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Greater dispersal corresponds to a larger entropy value."},{"text":"A measure of the strength of chemical bonds between particles within a system.","correct":false,"explanation":"Bond strength concerns the energy required to break chemical bonds, not entropy. Entropy measures the dispersal of energy and particles among possible arrangements within a system."},{"text":"A measure of the total thermal energy stored by particles within a system.","correct":false,"explanation":"Total thermal energy depends on factors such as temperature and amount of substance, but it is not entropy. Entropy measures how dispersed energy and particles are."},{"text":"A measure of the rate at which particles react and form products within a system.","correct":false,"explanation":"Reaction rate describes how quickly reactants become products and is a kinetic property. Entropy instead measures the dispersal of energy and particles within a system."}]},{"id":"0d847b3d-f0a9-4f07-8872-ebe3b2fa3319","cardType":"cloze","points":null,"front":"In terms of arrangements, a high entropy value indicates that there are [ ] ways to arrange particles and energy.","back":"In terms of arrangements, a high entropy value indicates that there are more ways to arrange particles and energy.","mcqOptions":[{"text":"more","correct":true,"explanation":"A larger number of possible arrangements allows particles and energy to be distributed in more ways, so the system has a higher entropy."},{"text":"fewer","correct":false,"explanation":"Fewer arrangements correspond to lower entropy because particles and energy are less widely dispersed. A high entropy value indicates more possible arrangements."},{"text":"equal","correct":false,"explanation":"Having the same number of arrangements would imply no change in entropy. High entropy is associated with more possible arrangements of particles and energy."},{"text":"single","correct":false,"explanation":"A single possible arrangement represents maximal order and therefore very low entropy. High entropy arises when more arrangements are available."}]},{"id":"5b10f76f-505b-43e0-930f-551a3931aae5","cardType":"basic","points":1,"front":"Rank the three physical states of a substance from lowest to highest entropy.","back":"

Solid (lowest)
Liquid
Gas (highest)

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

Solid (lowest)
Liquid
Gas (highest)

","correct":true,"explanation":"Particles are most constrained in a solid, less constrained in a liquid and widely dispersed in a gas. The number of possible arrangements therefore increases in this order."},{"text":"

Liquid (lowest)
Solid
Gas (highest)

","correct":false,"explanation":"This incorrectly places a liquid below a solid. Solid particles have the fewest possible arrangements, followed by liquids, while gases have the greatest entropy."},{"text":"

Solid (lowest)
Gas
Liquid (highest)

","correct":false,"explanation":"This incorrectly gives liquids the greatest entropy. Gas particles have much greater freedom and many more possible arrangements, so the correct order is solid, liquid, gas."},{"text":"

Gas (lowest)
Liquid
Solid (highest)

","correct":false,"explanation":"This reverses the correct trend by treating greater particle freedom as lower entropy. Entropy increases from solid to liquid to gas as possible arrangements increase."}]},{"id":"2810beaa-79a8-4335-872d-6f7035fbacc2","cardType":"basic","points":2,"front":"Why do gases have a much higher entropy than solids?","back":"Gas particles are far apart and move randomly, allowing many more possible arrangements.","mcqOptions":[{"text":"Gas particles are far apart and move randomly, allowing many more possible arrangements.","correct":true,"explanation":"The large separation and unrestricted random motion of gas particles provide many possible positions and energy distributions, giving gases much greater entropy than solids."},{"text":"Gas particles are close together and vibrate about fixed positions, allowing very few possible arrangements.","correct":false,"explanation":"Closely packed particles vibrating about fixed positions describe a solid, not a gas. Gas particles are far apart and move randomly, creating many more possible arrangements."},{"text":"Gas particles have greater mass and form stronger bonds, allowing many more possible arrangements.","correct":false,"explanation":"Particle mass does not increase when a substance changes state, and gaseous particles generally have negligible intermolecular bonding. Their high entropy comes from separation and random motion."},{"text":"Gas particles have more chemical energy and collide frequently, allowing many more possible arrangements.","correct":false,"explanation":"Frequent collisions are not the fundamental reason for higher entropy, and gases do not necessarily possess more chemical energy. Their separation and random movement permit more arrangements."}]},{"id":"7728dab4-0300-4309-83a6-88f491a0ff41","cardType":"cloze","points":null,"front":"The entropy of a system [ ] when a solid dissolves or when the number of gas particles increases.","back":"The entropy of a system increases when a solid dissolves or when the number of gas particles increases.","mcqOptions":[{"text":"increases","correct":true,"explanation":"Dissolving generally disperses particles through a solvent, while additional gas particles create more possible arrangements. Both changes therefore raise the system’s entropy."},{"text":"decreases","correct":false,"explanation":"Decreasing entropy would mean fewer possible arrangements. Dissolution and an increase in gas particles generally produce greater dispersal, so the entropy increases."},{"text":"stays constant","correct":false,"explanation":"The entropy does not stay constant because both changes increase the number of possible particle arrangements. It therefore increases rather than remaining unchanged."},{"text":"doubles","correct":false,"explanation":"Doubling is not a general entropy relationship because the size of the change depends on the substances and quantities involved. The valid qualitative prediction is that entropy increases."}]},{"id":"acf112fe-f4f2-4e23-80fe-ba7f78380c20","cardType":"basic","points":2,"front":"Predict the sign of $\\Delta S_{\\text{system}}$ for: $2\\text{H}_2(g) + \\text{O}_2(g) \\to 2\\text{H}_2\\text{O}(l)$. Give one reason.","back":"

