [ ].","back":"Standard conditions are usually 100 kPa and 298 K.","mcqOptions":[{"text":"Standard conditions are usually 100 kPa and 298 K.","correct":true,"explanation":"Standard enthalpy changes are normally defined with substances at a standard pressure of 100 kPa and a specified temperature, conventionally 298 K."},{"text":"Standard conditions are usually 100 Pa and 298 K.","correct":false,"explanation":"A pressure of 100 Pa is one thousand times too small for the standard state. The usual standard conditions are 100 kPa and 298 K."},{"text":"Standard conditions are usually 100 kPa and 273 K.","correct":false,"explanation":"Although 273 K is associated with the freezing point of water and older definitions of STP, standard enthalpy data usually use 100 kPa and 298 K."},{"text":"Standard conditions are usually 1 kPa and 298 °C.","correct":false,"explanation":"Room pressure is not 1 kPa, and 298 °C confuses kelvin with degrees Celsius. Standard enthalpy conditions are usually 100 kPa and 298 K."}]},{"id":"513655cd-eba3-4ffd-af02-a5d75bd8f10b","cardType":"basic","points":1,"front":"For ΔH°_r, which amounts does the enthalpy change apply to?","back":"The molar quantities shown in the equation.","mcqOptions":[{"text":"The molar quantities shown in the equation.","correct":true,"explanation":"A standard enthalpy change of reaction applies to the reaction occurring exactly as written, including all stoichiometric coefficients and their corresponding molar quantities."},{"text":"The amount of product actually obtained.","correct":false,"explanation":"Reaction enthalpy is defined using the stoichiometric quantities in the equation, not the experimental yield. The actual amount of product obtained may be reduced by incomplete reaction."},{"text":"The one mole of every substance present.","correct":false,"explanation":"Stoichiometric coefficients are not automatically all equal to one. ΔH°_r applies to the particular molar quantities represented by the balanced equation’s coefficients."},{"text":"The one mole of limiting reactant only.","correct":false,"explanation":"The limiting reactant controls experimental yield but does not define the quoted reaction enthalpy. ΔH°_r applies to all molar quantities shown in the equation."}]},{"id":"9ffe107a-097f-49da-a6f4-3e69aed63c11","cardType":"basic","points":1,"front":"In ΔH°_f, what does “1 mole” refer to?","back":"1 mol of the compound formed.","mcqOptions":[{"text":"1 mol of the compound formed.","correct":true,"explanation":"Standard enthalpy of formation is the enthalpy change when exactly 1 mol of a compound is formed from its constituent elements in their standard states."},{"text":"1 mol of the elements consumed.","correct":false,"explanation":"The quantities of constituent elements depend on the compound’s formula and need not total 1 mol. The definition requires 1 mol of the compound to be formed."},{"text":"1 mol of the gaseous product formed.","correct":false,"explanation":"Formation enthalpy is not restricted to gaseous compounds; the product must be in its specified standard state. The 1 mol refers to the compound formed."},{"text":"1 mol of the oxygen gas consumed.","correct":false,"explanation":"Oxygen is not used in every formation reaction, and its coefficient depends on the compound’s composition. The definition refers to 1 mol of the compound formed."}]},{"id":"81d4c73f-129d-4abb-954b-3c6df1efd709","cardType":"basic","points":2,"front":"Write the formation equation for ethanol, C₂H₅OH(l).","back":"2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)","mcqOptions":[{"text":"2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)","correct":true,"explanation":"This forms exactly 1 mol of ethanol from carbon, hydrogen and oxygen in their standard states. The coefficients also balance two carbon, six hydrogen and one oxygen atom."},{"text":"2C(s, diamond) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)","correct":false,"explanation":"Diamond is not carbon’s standard state under standard conditions; graphite must be used in a standard formation equation. All other coefficients remain those needed for 1 mol of ethanol."},{"text":"2C(s, graphite) + 3H₂(g) + O₂(g) → C₂H₅OH(l)","correct":false,"explanation":"Using 1 mol of O₂ supplies two oxygen atoms, but one ethanol molecule contains only one. The balanced formation equation therefore requires ½O₂(g)."},{"text":"2C(s, graphite) + 2H₂(g) + ½O₂(g) → C₂H₅OH(l)","correct":false,"explanation":"This equation is not balanced because its reactants contain four hydrogen atoms while ethanol contains six. Three moles of H₂ are required to form 1 mol of ethanol."}]},{"id":"5c9fdfe6-fdd6-4f2f-a93a-eeff8e4ecb3a","cardType":"basic","points":1,"front":"In ΔH°_c, what does “1 mole” refer to?","back":"1 mol of the substance burned completely in oxygen.","mcqOptions":[{"text":"1 mol of the substance burned completely in oxygen.","correct":true,"explanation":"The one mole in the definition is the amount of the substance being burned completely in oxygen, under the stated standard conditions. It does not refer to one mole of oxygen consumed."