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What type of reaction is it?","back":"Exothermic.","mcqOptions":[{"text":"Exothermic.","correct":true,"explanation":"A reaction that warms its surroundings releases energy to them, so the reaction is classified as exothermic."},{"text":"Endothermic.","correct":false,"explanation":"A colder mixture indicates that the reaction absorbs energy from its surroundings. The correct classification is exothermic because the mixture becomes warmer."},{"text":"Combustion.","correct":false,"explanation":"Combustion is often exothermic, but it is not the general classification required here. A reaction mixture that gets warmer is exothermic."},{"text":"Neutral.","correct":false,"explanation":"A temperature rise does not show that the reaction is neutral. The correct term is exothermic because energy is released, warming the mixture."}]},{"id":"b24bba87-aeb5-4f60-b674-c3464e2b0800","cardType":"basic","points":1,"front":"A reaction mixture gets colder. What type of reaction is it?","back":"Endothermic.","mcqOptions":[{"text":"Endothermic.","correct":true,"explanation":"A reaction that cools its surroundings absorbs energy from them, so the reaction is classified as endothermic."},{"text":"Exothermic.","correct":false,"explanation":"A warmer mixture indicates that energy is released to the surroundings. The correct classification is endothermic because this mixture becomes colder."},{"text":"Freezing.","correct":false,"explanation":"Freezing can be exothermic rather than a general classification for a cooling reaction. The correct term here is endothermic because energy is absorbed."},{"text":"Neutral.","correct":false,"explanation":"A temperature fall does not make the reaction neutral. The correct term is endothermic because the reaction takes in energy from its surroundings."}]},{"id":"0a7b59ee-5830-470d-8de8-9baa2ee0f1a5","cardType":"cloze","points":null,"front":"Calorimetry estimates heat energy changes by measuring the [ ] of water or a solution.","back":"Calorimetry estimates heat energy changes by measuring the temperature change of water or a solution.","mcqOptions":[{"text":"temperature change","correct":true,"explanation":"Calorimetry determines heat transfer from the water’s temperature change, together with its mass and specific heat capacity, using the measured temperature difference."},{"text":"mass change","correct":false,"explanation":"Mass is needed in the heat-energy calculation, but it is not the change measured to estimate the energy transfer. The correct measurement is temperature change."},{"text":"volume change","correct":false,"explanation":"The solution’s volume may help determine its mass, but calorimetry directly measures how its temperature changes. The correct term is temperature change."},{"text":"heat capacity","correct":false,"explanation":"Heat capacity is a property used in interpreting results, not the quantity directly measured during the experiment. The correct term is temperature change."}]},{"id":"141a9e69-fc5e-494d-9dfe-f9d4cd089e32","cardType":"basic","points":2,"front":"Why use an insulated polystyrene cup in solution calorimetry?","back":"It reduces heat transfer to the room, making ΔT closer to the true value.","mcqOptions":[{"text":"It reduces heat transfer to the room, making ΔT closer to the true value.","correct":true,"explanation":"Polystyrene is a good thermal insulator, so it reduces energy lost to the surroundings and makes the measured temperature change closer to the true value."},{"text":"It increases heat transfer to the room, making ΔT closer to the true value.","correct":false,"explanation":"Allowing more heat into the room would increase experimental energy loss and make the measured ΔT too small. Insulation reduces heat transfer to the room."},{"text":"It prevents all heat transfer to the room, making ΔT equal to the true value.","correct":false,"explanation":"The cup does not prevent heat transfer completely, and claiming that it does overstates its insulating effect. It reduces heat transfer to the room."