Negative
Fewer moles of gas are present in the products.

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

Negative
Fewer moles of gas are present in the products.

","correct":true,"explanation":"The reactants contain three moles of gas, whereas the product is liquid water with no gaseous particles. This reduction in gaseous arrangements makes the system entropy decrease."},{"text":"

Positive
More moles of gas are present in the products.

","correct":false,"explanation":"This reverses the change in gaseous amount. Three moles of gaseous reactants form liquid water, so gaseous arrangements decrease and the system entropy change is negative."},{"text":"

Zero
Equal moles of gas are present on both sides.

","correct":false,"explanation":"The gaseous amounts are not equal: the reactants have three moles of gas and the products have none. The resulting loss of gaseous freedom gives a negative entropy change."},{"text":"

Positive
Particles in a liquid have more possible arrangements.

","correct":false,"explanation":"Liquids have fewer possible particle arrangements than gases, not more. Converting three moles of gaseous reactants into liquid water therefore gives a negative system entropy change."}]},{"id":"0fa5cd0f-ce45-4949-ad6d-1d3e4c0fead1","cardType":"basic","points":1,"front":"What is the entropy of a perfect crystal at 0 K?","back":"Zero.","mcqOptions":[{"text":"Zero.","correct":true,"explanation":"By the third law of thermodynamics, a perfect crystal at absolute zero has a single completely ordered arrangement and therefore has no configurational entropy."},{"text":"Positive.","correct":false,"explanation":"A perfect crystal at absolute zero has only one possible arrangement, so its entropy is zero by the third law of thermodynamics, not positive."},{"text":"Undefined.","correct":false,"explanation":"Entropy is well defined for the idealised perfect crystal at absolute zero. The third law assigns it an entropy of zero because only one arrangement is available."},{"text":"Negative.","correct":false,"explanation":"Entropy is not negative for a perfect crystal at absolute zero. Its single perfectly ordered arrangement gives an entropy of exactly zero."}]},{"id":"b05b9864-c40d-439f-ad6f-865e1a4eb22c","cardType":"cloze","points":null,"front":"Standard molar entropy ($S^\\circ$) is measured at a temperature of [ ] and a pressure of 100 kPa.","back":"Standard molar entropy ($S^\\circ$) is measured at a temperature of 298 K and a pressure of 100 kPa.","mcqOptions":[{"text":"298 K","correct":true,"explanation":"Standard molar entropy values used in thermochemical data tables are conventionally quoted at 298 K and a standard pressure of 100 kPa."},{"text":"273 K","correct":false,"explanation":"A temperature of 273 K corresponds approximately to the freezing point of water, not the conventional temperature used for standard molar entropy data, which is 298 K."},{"text":"298 °C","correct":false,"explanation":"This confuses kelvin and degrees Celsius: 298 °C equals about 571 K. Standard molar entropy values are conventionally measured at 298 K."},{"text":"273 °C","correct":false,"explanation":"A temperature of 273 °C equals about 546 K, so this is neither 273 K nor the conventional standard temperature. The required value is 298 K."}]},{"id":"0e045f17-a0d7-4cfc-a5da-6961828c86db","cardType":"cloze","points":null,"front":"Standard molar entropy ($S^\\circ$) is measured at a temperature of 298 K and a pressure of [ ].","back":"Standard molar entropy ($S^\\circ$) is measured at a temperature of 298 K and a pressure of 100 kPa.","mcqOptions":[{"text":"100 kPa","correct":true,"explanation":"The standard pressure used for standard molar entropy data is 100 kPa, alongside the specified standard temperature of 298 K."},{"text":"101 kPa","correct":false,"explanation":"This approximates atmospheric pressure and reflects the older convention of using 1 atm. The current standard pressure for molar entropy data is 100 kPa."},{"text":"1 kPa","correct":false,"explanation":"This confuses the numerical value in bar with pressure in kilopascals. Standard pressure is 1 bar, which is equal to 100 kPa."},{"text":"100 Pa","correct":false,"explanation":"This results from confusing pascals with kilopascals, making the pressure one thousand times too small. The required standard pressure is 100 kPa."