},{"text":"1 mol of the oxygen consumed in the reaction.","correct":false,"explanation":"The definition does not count one mole of oxygen, because the oxygen amount varies with the fuel. The correct reference is 1 mol of the substance burned completely in oxygen."},{"text":"1 mol of the carbon dioxide produced in the reaction.","correct":false,"explanation":"Combustion enthalpy is defined per mole of the substance burned, not per mole of carbon dioxide produced. The correct wording refers to 1 mol of the substance burned completely in oxygen."},{"text":"1 mol of the water formed during combustion.","correct":false,"explanation":"The amount of water formed is relevant to neutralisation enthalpy, not combustion enthalpy. For ΔH°c, the correct reference is 1 mol of the substance burned completely in oxygen."}]},{"id":"9383856e-9e93-4c06-b1ee-55ec676c36d8","cardType":"basic","points":1,"front":"In ΔH°_neut, what does “1 mole” refer to?","back":"1 mol of water formed.","mcqOptions":[{"text":"1 mol of water formed.","correct":true,"explanation":"Neutralisation enthalpy is defined per mole of water formed. In a strong acid-strong alkali reaction, H+ reacts with OH− to produce water."},{"text":"1 mol of acid neutralised.","correct":false,"explanation":"One mole of acid is not always equivalent to one mole of water, especially for polyprotic acids. The definition specifically uses 1 mol of water formed."},{"text":"1 mol of alkali neutralised.","correct":false,"explanation":"The alkali concentration and stoichiometry determine how much hydroxide reacts, but the enthalpy is reported per mole of water formed. That is the defined reference amount."},{"text":"1 mol of hydrogen ions reacted.","correct":false,"explanation":"Water is a product of neutralisation, whereas hydrogen ions are reactants. The correct definition expresses the enthalpy change for forming 1 mol of water."}]},{"id":"20d546e2-c264-4414-8716-b8e3ce9a4cbe","cardType":"cloze","points":null,"front":"For strong acid-strong alkali neutralisation, ΔH°_neut is close to [ ] because the ionic equation is [ ].","back":"For strong acid-strong alkali neutralisation, ΔH°_neut is close to −57 kJ mol⁻¹ because the ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).","mcqOptions":[{"text":"For strong acid-strong alkali neutralisation, ΔH°_neut is close to −57 kJ mol⁻¹ because the ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).","correct":true,"explanation":"Strong acids and alkalis fully dissociate, so every neutralisation involves H+ combining with OH− to form liquid water. This gives a value close to −57 kJ mol−1."},{"text":"For strong acid-strong alkali neutralisation, ΔH°_neut is close to +57 kJ mol⁻¹ because the ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).","correct":false,"explanation":"The positive sign incorrectly indicates that heat is absorbed. Neutralisation is exothermic, and the correct ionic equation is H+ + OH− → H2O(l), giving about −57 kJ mol−1."},{"text":"For strong acid-strong alkali neutralisation, ΔH°_neut is close to −286 kJ mol⁻¹ because the ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).","correct":false,"explanation":"The magnitude is often confused with the enthalpy of forming water vapour, but neutralisation forms liquid water. The correct value is about −57 kJ mol−1 for H+ + OH− → H2O(l)."},{"text":"For strong acid-strong alkali neutralisation, ΔH°_neut is close to −57 kJ mol⁻¹ because the ionic equation is Na⁺(aq) + Cl⁻(aq) → NaCl(aq).","correct":false,"explanation":"Including spectator ions such as Na+ and Cl− obscures the net ionic reaction. The correct ionic equation contains only H+ and OH−, with a value close to −57 kJ mol−1."}]},{"id":"31fdae24-3b03-4ff8-a05b-92e26d67d315","cardType":"cloze","points":null,"front":"The heat transferred in solution calorimetry is calculated using [ ].","back":"The heat transferred in solution calorimetry is calculated using $q = mc\\Delta T$.","mcqOptions":[{"text":"$$q = mc\\Delta T$","correct":true,"explanation":"The expression q = mcΔT calculates the thermal energy transferred using the solution’s mass, specific heat capacity and temperature change. Consistent units give q in joules."},{"text":"$$q = m\\Delta T$","correct":false,"explanation":"Using q = mΔT omits the specific heat capacity, so it cannot represent the energy needed per gram per degree. The correct calorimetry expression is q = mcΔT."},{"text":"$$q = c\\Delta T$","correct":false,"explanation":"Using q = cΔT omits the mass, although the energy transferred depends on how much solution is heated. The correct expression includes both m and c: q = mcΔT."},{"text":"$$q = \\frac{mc}{\\Delta T}$","correct":false,"explanation":"Multiplying by the reciprocal of the temperature change gives the wrong dependence on temperature. Heat transferred is directly proportional to ΔT, so q = mcΔT."