},{"text":"It increases heat transfer from the room, making ΔT closer to the true value.","correct":false,"explanation":"A polystyrene cup is chosen for insulation, not because it conducts heat rapidly. Its purpose is to reduce heat transfer to the room."}]},{"id":"b4709a9a-b983-45c1-8ae5-692cae335953","cardType":"basic","points":1,"front":"In solution calorimetry, what temperature reading is used after stirring?","back":"The highest or lowest temperature reached.","mcqOptions":[{"text":"The highest or lowest temperature reached.","correct":true,"explanation":"Stirring makes the solution more uniform, so the relevant reading is the extreme temperature reached: the highest for warming or the lowest for cooling."},{"text":"The final temperature after stirring.","correct":false,"explanation":"The final temperature may miss the true extreme because heat can begin escaping before the reading is taken. The highest or lowest temperature reached is used."},{"text":"The initial temperature before stirring.","correct":false,"explanation":"The initial temperature provides the baseline but not the temperature change caused by the reaction. The highest or lowest temperature reached is required."},{"text":"The average temperature during stirring.","correct":false,"explanation":"An average temperature can hide the maximum warming or cooling and gives an inaccurate ΔT. The highest or lowest temperature reached is used instead."}]},{"id":"05cedfe2-33d0-4e86-8821-76ada851c0a0","cardType":"basic","points":2,"front":"How is the mass of fuel burned found in combustion calorimetry?","back":"Initial burner mass - final burner mass.","mcqOptions":[{"text":"Initial burner mass - final burner mass.","correct":true,"explanation":"The fuel mass burned is the decrease in burner mass, so subtracting the final mass from the initial mass gives the amount consumed."},{"text":"Final burner mass - initial burner mass.","correct":false,"explanation":"Reversing the subtraction gives a negative value because the burner becomes lighter. The mass burned is initial burner mass minus final burner mass."},{"text":"Initial burner mass + final burner mass.","correct":false,"explanation":"Adding the readings includes the burner mass twice rather than finding its decrease. The correct calculation subtracts final burner mass from initial burner mass."},{"text":"Final burner mass.","correct":false,"explanation":"The final mass alone is the remaining burner and fuel mass, not the fuel consumed. The burned mass is found from initial minus final burner mass."}]},{"id":"c394aa69-5fb2-4844-a839-3bbb9d597bff","cardType":"basic","points":1,"front":"In combustion calorimetry, what effect does heat escaping around the calorimeter have?","back":"The calculated energy released is too small.","mcqOptions":[{"text":"The calculated energy released is too small.","correct":true,"explanation":"Escaped heat is not included in the measured temperature rise, so the calculation underestimates the energy released by the fuel."},{"text":"The calculated energy released is too large.","correct":false,"explanation":"Heat escaping means some released energy is missed, not added to the measured value. Therefore, the calculated energy released is too small."},{"text":"The calculated energy released is unchanged.","correct":false,"explanation":"Escaping heat reduces the observed temperature change, so it does affect the result. The calculated energy released is too small."},{"text":"The calculated energy released is negative.","correct":false,"explanation":"Heat loss does not reverse the energy release; it makes less of that energy heat the water. The calculated energy released is too small."}]},{"id":"f5b4ab2a-c67c-4850-b96e-3c8ae8bc0b88","cardType":"cloze","points":null,"front":"Use [ ] to calculate heat energy.","back":"Use $Q=mc\\Delta T$ to calculate heat energy.","mcqOptions":[{"text":"$$Q=mc\\Delta T$","correct":true,"explanation":"The equation combines mass, specific heat capacity and temperature change to calculate the heat energy transferred: $Q=mc\\Delta T$."},{"text":"$$Q=\\frac{mc}{\\Delta T}$","correct":false,"explanation":"Dividing by temperature change reverses the relationship and gives the wrong quantity. Heat energy is calculated by multiplying mass, specific heat capacity and temperature change: $Q=mc\\Delta T$."