}]},{"id":"8304ea97-a080-43d1-9981-46859890b3be","cardType":"basic","points":1,"front":"State the formula for the standard entropy change of a system ($\\Delta S^\\circ_{\\text{system}}$).","back":"$$$\\sum S^\\circ_{\\text{products}} - \\sum S^\\circ_{\\text{reactants}}$$","mcqOptions":[{"text":"$$$\\sum S^\\circ_{\\text{products}} - \\sum S^\\circ_{\\text{reactants}}$$","correct":true,"explanation":"The system’s standard entropy change is calculated by subtracting the total standard entropies of the reactants from those of the products, including stoichiometric coefficients."},{"text":"$$$\\sum S^\\circ_{\\text{reactants}} - \\sum S^\\circ_{\\text{products}}$$","correct":false,"explanation":"This reverses the subtraction and therefore gives the negative of the required entropy change. The correct order is products minus reactants."},{"text":"$$$\\sum S^\\circ_{\\text{products}} + \\sum S^\\circ_{\\text{reactants}}$$","correct":false,"explanation":"Adding both totals does not represent the change between initial and final states. Subtract the reactants’ total entropy from the products’ total entropy."},{"text":"$$$\\sum H^\\circ_{\\text{products}} - \\sum H^\\circ_{\\text{reactants}}$$","correct":false,"explanation":"This calculates a standard enthalpy change rather than a standard entropy change. The calculation must use standard molar entropies, $S^\\circ$, for products minus reactants."}]},{"id":"238f5d9d-bdc3-429f-994a-099c8b0979e8","cardType":"cloze","points":null,"front":"The units for standard molar entropy are $\\text{[...]}$.","back":"The units for standard molar entropy are $\\boxed{\\text{J K}^{-1}\\text{ mol}^{-1}}$.","mcqOptions":[{"text":"$$\\text{J K}^{-1}\\text{ mol}^{-1}$","correct":true,"explanation":"Standard molar entropy is energy dispersed per unit temperature per mole, so its units are joules per kelvin per mole."},{"text":"$$\\text{kJ K}^{-1}\\text{ mol}^{-1}$","correct":false,"explanation":"Standard molar entropy data are conventionally quoted in joules, not kilojoules. The correct units are $\\text{J K}^{-1}\\text{ mol}^{-1}$."},{"text":"$$\\text{J K mol}^{-1}$","correct":false,"explanation":"Temperature must appear with an exponent of $-1$ because entropy is energy divided by temperature. The correct units are $\\text{J K}^{-1}\\text{ mol}^{-1}$."},{"text":"$$\\text{J K}^{-1}\\text{ mol}$","correct":false,"explanation":"Molar entropy is measured per mole, so the mole exponent must be $-1$. The correct units are $\\text{J K}^{-1}\\text{ mol}^{-1}$."}]},{"id":"947ce404-58d1-47dc-a0c5-a83121bd7af2","cardType":"basic","points":1,"front":"State the formula for the entropy change of the surroundings ($\\Delta S_{\\text{surroundings}}$).","back":"$$$\\Delta S_{\\text{surroundings}} = \\frac{-\\Delta H}{T}$$","mcqOptions":[{"text":"$$$\\Delta S_{\\text{surroundings}} = \\frac{-\\Delta H}{T}$$","correct":true,"explanation":"Heat transferred to the surroundings is the negative of the system’s enthalpy change. Dividing this heat by absolute temperature gives the surroundings’ entropy change."},{"text":"$$$\\Delta S_{\\text{surroundings}} = \\frac{\\Delta H}{T}$$","correct":false,"explanation":"This misses the sign reversal between the system and surroundings. Heat released by the system enters the surroundings, so the correct numerator is $-\\Delta H$."},{"text":"$$$\\Delta S_{\\text{surroundings}} = \\frac{-T}{\\Delta H}$$","correct":false,"explanation":"This inverts the required relationship and gives incorrect units. Entropy change is heat divided by absolute temperature, so the formula is $-\\Delta H/T$."},{"text":"$$$\\Delta S_{\\text{surroundings}} = -\\Delta H T$$","correct":false,"explanation":"Multiplying by temperature gives the wrong dimensions and relationship. The heat transferred to the surroundings must be divided by absolute temperature: $\\Delta S_{\\text{surroundings}}=-\\Delta H/T$."}]},{"id":"ecf4cefe-d16c-4a2c-b17f-955bd3edd736","cardType":"basic","points":2,"front":"How does an exothermic reaction affect the entropy of the surroundings?","back":"Increases it (heat is released, increasing the dispersal of energy in the surroundings).","mcqOptions":[{"text":"Increases it (heat is released, increasing the dispersal of energy in the surroundings).","correct":true,"explanation":"An exothermic reaction releases heat into the surroundings. This disperses more energy among the surroundings and therefore produces a positive surroundings entropy change."},{"text":"Decreases it (heat is released, reducing the dispersal of energy in the surroundings).","correct":false,"explanation":"This reverses the effect of released heat. Energy entering the surroundings becomes more widely dispersed, so an exothermic reaction increases the entropy of the surroundings."