}]},{"id":"e7990ad9-f763-4add-89aa-879d4a349247","cardType":"basic","points":1,"front":"What value of c is normally used for aqueous solutions in calorimetry?","back":"4.18 J g⁻¹ K⁻¹.","mcqOptions":[{"text":"4.18 J g⁻¹ K⁻¹.","correct":true,"explanation":"Aqueous solutions are usually approximated as having the same specific heat capacity as water, 4.18 J g−1 K−1. This value links energy transfer to mass and temperature change."},{"text":"8.36 J g⁻¹ K⁻¹.","correct":false,"explanation":"4.18 is the accepted approximate value for water, whereas 8.36 doubles the heat capacity and would overestimate the energy transferred. The standard value is 4.18 J g−1 K−1."},{"text":"0.418 J g⁻¹ K⁻¹.","correct":false,"explanation":"This value is close to the specific heat capacity of some metals, not aqueous solutions. Calorimetry normally treats dilute aqueous solution as water, using 4.18 J g−1 K−1."},{"text":"4.18 J kg⁻¹ K⁻¹.","correct":false,"explanation":"The units have the wrong mass scale for the standard aqueous approximation. The value used is 4.18 J g−1 K−1, with mass measured in grams."}]},{"id":"4e2b7027-a879-40e1-bcfd-f3bed7ecc78c","cardType":"basic","points":1,"front":"How is mass estimated from volume for dilute aqueous solutions?","back":"Use density = 1.00 g cm⁻³, so cm³ ≈ g.","mcqOptions":[{"text":"Use density = 1.00 g cm⁻³, so cm³ ≈ g.","correct":true,"explanation":"For dilute aqueous solutions, density is approximated as 1.00 g cm−3. Therefore, a measured volume in cm3 has approximately the same numerical value as its mass in grams."},{"text":"Use density = 1.00 g dm⁻³, so cm³ ≈ g.","correct":false,"explanation":"A density of 1.00 g dm−3 is one thousand times too small for water and mixes the units incorrectly. The correct approximation is 1.00 g cm−3, so cm3 ≈ g."},{"text":"Use density = 0.100 g cm⁻³, so cm³ ≈ g.","correct":false,"explanation":"Water is not approximated as having a density of 0.100 g cm−3; that would make the estimated mass ten times too small. Use 1.00 g cm−3 instead."},{"text":"Use density = 100 g cm⁻³, so cm³ ≈ g.","correct":false,"explanation":"Using 100 g cm−3 makes the estimated mass one hundred times too large. Dilute aqueous solutions are treated as having density 1.00 g cm−3, giving cm3 approximately equal to grams."}]},{"id":"0b2ad526-69bf-4cb9-b324-cdd19a5382f1","cardType":"basic","points":1,"front":"In n = cV, how must cm³ be converted when concentration is mol dm⁻³?","back":"Convert to dm³ by dividing by 1000.","mcqOptions":[{"text":"Convert to dm³ by dividing by 1000.","correct":true,"explanation":"Concentration in mol dm−3 must be multiplied by volume in dm3. Since 1 dm3 equals 1000 cm3, convert a cm3 volume by dividing by 1000."},{"text":"Convert to dm³ by multiplying by 1000.","correct":false,"explanation":"Multiplying cm3 by 1000 makes the numerical volume larger instead of converting it to the smaller dm3 value. The required conversion is division by 1000."},{"text":"Convert to cm³ by dividing by 1000.","correct":false,"explanation":"Converting to litres is not the stated unit conversion for concentration in mol dm−3, and multiplying by 1000 gives the wrong numerical direction. Divide by 1000 to obtain dm3."},{"text":"Convert to dm³ by dividing by 100.","correct":false,"explanation":"Dividing by 100 changes cm3 to an incorrect intermediate scale. Because 1 dm3 contains 1000 cm3, divide the volume by 1000 before using n = cV."}]},{"id":"51315564-c5dc-49e1-81bc-652ad8cecf56","cardType":"cloze","points":null,"front":"In calorimetry, [ ] because the solution is the [ ].","back":"In calorimetry, q_reaction = −q_solution because the solution is the surroundings.","mcqOptions":[{"text":"In calorimetry, q_reaction = −q_solution because the solution is the surroundings.","correct":true,"explanation":"The reaction and solution exchange equal amounts of energy in opposite directions, so their heat changes have opposite signs. The solution is the surroundings, hence qreaction = −qsolution."},{"text":"In calorimetry, q_reaction = q_solution because the solution is the surroundings.","correct":false,"explanation":"The reaction and solution do not have the same signed heat change: one loses exactly what the other gains. The solution is the surroundings, so qreaction = −qsolution."},{"text":"In calorimetry, q_reaction = −q_solution because the solution is the system.","correct":false,"explanation":"The reaction is the system, not the surroundings, while the solution surrounding it absorbs or releases the measured heat. Therefore qreaction = −qsolution because the solution is the surroundings."},{"text":"In calorimetry, q_reaction = +q_solution because the solution is the system.","correct":false,"explanation":"The minus sign is essential because energy transferred from the reaction to the solution is exothermic for the reaction but positive for the solution. The solution is the surroundings."}]},{"id":"5f2652a6-380b-4224-9af2-41b65064dc75","cardType":"basic","points":2,"front":"In solution calorimetry, what effect does heat loss have on an exothermic ΔH value?","back":"

