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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/chemistry/aqa/enthalpy-change/flashcards","label":"Enthalpy change","subjectCode":"a_level_chemistry__aqa","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"64dd3125-0bdc-4f2b-9b97-3289a5c9d072","title":"Enthalpy change Flashcards","topicId":"1f9e615e-7af8-5aa7-913e-c50d03cbd32b","topicSlug":"enthalpy-change","cards":[{"id":"ba8a8da2-b1b5-4df8-a6ae-e6924ed6b1f5","cardType":"basic","points":1,"front":"In energetics, what is the system?","back":"The chemical reaction being studied.","mcqOptions":[{"text":"The chemical reaction being studied.","correct":true,"explanation":"The system is the chemical reaction selected for investigation. Energy transfers are described relative to this system and everything outside it is treated as the surroundings."},{"text":"The laboratory apparatus used for measurement.","correct":false,"explanation":"The apparatus forms part of the surroundings, although it may absorb heat released by the reaction. The system is the chemical reaction being studied."},{"text":"The external environment around the reaction.","correct":false,"explanation":"The external environment belongs to the surroundings, not the system. The system is the chemical reaction whose energy change is being investigated."},{"text":"The measuring instrument recording the temperature.","correct":false,"explanation":"A measuring instrument is part of the surroundings and merely detects changes caused by energy transfer. The system is the chemical reaction being studied."}]},{"id":"a1d8faad-527c-47a6-9a0f-4d28918a3bab","cardType":"basic","points":1,"front":"In an exothermic reaction, in which direction does heat transfer occur?","back":"From the system to the surroundings.","mcqOptions":[{"text":"From the system to the surroundings.","correct":true,"explanation":"An exothermic reaction releases heat from the reacting system into the surroundings, usually causing the surroundings and reaction mixture to increase in temperature."},{"text":"From the surroundings to the system.","correct":false,"explanation":"This reverses the direction of heat transfer and describes an endothermic reaction. In an exothermic reaction, heat passes from the system to the surroundings."},{"text":"From the reactants to the products.","correct":false,"explanation":"Reactants and products are both parts of the reacting system, so this does not describe transfer across its boundary. Heat flows from the system to the surroundings."},{"text":"From the products to the reactants.","correct":false,"explanation":"This treats heat transfer as occurring between substances within the system. An exothermic reaction instead transfers heat from the system to the surroundings."}]},{"id":"26aaa34f-04a5-4893-b46d-1954fc48acbb","cardType":"basic","points":1,"front":"In an endothermic reaction, in which direction does heat transfer occur?","back":"From the surroundings into the system.","mcqOptions":[{"text":"From the surroundings into the system.","correct":true,"explanation":"An endothermic reaction absorbs heat from the surroundings into the reacting system, which commonly causes the surroundings and reaction mixture to decrease in temperature."},{"text":"From the system into the surroundings.","correct":false,"explanation":"This is the direction of heat transfer in an exothermic reaction. An endothermic reaction absorbs heat from the surroundings into the system."},{"text":"From the reactants into the products.","correct":false,"explanation":"Reactants and products are both within the system, so this is not heat transfer across the system boundary. Heat enters the system from the surroundings."},{"text":"From the products into the reactants.","correct":false,"explanation":"This incorrectly describes transfer between substances within the reaction. During an endothermic reaction, heat is transferred from the surroundings into the system."}]},{"id":"988b2046-f7b0-42bf-bba3-a2f0a3ecd6a0","cardType":"basic","points":2,"front":"What is the sign of $\\Delta H$ for an exothermic reaction, and why?","back":"

Negative ($\\Delta H < 0$)
Products have less enthalpy than reactants.

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

Negative ($\\Delta H < 0$)
Products have less enthalpy than reactants.

","correct":true,"explanation":"An exothermic reaction releases energy, leaving the products at lower enthalpy than the reactants. Therefore $H_{\\text{products}}-H_{\\text{reactants}}$ is negative."},{"text":"

Positive ($\\Delta H > 0$)
Products have less enthalpy than reactants.

","correct":false,"explanation":"Lower-enthalpy products make $H_{\\text{products}}-H_{\\text{reactants}}$ negative, not positive. Exothermic reactions therefore have $\\Delta H<0$ because energy is released."},{"text":"

Negative ($\\Delta H < 0$)
Products have more enthalpy than reactants.