},{"text":"$$Q=\\frac{m\\Delta T}{c}$","correct":false,"explanation":"Dividing by specific heat capacity is an algebraic error. The correct equation multiplies all three quantities: $Q=mc\\Delta T$."},{"text":"$$Q=m+c+\\Delta T$","correct":false,"explanation":"Adding the quantities ignores the multiplicative relationship in the heat equation. The correct formula is $Q=mc\\Delta T$."}]},{"id":"ab18bd31-b25b-4a46-9678-e0d2c7311416","cardType":"basic","points":1,"front":"For water and dilute solutions, what value of $c$ is used in $Q=mc\\Delta T$?","back":"4.18 J/g °C.","mcqOptions":[{"text":"4.18 J/g °C.","correct":true,"explanation":"Water has a specific heat capacity of approximately 4.18 J/g °C, and dilute aqueous solutions are assumed to have the same value in calorimetry calculations."},{"text":"1.00 J/g °C.","correct":false,"explanation":"This is close to water’s value in J/g K but is numerically too small. The standard value used for water and dilute solutions is 4.18 J/g °C."},{"text":"0.90 J/g °C.","correct":false,"explanation":"This confuses the specific heat capacity of water with a lower value typical of some metals. For dilute aqueous solutions, use 4.18 J/g °C."},{"text":"41.8 J/g °C.","correct":false,"explanation":"This is ten times the accepted value, usually caused by a decimal-place error. Water and dilute solutions use a specific heat capacity of 4.18 J/g °C."}]},{"id":"b364e23b-213e-4480-835d-e5699eb69f86","cardType":"cloze","points":null,"front":"For dilute aqueous solutions, assume density [ ], so 50 cm³ has mass [ ].","back":"For dilute aqueous solutions, assume density 1 g/cm³, so 50 cm³ has mass 50 g.","mcqOptions":[{"text":"For dilute aqueous solutions, assume density 1 g/cm³, so 50 cm³ has mass 50 g.","correct":true,"explanation":"Dilute aqueous solutions are treated like water, whose density is approximately 1 g/cm³. Therefore, 50 cm³ is assumed to have a mass of 50 g."},{"text":"For dilute aqueous solutions, assume density 0.5 g/cm³, so 50 cm³ has mass 25 g.","correct":false,"explanation":"Using 0.5 g/cm³ incorrectly halves the assumed density and mass. Dilute aqueous solutions use 1 g/cm³, making the mass of 50 cm³ equal to 50 g."},{"text":"For dilute aqueous solutions, assume density 2 g/cm³, so 50 cm³ has mass 100 g.","correct":false,"explanation":"This treats each cubic centimetre as two grams, doubling the mass. The assumed density is 1 g/cm³, so 50 cm³ corresponds to 50 g."},{"text":"For dilute aqueous solutions, assume density 1 g/cm³, so 50 cm³ has mass 500 g.","correct":false,"explanation":"The density assumption is correct, but the volume-to-mass conversion has gained an extra zero. At 1 g/cm³, 50 cm³ has a mass of 50 g."}]},{"id":"0116d6cc-b929-42d9-967a-0e4032e9ab52","cardType":"basic","points":1,"front":"Which mass should be used in $Q=mc\\Delta T$ for solution calorimetry?","back":"The mass of the water or solution being heated.","mcqOptions":[{"text":"The mass of the water or solution being heated.","correct":true,"explanation":"In solution calorimetry, $Q=mc\\Delta T$ calculates the energy transferred to or from the material whose temperature changes, namely the water or solution being heated."},{"text":"The mass of the solute dissolved in the water.","correct":false,"explanation":"Using only the solute mass ignores most of the material absorbing the heat. The required mass is the total water or solution whose temperature change is measured."},{"text":"The mass of the reactant used in the reaction.","correct":false,"explanation":"The reacting substance’s mass is not necessarily the mass undergoing the measured temperature change. Use the mass of the water or solution being heated in the calorimeter."},{"text":"The mass of the calorimeter containing the solution.","correct":false,"explanation":"The calorimeter’s mass is excluded when the calculation assumes the solution alone absorbs the heat. Use the mass of the water or solution being heated."