ΔT is too small.
ΔH is less exothermic than the true value.

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

ΔT is too small.
ΔH is less exothermic than the true value.

","correct":true,"explanation":"Heat escaping to the surroundings reduces the measured temperature rise, so the calculated heat released is too small. The resulting ΔH is less negative, or less exothermic, than the true value."},{"text":"

ΔT is too large.
ΔH is more exothermic than the true value.

","correct":false,"explanation":"Heat loss does not increase the measured temperature rise. It makes ΔT too small, so the calculated heat released is underestimated and ΔH is less exothermic than the true value."},{"text":"

ΔT is too small.
ΔH is more exothermic than the true value.

","correct":false,"explanation":"Although heat loss makes ΔT too small, it reduces the calculated magnitude of the heat released rather than increasing it. Therefore, ΔH is less exothermic than the true value."},{"text":"

ΔT is too large.
ΔH is less exothermic than the true value.

","correct":false,"explanation":"This reverses the effect on temperature: heat escaping from the calorimeter makes the measured ΔT too small. Consequently, the calculated ΔH is less exothermic than the true value."}]}]}],"examCountdown":{"examDate":"2027-05-11","paperLabel":"Paper 1"},"heading":"Enthalpy changes","title":"Enthalpy changes Flashcards","topbar":{"backHref":"/as-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"1bbf2b57-6ecc-590f-ba5a-1b868b5a58e5","walled":"$L38"}],"$L39","$L3a"]