","correct":false,"explanation":"Products with more enthalpy would give a positive change and describe an endothermic reaction. Exothermic products have less enthalpy than reactants, giving $\\Delta H<0$."},{"text":"

Zero ($\\Delta H = 0$)
Products have equal enthalpy to reactants.

","correct":false,"explanation":"A zero enthalpy change would mean no net energy difference, so the reaction would not be exothermic. Exothermic products have less enthalpy and $\\Delta H<0$."}]},{"id":"1db8d429-fa3a-402a-8bb0-8447c4202d99","cardType":"basic","points":2,"front":"What is the sign of $\\Delta H$ for an endothermic reaction, and why?","back":"

Positive ($\\Delta H > 0$)
Products have more enthalpy than reactants.

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

Positive ($\\Delta H > 0$)
Products have more enthalpy than reactants.

","correct":true,"explanation":"An endothermic reaction absorbs energy, leaving the products at higher enthalpy than the reactants. Therefore $H_{\\text{products}}-H_{\\text{reactants}}$ is positive."},{"text":"

Negative ($\\Delta H < 0$)
Products have more enthalpy than reactants.

","correct":false,"explanation":"Higher-enthalpy products make $H_{\\text{products}}-H_{\\text{reactants}}$ positive, not negative. Endothermic reactions absorb energy and therefore have $\\Delta H>0$."},{"text":"

Positive ($\\Delta H > 0$)
Products have less enthalpy than reactants.

","correct":false,"explanation":"Products with less enthalpy would give a negative change and describe an exothermic reaction. Endothermic products have more enthalpy than reactants, giving $\\Delta H>0$."},{"text":"

Zero ($\\Delta H = 0$)
Products have equal enthalpy to reactants.

","correct":false,"explanation":"A zero enthalpy change means there is no net energy difference between reactants and products. Endothermic reactions absorb energy, so products have more enthalpy and $\\Delta H>0$."}]},{"id":"8e49bed5-fcc9-4027-9aa2-6e611db36c11","cardType":"cloze","points":null,"front":"Enthalpy change ($\\Delta H$) is the heat energy change measured under conditions of [ ].","back":"Enthalpy change ($\\Delta H$) is the heat energy change measured under conditions of constant pressure.","mcqOptions":[{"text":"constant pressure","correct":true,"explanation":"Enthalpy change is defined as the heat energy transferred at constant pressure. This condition allows the heat exchanged to be identified with $\\Delta H$."},{"text":"constant volume","correct":false,"explanation":"Constant volume applies when relating heat transfer directly to internal energy change, not enthalpy change. Enthalpy change is the heat energy change measured at constant pressure."},{"text":"constant temperature","correct":false,"explanation":"Temperature does not generally remain constant while a reaction releases or absorbs heat. Enthalpy change is defined using heat transfer under conditions of constant pressure."},{"text":"constant concentration","correct":false,"explanation":"Reactant and product concentrations normally change as a reaction proceeds, so constant concentration is not the defining condition. Enthalpy change is measured at constant pressure."}]},{"id":"85fe1774-dadf-451d-a746-01fdff79493a","cardType":"cloze","points":null,"front":"$$$\\Delta H = \\text{[...]}$$","back":"$$$\\Delta H = \\boxed{H_{\\text{products}} - H_{\\text{reactants}}}$$","mcqOptions":[{"text":"$$H_{\\text{products}} - H_{\\text{reactants}}$","correct":true,"explanation":"Enthalpy change is calculated as the enthalpy of the products minus the enthalpy of the reactants, so the order determines the correct sign of $\\Delta H$."},{"text":"$$H_{\\text{reactants}} - H_{\\text{products}}$","correct":false,"explanation":"This reverses the subtraction and therefore reverses the sign of the enthalpy change. The correct expression is $H_{\\text{products}}-H_{\\text{reactants}}$."},{"text":"$$H_{\\text{products}} + H_{\\text{reactants}}$","correct":false,"explanation":"Adding the two enthalpies gives their total rather than the change between initial and final states. The correct calculation is products minus reactants."},{"text":"$$H_{\\text{reactants}} + H_{\\text{products}}$","correct":false,"explanation":"Changing the order of an addition does not produce an enthalpy difference. Enthalpy change is found by subtracting reactant enthalpy from product enthalpy."}]},{"id":"9c2ee50d-85c6-4a36-967c-d7357e55f006","cardType":"basic","points":1,"front":"What are the numerical standard conditions for enthalpy changes?","back":"