}]},{"id":"17cf8ff0-0364-493b-a6ee-c6679f4dd69e","cardType":"basic","points":2,"front":"100 g of solution warms by 6.8 °C. Using c = 4.18 J/g °C, calculate $Q$ in kJ.","back":"2.84 kJ.","mcqOptions":[{"text":"2.84 kJ.","correct":true,"explanation":"Using $Q=mc\\Delta T$ gives $100\\times4.18\\times6.8=2842.4$ J. Dividing by 1000 converts this to 2.84 kJ to three significant figures."},{"text":"28.4 kJ.","correct":false,"explanation":"This is ten times too large, indicating a decimal-place slip after calculating the energy. The result is 2842.4 J, which converts to 2.84 kJ."},{"text":"0.284 kJ.","correct":false,"explanation":"This is ten times too small, usually from an unnecessary extra division during unit conversion. The calculated 2842.4 J is equal to 2.84 kJ."},{"text":"284 kJ.","correct":false,"explanation":"This fails to convert joules to kilojoules correctly and leaves a value one hundred times too large. Dividing 2842.4 J by 1000 gives 2.84 kJ."}]},{"id":"9ea5984a-5e57-4786-a601-930157451897","cardType":"cloze","points":null,"front":"Molar enthalpy change, $\\Delta H$, is heat energy change per [ ] and is usually measured in [ ].","back":"Molar enthalpy change, $\\Delta H$, is heat energy change per mole and is usually measured in kJ/mol.","mcqOptions":[{"text":"Molar enthalpy change, $\\Delta H$, is heat energy change per mole and is usually measured in kJ/mol.","correct":true,"explanation":"A molar enthalpy change states the energy change associated with one mole of a specified substance or reaction, so its standard unit is kilojoules per mole."},{"text":"Molar enthalpy change, $\\Delta H$, is heat energy change per gram and is usually measured in kJ/mol.","correct":false,"explanation":"This confuses molar enthalpy with an energy change per unit mass. Molar enthalpy is defined per mole and is therefore usually expressed in kJ/mol."},{"text":"Molar enthalpy change, $\\Delta H$, is heat energy change per mole and is usually measured in kJ.","correct":false,"explanation":"Kilojoules alone measure an amount of energy, not energy per amount of substance. Molar enthalpy is energy per mole and is usually measured in kJ/mol."},{"text":"Molar enthalpy change, $\\Delta H$, is heat energy change per gram and is usually measured in kJ/g.","correct":false,"explanation":"Energy per gram is a specific energy change rather than a molar enthalpy change. Molar enthalpy is defined per mole and is usually measured in kJ/mol."}]},{"id":"63fe605e-fdd1-4ed6-a5d4-97259c04bf26","cardType":"basic","points":1,"front":"A reaction mixture warms up. What sign does $\\Delta H$ have?","back":"Negative.","mcqOptions":[{"text":"Negative.","correct":true,"explanation":"A temperature rise means the reaction releases heat to the surroundings, so it is exothermic. By convention, an exothermic reaction has a negative enthalpy change."},{"text":"Positive.","correct":false,"explanation":"A positive enthalpy change describes an endothermic reaction that absorbs heat. Since the mixture warms, the reaction released heat and therefore has a negative enthalpy change."},{"text":"Zero.","correct":false,"explanation":"A zero enthalpy change would mean no net heat energy change. The observed warming shows that heat was released, so the enthalpy change is negative."},{"text":"Undefined.","correct":false,"explanation":"The sign can be determined from the temperature change: warming indicates heat release. The reaction is exothermic, so its enthalpy change is negative rather than undefined."}]},{"id":"078a10df-3ac4-43a3-b731-ce653d69c9ea","cardType":"basic","points":1,"front":"For an exothermic solution reaction, what formula links $\\Delta H$, $Q$ and $n$?","back":"$\\Delta H=-\\frac{Q}{n}$.","mcqOptions":[{"text":"$\\Delta H=-\\frac{Q}{n}$.","correct":true,"explanation":"For an exothermic reaction, the positive heat gained by the solution corresponds to heat lost by the reaction. Dividing by moles and reversing the sign gives $\\Delta H=-\\frac{Q}{n}$."},{"text":"$\\Delta H=\\frac{Q}{n}$.","correct":false,"explanation":"Omitting the minus sign would assign a positive enthalpy change to heat released by an exothermic reaction. The correct relationship is $\\Delta H=-\\frac{Q}{n}$."