Pressure: 100 kPa
Temperature: 298 K (stated temperature)

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

Pressure: 100 kPa
Temperature: 298 K (stated temperature)

","correct":true,"explanation":"Standard enthalpy changes use a standard pressure of 100 kPa and are usually quoted at 298 K. The temperature must be stated if another value is used."},{"text":"

Pressure: 101 kPa
Temperature: 273 K (stated temperature)

","correct":false,"explanation":"This confuses standard enthalpy conditions with approximate atmospheric pressure and the freezing point of water. The specified values are 100 kPa and usually 298 K."},{"text":"

Pressure: 100 kPa
Temperature: 273 K (stated temperature)

","correct":false,"explanation":"The pressure is correct, but 273 K is the freezing point of water rather than the usual standard temperature. Standard enthalpy data are normally given at 298 K."},{"text":"

Pressure: 101 kPa
Temperature: 298 K (stated temperature)

","correct":false,"explanation":"The temperature is the usual value, but 101 kPa approximates one atmosphere rather than the defined standard pressure. Standard enthalpy changes use 100 kPa."}]},{"id":"69167161-451b-46c0-a3f9-70eff9ed8451","cardType":"basic","points":1,"front":"What is meant by a standard state?","back":"The physical state of a substance under standard conditions (100 kPa and 298 K).","mcqOptions":[{"text":"The physical state of a substance under standard conditions (100 kPa and 298 K).","correct":true,"explanation":"A standard state specifies the physical state of a substance at the standard conditions of 100 kPa and 298 K, allowing enthalpy values to be compared consistently."},{"text":"The temperature of a substance under standard conditions (100 kPa and 298 K).","correct":false,"explanation":"Standard state describes physical state, not temperature. The correct answer is the physical state of a substance under 100 kPa and 298 K; temperature is only one standard condition."},{"text":"The pressure of a substance under standard conditions (100 kPa and 298 K).","correct":false,"explanation":"Pressure is part of the standard conditions but does not define the substance’s state. The correct answer identifies whether it is solid, liquid or gas at 100 kPa and 298 K."},{"text":"The amount of a substance under standard conditions (100 kPa and 298 K).","correct":false,"explanation":"Standard state is not defined by the amount of substance. The correct answer is its physical state, such as solid, liquid or gas, under 100 kPa and 298 K."}]},{"id":"7a58e2dd-d341-45eb-9cdc-40c651604398","cardType":"basic","points":1,"front":"What is the standard state of Carbon?","back":"Graphite (s).","mcqOptions":[{"text":"Graphite (s).","correct":true,"explanation":"Graphite is the standard state of carbon at 100 kPa and 298 K. It is therefore assigned a standard enthalpy of formation of zero."},{"text":"Diamond (s).","correct":false,"explanation":"Diamond is an allotrope of carbon but not its standard state under these conditions. Graphite is the standard state because it is the more stable allotrope at 100 kPa and 298 K."},{"text":"Graphene (s).","correct":false,"explanation":"Graphene is another carbon allotrope, but it is not used as carbon’s standard state in enthalpy definitions. The accepted standard state is graphite at 100 kPa and 298 K."},{"text":"Fullerene (s).","correct":false,"explanation":"Fullerene is an allotrope of carbon, but it is not carbon’s standard state. Graphite is the most stable accepted form under standard conditions and is the correct answer."}]},{"id":"b7595623-a82d-4a5a-b40b-e88d2151a2e9","cardType":"basic","points":3,"front":"Define the standard enthalpy of combustion ($\\Delta_c H^\\ominus$).","back":"The enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, with all substances in their standard states.","mcqOptions":[{"text":"The enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, with all substances in their standard states.","correct":true,"explanation":"This is the standard enthalpy of combustion definition: one mole of substance is completely burned in oxygen under standard conditions, with all substances in their standard states."},{"text":"The enthalpy change when one mole of a substance is partially burned in oxygen under standard conditions, with all substances in their standard states.","correct":false,"explanation":"Incomplete or partial burning does not define standard enthalpy of combustion. The correct definition requires one mole of substance to be completely burned in oxygen."},{"text":"The enthalpy change when one mole of a substance is completely burned in air under standard conditions, with all substances in their standard states.","correct":false,"explanation":"Combustion enthalpy is defined using oxygen, not air, even though combustion experiments may use air. The correct answer specifies complete burning in oxygen with standard-state substances."},{"text":"The enthalpy change when one mole of a substance is completely formed from oxygen under standard conditions, with all substances in their standard states.","correct":false,"explanation":"Formation from oxygen confuses combustion with formation and does not describe burning. The correct definition requires one mole of substance to be completely burned in oxygen."}]},{"id":"0bf767bd-1511-4723-8643-2b720fb15718","cardType":"cloze","points":null,"front":"The standard enthalpy of combustion of methane is represented by the equation: $\\text{[...]}$","back":"The standard enthalpy of combustion of methane is represented by the equation: $\\boxed{\\text{CH}_4(g) + 2\\text{O}_2(g) \\to \\text{CO}_2(g) + 2\\text{H}_2\\text{O}(l)}$","mcqOptions":[{"text":"$$\\text{CH}_4(g) + 2\\text{O}_2(g) \\to \\text{CO}_2(g) + 2\\text{H}_2\\text{O}(l)$","correct":true,"explanation":"Methane combustion produces carbon dioxide and liquid water. Balancing carbon, hydrogen and oxygen gives one methane, two oxygen molecules, one carbon dioxide and two water molecules."},{"text":"$$\\text{CH}_4(g) + 2\\text{O}_2(g) \\to \\text{CO}(g) + 2\\text{H}_2\\text{O}(l)$","correct":false,"explanation":"Carbon monoxide indicates incomplete combustion, not complete combustion. The correct equation forms carbon dioxide, with two oxygen molecules reacting with one methane molecule."},{"text":"$$\\text{CH}_4(g) + \\text{O}_2(g) \\to \\text{CO}_2(g) + 2\\text{H}_2\\text{O}(l)$","correct":false,"explanation":"This equation has too few oxygen atoms on the reactant side. Balancing oxygen requires 2O₂, giving the correct complete-combustion equation with CO₂ and liquid water."},{"text":"$$\\text{CH}_4(g) + 2\\text{O}_2(g) \\to \\text{CO}_2(g) + 2\\text{H}_2\\text{O}(g)$","correct":false,"explanation":"Standard enthalpy of combustion uses water in its standard state, which is liquid at 298 K. Water vapour would describe a different enthalpy change."}]},{"id":"89ad7797-39b7-4670-b44f-f8f7ced0e32f","cardType":"basic","points":3,"front":"Define the standard enthalpy of formation ($\\Delta_f H^\\ominus$).","back":"The enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions, with all substances in their standard states.","mcqOptions":[{"text":"The enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions, with all substances in their standard states.","correct":true,"explanation":"Standard enthalpy of formation is the enthalpy change for forming exactly one mole of a compound from its constituent elements in their standard states under standard conditions."},{"text":"The enthalpy change when one mole of an element is formed from its constituent compounds under standard conditions, with all substances in their standard states.","correct":false,"explanation":"Formation enthalpy concerns a compound being made from its elements, not an element being made from compounds. It also requires exactly one mole of compound to form."},{"text":"The enthalpy change when one mole of a compound is formed from its constituent compounds under standard conditions, with all substances in their standard states.","correct":false,"explanation":"The reactants must be the compound’s constituent elements, not other compounds. The correct definition forms exactly one mole of compound from elements in standard states."},{"text":"The enthalpy change when one mole of a compound is formed from its constituent elements under any conditions, with all substances in their standard states.","correct":false,"explanation":"Standard conditions are essential to the definition, so ‘under any conditions’ is incorrect. The correct answer specifies formation under standard conditions and standard states."}]},{"id":"2b5fd64e-8e3b-4936-8332-70b138e9dbb6","cardType":"basic","points":1,"front":"What is the value of $\\Delta_f H^\\ominus$ for any element in its standard state?","back":"Zero (0 kJ mol⁻¹).","mcqOptions":[{"text":"Zero (0 kJ mol⁻¹).","correct":true,"explanation":"An element in its standard state is the reference form for enthalpy of formation data, so its standard enthalpy of formation is defined as zero."},{"text":"One (1 kJ mol⁻¹).","correct":false,"explanation":"The reference value is defined as zero, not one. Any element in its standard state has ΔfH° = 0 kJ mol⁻¹ by convention."},{"text":"Zero (0 kJ).","correct":false,"explanation":"The numerical value is zero, but the units must be kJ mol⁻¹ because standard enthalpy of formation is defined per mole. The correct answer includes those units."