},{"text":"$\\Delta H=-\\frac{n}{Q}$.","correct":false,"explanation":"This inverts the required division and gives inappropriate units of mol/kJ. Molar enthalpy is energy per mole, so the correct formula is $\\Delta H=-\\frac{Q}{n}$."},{"text":"$\\Delta H=-Qn$.","correct":false,"explanation":"Multiplying energy by amount does not produce energy per mole. Molar enthalpy requires division by the number of moles, with a negative sign for an exothermic reaction."}]},{"id":"855a21ba-88c7-4361-a6f8-7854e86b1056","cardType":"basic","points":2,"front":"$$Q=2.84$ kJ is released by 0.0500 mol in a neutralisation. Calculate $\\Delta H$.","back":"-56.8 kJ/mol.","mcqOptions":[{"text":"-56.8 kJ/mol.","correct":true,"explanation":"Dividing the released energy by the amount gives $2.84\\div0.0500=56.8$ kJ/mol. Neutralisation releases heat, so the enthalpy change is -56.8 kJ/mol."},{"text":"+56.8 kJ/mol.","correct":false,"explanation":"The magnitude is calculated correctly, but the sign is wrong. Because the neutralisation releases heat, it is exothermic and its enthalpy change is -56.8 kJ/mol."},{"text":"-0.142 kJ/mol.","correct":false,"explanation":"This multiplies $Q$ by the number of moles instead of dividing, producing 0.142 rather than a molar value. The correct calculation gives -56.8 kJ/mol."},{"text":"-568 kJ/mol.","correct":false,"explanation":"This is ten times too large because of a decimal-place error when dividing by 0.0500 mol. The correct exothermic molar enthalpy is -56.8 kJ/mol."}]},{"id":"006e83f0-c576-42c3-a974-033d327eb6a5","cardType":"cloze","points":null,"front":"In an exothermic energy level diagram, reactants are [ ] than products and $\\Delta H$ is [ ].","back":"In an exothermic energy level diagram, reactants are higher than products and $\\Delta H$ is negative.","mcqOptions":[{"text":"In an exothermic energy level diagram, reactants are higher than products and $\\Delta H$ is negative.","correct":true,"explanation":"Exothermic reactions transfer energy to the surroundings, so the reactants have more energy than the products. The decrease in enthalpy makes ΔH negative."},{"text":"In an exothermic energy level diagram, reactants are lower than products and $\\Delta H$ is negative.","correct":false,"explanation":"Placing the reactants lower reverses the energy levels while retaining the exothermic sign. In an exothermic reaction, reactants are higher than products and ΔH is negative."},{"text":"In an exothermic energy level diagram, reactants are higher than products and $\\Delta H$ is positive.","correct":false,"explanation":"A positive ΔH describes an endothermic reaction, not an exothermic one. Reactants are higher than products in an exothermic diagram, and the resulting ΔH is negative."},{"text":"In an exothermic energy level diagram, reactants are lower than products and $\\Delta H$ is positive.","correct":false,"explanation":"These are the energy levels and sign for an endothermic reaction. An exothermic reaction has higher-energy reactants, lower-energy products and a negative ΔH."}]},{"id":"56144a7e-a0ef-48a5-845a-ba6ca4e4c11c","cardType":"cloze","points":null,"front":"In an endothermic energy level diagram, products are [ ] than reactants and $\\Delta H$ is [ ].","back":"In an endothermic energy level diagram, products are higher than reactants and $\\Delta H$ is positive.","mcqOptions":[{"text":"In an endothermic energy level diagram, products are higher than reactants and $\\Delta H$ is positive.","correct":true,"explanation":"Endothermic reactions take in energy from the surroundings, leaving the products at a higher energy level than the reactants. This enthalpy increase gives a positive ΔH."},{"text":"In an endothermic energy level diagram, products are lower than reactants and $\\Delta H$ is positive.","correct":false,"explanation":"Placing products below reactants describes an energy decrease, which is not endothermic. Endothermic products are higher than reactants, and the enthalpy increase makes ΔH positive."