},{"text":"Zero (0 kJ mol⁻²).","correct":false,"explanation":"The exponent on the mole unit is incorrect. Enthalpy of formation is measured per mole, so the correct units are kJ mol⁻¹, with a value of zero."}]},{"id":"3c3d7c21-695b-4ee8-8e58-88bce94e21a5","cardType":"basic","points":1,"front":"Why is the $\\Delta_f H^\\ominus$ of diamond not zero?","back":"Diamond is not the standard state of carbon at 100 kPa and 298 K.","mcqOptions":[{"text":"Diamond is not the standard state of carbon at 100 kPa and 298 K.","correct":true,"explanation":"Graphite, not diamond, is carbon’s standard state at 100 kPa and 298 K. Therefore diamond has a non-zero standard enthalpy of formation relative to graphite."},{"text":"Diamond is not an allotrope of carbon at 100 kPa and 298 K.","correct":false,"explanation":"Diamond is an allotrope of carbon, so this statement is false. Its enthalpy of formation is not zero because graphite, rather than diamond, is the standard state."},{"text":"Diamond is not a solid form of carbon at 100 kPa and 298 K.","correct":false,"explanation":"Diamond is a solid at these conditions. Its non-zero value comes from not being carbon’s standard state; graphite is the standard state used as the zero reference."},{"text":"Diamond is not made from carbon at 100 kPa and 298 K.","correct":false,"explanation":"Diamond is composed of carbon, so its composition is not the issue. Its standard enthalpy of formation is non-zero because diamond is not carbon’s standard state."}]},{"id":"a92d57fa-ebff-4b1a-8c72-1d00c2f2a3b0","cardType":"basic","points":2,"front":"Why are fractions (e.g. $\\frac{1}{2}O_2$) allowed in enthalpy of formation equations?","back":"The definition requires that exactly one mole of product is formed.","mcqOptions":[{"text":"The definition requires that exactly one mole of product is formed.","correct":true,"explanation":"Formation enthalpy is defined for producing exactly one mole of the compound. Fractions such as ½O₂ are therefore needed to balance the elements while making one mole of product."},{"text":"The definition requires that exactly one mole of reactant is used.","correct":false,"explanation":"The defining quantity is one mole of product, not one mole of reactant. Fractions are allowed so the equation forms exactly one mole of the compound from its elements."},{"text":"The definition requires that exactly two moles of product are formed.","correct":false,"explanation":"Formation enthalpy is defined for one mole of product, not two. Using two moles would double the enthalpy change and would require rescaling the equation."},{"text":"The definition requires that exactly one mole of element is formed.","correct":false,"explanation":"The product is one mole of compound, while the elements are reactants. Fractions in equations such as ½O₂ ensure the compound’s formula is balanced for one mole of product."}]},{"id":"4b2bd73c-09ad-41d2-9801-5ac67bf23ec8","cardType":"cloze","points":null,"front":"The equation representing the standard enthalpy of formation of liquid water is $\\text{[...]}$.","back":"The equation representing the standard enthalpy of formation of liquid water is $\\boxed{\\text{H}_2(g) + \\frac{1}{2}\\text{O}_2(g) \\to \\text{H}_2\\text{O}(l)}$.","mcqOptions":[{"text":"$$\\text{H}_2(g) + \\frac{1}{2}\\text{O}_2(g) \\to \\text{H}_2\\text{O}(l)$","correct":true,"explanation":"Standard enthalpy of formation uses exactly one mole of compound formed from its elements in their standard states. Liquid water forms from hydrogen gas and oxygen gas in the 1:2 atom ratio."},{"text":"$$\\text{H}_2(g) + \\text{O}_2(g) \\to \\text{H}_2\\text{O}(l)$","correct":false,"explanation":"This uses one mole of oxygen molecules, giving twice the oxygen needed for one mole of water. The balanced formation equation requires $\\frac{1}{2}\\text{O}_2(g)$."},{"text":"$$\\text{H}_2\\text{O}(l) \\to \\text{H}_2(g) + \\frac{1}{2}\\text{O}_2(g)$","correct":false,"explanation":"The substances and direction are reversed, so this represents decomposition rather than formation. Formation must produce one mole of liquid water from its elements."},{"text":"$$\\text{H}_2(g) + \\frac{1}{2}\\text{O}_2(l) \\to \\text{H}_2\\text{O}(l)$","correct":false,"explanation":"Oxygen is incorrectly shown as a liquid. In its standard state under standard conditions, oxygen exists as $\\text{O}_2(g)$, so the gaseous state must be used."}]},{"id":"87d55094-bfd7-476a-8fe1-a9a8fb6b2ef6","cardType":"basic","points":2,"front":"A reaction causes the temperature of the surrounding water to rise. Is this exothermic or endothermic?","back":"

Exothermic.
The system has lost energy to the surroundings.

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

Exothermic.
The system has lost energy to the surroundings.

","correct":true,"explanation":"A rise in the surrounding water’s temperature means energy has been transferred from the reacting system to its surroundings. A reaction that releases energy in this way is exothermic."},{"text":"

Endothermic.
The system has gained energy from the surroundings.

","correct":false,"explanation":"Endothermic reactions absorb energy and cool their surroundings. Because the water temperature rises, the reaction is exothermic and the system has lost energy to the surroundings."},{"text":"

Endothermic.
The system has lost energy to the surroundings.

","correct":false,"explanation":"The energy direction describes an exothermic reaction, not an endothermic one. Energy leaving the system warms the water, so the correct classification is exothermic."},{"text":"

Exothermic.
The system has gained energy from the surroundings.

","correct":false,"explanation":"The classification is correctly identified, but the energy direction is reversed. An exothermic system loses energy, which is transferred to and warms the surrounding water."}]},{"id":"c22a43d0-3291-4278-8645-532ceb2f7b17","cardType":"basic","points":1,"front":"Calculate $\\Delta H$ if reactants are at $250\\text{ kJ mol}^{-1}$ and products are at $90\\text{ kJ mol}^{-1}$.","back":"$$90 - 250 = \\mathbf{-160\\text{ kJ mol}^{-1}}$","mcqOptions":[{"text":"$$90 - 250 = \\mathbf{-160\\text{ kJ mol}^{-1}}$","correct":true,"explanation":"Enthalpy change is calculated as enthalpy of products minus enthalpy of reactants: $90 - 250 = -160\\text{ kJ mol}^{-1}$. The negative value indicates an exothermic change."},{"text":"$$250 - 90 = \\mathbf{160\\text{ kJ mol}^{-1}}$","correct":false,"explanation":"The reactant and product values have been subtracted in the wrong order. Using products minus reactants gives $90 - 250 = -160\\text{ kJ mol}^{-1}$, indicating an exothermic change."},{"text":"$$90 - 250 = \\mathbf{160\\text{ kJ mol}^{-1}}$","correct":false,"explanation":"The subtraction gives a negative result, not a positive one; this is a sign error. Products minus reactants is $90 - 250 = -160\\text{ kJ mol}^{-1}$."},{"text":"$$250 + 90 = \\mathbf{340\\text{ kJ mol}^{-1}}$","correct":false,"explanation":"Adding the reactant and product enthalpies does not calculate an enthalpy change. The required calculation is products minus reactants: $90 - 250 = -160\\text{ kJ mol}^{-1}$."}]},{"id":"e82764ee-1a05-4346-9bf8-a40c8f41293b","cardType":"basic","points":1,"front":"What is the standard concentration for solutions in standard enthalpy changes?","back":"1.00 mol dm⁻³.","mcqOptions":[{"text":"1.00 mol dm⁻³.","correct":true,"explanation":"The standard concentration used for solutions in standard enthalpy change definitions is 1.00 mol dm⁻³. This provides the specified standard-state concentration for dissolved substances."},{"text":"0.100 mol dm⁻³.","correct":false,"explanation":"This is a common tenfold dilution error. The standard concentration is 1.00 mol dm⁻³, whereas 0.100 mol dm⁻³ is below the defined standard-state concentration."},{"text":"1.00 mol cm⁻³.","correct":false,"explanation":"The numerical value is paired with the wrong volume unit. Standard concentration is expressed as 1.00 mol dm⁻³, not mol cm⁻³."},{"text":"100 mol dm⁻³.","correct":false,"explanation":"This value is 100 times larger than the standard concentration and may result from confusing cubic centimetres with cubic decimetres. The correct value is 1.00 mol dm⁻³."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Enthalpy change","title":"Enthalpy change Flashcards","topbar":{"backHref":"/a-level/chemistry/aqa/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"1f9e615e-7af8-5aa7-913e-c50d03cbd32b","walled":"$L39"}],"$L3a","$L3b"]