},{"text":"In an endothermic energy level diagram, products are higher than reactants and $\\Delta H$ is negative.","correct":false,"explanation":"The energy levels correctly show an endothermic change, but the sign has been reversed. Products are higher than reactants, so the enthalpy change is positive."},{"text":"In an endothermic energy level diagram, products are lower than reactants and $\\Delta H$ is negative.","correct":false,"explanation":"These are the energy levels and sign for an exothermic reaction. In an endothermic reaction, products are higher than reactants and ΔH is positive."}]},{"id":"92329cac-65a1-42bb-85b4-a1ab574170bf","cardType":"basic","points":1,"front":"On an energy level diagram, what does the vertical gap between reactants and products show?","back":"The enthalpy change, ΔH.","mcqOptions":[{"text":"The enthalpy change, ΔH.","correct":true,"explanation":"The vertical separation measures the energy difference between the reactants and products. This difference is the reaction’s overall enthalpy change, represented by ΔH."},{"text":"The activation energy, Ea.","correct":false,"explanation":"Activation energy is measured from the reactants’ level to the top of the hump, not between reactants and products. The vertical gap between those levels is the enthalpy change, ΔH."},{"text":"The average bond energy.","correct":false,"explanation":"Bond energy refers to the energy needed to break a particular bond and is not represented directly by this gap. The reactant-product gap shows the enthalpy change, ΔH."},{"text":"The rate of reaction.","correct":false,"explanation":"Reaction rate describes how quickly reactants become products and cannot be read from a vertical energy difference. The gap between reactants and products represents the enthalpy change, ΔH."}]},{"id":"a139bf68-3e18-4137-921c-5bb7edf10b4b","cardType":"basic","points":1,"front":"On an energy level diagram, what does the hump show?","back":"The activation energy.","mcqOptions":[{"text":"The activation energy.","correct":true,"explanation":"The hump rises from the reactants’ energy level to the transition state. This energy barrier is the activation energy that reacting particles must overcome."},{"text":"The enthalpy change.","correct":false,"explanation":"The enthalpy change is the vertical difference between reactant and product energy levels, not the hump. The hump represents the activation energy barrier."},{"text":"The average bond energy.","correct":false,"explanation":"Average bond energy is a measured value for breaking a particular type of bond, rather than the diagram’s hump. The hump shows the activation energy."},{"text":"The energy released.","correct":false,"explanation":"The overall energy released is related to a downward enthalpy change between reactants and products. The hump instead shows the activation energy needed to begin the reaction."}]},{"id":"65a95381-cc45-48d9-97a5-6a50157814d6","cardType":"basic","points":2,"front":"Why is bond-breaking endothermic?","back":"Energy is needed to break attractions between atoms.","mcqOptions":[{"text":"Energy is needed to break attractions between atoms.","correct":true,"explanation":"A chemical bond is an electrostatic attraction between atoms. Supplying energy is necessary to overcome this attraction and separate the bonded atoms, so bond-breaking is endothermic."},{"text":"Energy is released as attractions form between atoms.","correct":false,"explanation":"Breaking a bond does not form an attraction; it overcomes an existing one. Energy must be supplied to separate the attracted atoms, which makes bond-breaking endothermic."},{"text":"Energy is released as atoms separate from each other.","correct":false,"explanation":"Atoms do not release energy simply because their separation increases. Separating bonded atoms requires an input of energy to overcome their mutual attraction, making bond-breaking endothermic."},{"text":"Energy is needed to increase attractions between atoms.","correct":false,"explanation":"Bond-breaking involves overcoming attractions rather than increasing them. Energy must be absorbed to separate bonded atoms against their attraction, which is why the process is endothermic."}]},{"id":"81393de3-0243-4142-9232-be5789cfa694","cardType":"basic","points":2,"front":"Why is bond-making exothermic?","back":"Energy is released when new attractions form.","mcqOptions":[{"text":"Energy is released when new attractions form.","correct":true,"explanation":"When atoms form a bond, they move into a more stable, lower-energy arrangement. The lost energy is transferred to the surroundings, making bond-making exothermic."},{"text":"Energy is absorbed when old attractions break.","correct":false,"explanation":"Energy is required when attractions are broken, not when they form. Forming new bonds lowers the atoms’ energy and releases energy to the surroundings."},{"text":"Energy is absorbed when new attractions form.","correct":false,"explanation":"Forming an attraction lowers the system’s energy rather than raising it. The energy lost by the bonding atoms is released to the surroundings, so bond-making is exothermic."},{"text":"Energy is released when old attractions break.","correct":false,"explanation":"Breaking existing attractions requires an energy input and is endothermic, so it cannot explain exothermic bond-making. Energy is released specifically when new attractions form."}]},{"id":"2466785a-0e36-42d0-bd3b-569067e227a4","cardType":"cloze","points":null,"front":"For bond energies, use [ ]: $\\Delta H=\\text{energy to break bonds}-\\text{energy released making bonds}$.","back":"For bond energies, use broken minus made: $\\Delta H=\\text{energy to break bonds}-\\text{energy released making bonds}$.","mcqOptions":[{"text":"broken minus made","correct":true,"explanation":"Bond breaking requires energy, while bond making releases energy. Subtracting the energy released in making bonds from the energy required to break bonds gives ΔH."},{"text":"made minus broken","correct":false,"explanation":"Reversing the subtraction reverses the sign of ΔH, incorrectly swapping exothermic and endothermic results. Bond-energy calculations use energy for bonds broken minus energy for bonds made."},{"text":"broken plus made","correct":false,"explanation":"Adding both quantities ignores that bond making releases energy and therefore contributes with the opposite sign. The required rule is broken minus made."},{"text":"made plus broken","correct":false,"explanation":"This both adds the energy quantities and places bond making first, so it cannot give the net enthalpy change. The correct bond-energy rule is broken minus made."}]},{"id":"ff922df3-7f90-40ff-9fb2-0c62a27f0b64","cardType":"basic","points":3,"front":"Use H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol and H-Cl = 431 kJ/mol. Calculate ΔH for $H_2+Cl_2\\rightarrow2HCl$.","back":"

Broken: 436 + 242 = 678 kJ/mol
Made: 2 × 431 = 862 kJ/mol
ΔH: 678 - 862 = -184 kJ/mol

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

Broken: 436 + 242 = 678 kJ/mol
Made: 2 × 431 = 862 kJ/mol
ΔH: 678 - 862 = -184 kJ/mol

","correct":true,"explanation":"One H-H bond and one Cl-Cl bond are broken, requiring 678 kJ/mol. Two H-Cl bonds release 862 kJ/mol, so ΔH = 678 - 862 = -184 kJ/mol."},{"text":"

Broken: 436 + 242 = 678 kJ/mol
Made: 2 × 431 = 862 kJ/mol
ΔH: 862 - 678 = +184 kJ/mol

","correct":false,"explanation":"The subtraction has been reversed, producing the wrong sign. Using broken minus made gives 678 - 862 = -184 kJ/mol, consistent with this exothermic reaction."},{"text":"

Broken: 436 + 242 = 678 kJ/mol
Made: 1 × 431 = 431 kJ/mol
ΔH: 678 - 431 = +247 kJ/mol

","correct":false,"explanation":"This forgets that the equation forms two H-Cl bonds, so the energy released must be doubled. The made energy is 862 kJ/mol and ΔH is -184 kJ/mol."},{"text":"

Broken: 436 + 2 × 242 = 920 kJ/mol
Made: 2 × 431 = 862 kJ/mol
ΔH: 920 - 862 = +58 kJ/mol

","correct":false,"explanation":"This incorrectly doubles the Cl-Cl bond energy even though only one Cl-Cl bond is broken. The broken total is 678 kJ/mol, giving ΔH = -184 kJ/mol."}]}]}],"examCountdown":{"examDate":"2027-05-19","paperLabel":"Paper 1"},"heading":"Energetics","title":"Energetics Flashcards","topbar":{"backHref":"/igcse/chemistry/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"8aa3440e-675a-51a9-8944-b2635e7e4126","walled":"$L38"}],"$L39","$L3a"]