},{"text":"Does not change it (heat transfer has no effect on the entropy of the surroundings).","correct":false,"explanation":"Entropy of the surroundings changes when heat is transferred. An exothermic reaction releases heat, increasing energy dispersal and therefore increasing the surroundings’ entropy."},{"text":"Makes zero it (the energy released is balanced by the energy in the surroundings).","correct":false,"explanation":"Conservation of energy does not mean entropy change is zero. Released heat becomes dispersed through the surroundings, so their entropy increases during an exothermic reaction."}]},{"id":"b87d4215-eb4b-4b34-b964-377bcb9662d2","cardType":"basic","points":1,"front":"State the formula for total entropy change ($\\Delta S_{\\text{total}}$).","back":"$$$\\Delta S_{\\text{system}} + \\Delta S_{\\text{surroundings}}$$","mcqOptions":[{"text":"$$$\\Delta S_{\\text{system}} + \\Delta S_{\\text{surroundings}}$$","correct":true,"explanation":"Total entropy change includes both parts of the thermodynamic universe, so it is the sum of the entropy changes of the system and the surroundings."},{"text":"$$$\\Delta S_{\\text{system}} - \\Delta S_{\\text{surroundings}}$$","correct":false,"explanation":"This incorrectly subtracts the surroundings’ contribution. Total entropy change accounts for both contributions by adding $\\Delta S_{\\text{system}}$ and $\\Delta S_{\\text{surroundings}}$."},{"text":"$$$\\Delta S_{\\text{surroundings}} - \\Delta S_{\\text{system}}$$","correct":false,"explanation":"This treats total entropy as a difference and reverses the terms. The total change is found by adding the system and surroundings entropy changes."},{"text":"$$$\\Delta S_{\\text{system}} \\times \\Delta S_{\\text{surroundings}}$$","correct":false,"explanation":"Multiplying entropy changes gives incorrect units and does not describe their combined effect. Total entropy change is the sum of the system and surroundings changes."}]},{"id":"bd1bbc45-2d8e-486d-aafc-aa96f6ceb36f","cardType":"cloze","points":null,"front":"A reaction is thermodynamically feasible if $\\Delta S_{\\text{total}}$ is [ ] (greater than zero).","back":"A reaction is thermodynamically feasible if $\\Delta S_{\\text{total}}$ is positive (greater than zero).","mcqOptions":[{"text":"positive","correct":true,"explanation":"A positive total entropy change means the entropy of the universe increases, satisfying the thermodynamic condition for a reaction to be feasible."},{"text":"negative","correct":false,"explanation":"A negative total entropy change means the entropy of the universe would decrease, so the reaction is not thermodynamically feasible. Feasibility requires a positive value."},{"text":"zero","correct":false,"explanation":"A zero total entropy change describes equilibrium rather than a spontaneous forward change. Thermodynamic feasibility in the forward direction requires $\\Delta S_{\\text{total}}>0$."},{"text":"non-negative","correct":false,"explanation":"This incorrectly includes zero, which corresponds to equilibrium with no net thermodynamic driving force. The stated condition of greater than zero requires the term positive."}]},{"id":"60e3b4e0-485d-4644-9be2-7a6500e9548b","cardType":"basic","points":1,"front":"Why might a feasible reaction not occur in practice?","back":"It has a high activation energy, meaning the rate of reaction is too slow.","mcqOptions":[{"text":"It has a high activation energy, meaning the rate of reaction is too slow.","correct":true,"explanation":"Thermodynamic feasibility describes whether a reaction is energetically possible, not how quickly it occurs. A high activation energy can make a feasible reaction negligibly slow."},{"text":"It has a small equilibrium constant, meaning the rate of reaction is too slow.","correct":false,"explanation":"This confuses equilibrium position with reaction rate. A feasible reaction may fail to occur observably because a high activation energy makes its rate extremely slow."},{"text":"It has a low activation energy, meaning the rate of reaction is too slow.","correct":false,"explanation":"A low activation energy generally allows more successful collisions and therefore a faster reaction. A feasible reaction may be imperceptibly slow when its activation energy is high."},{"text":"It has a negative total entropy change, meaning the reaction is too slow.","correct":false,"explanation":"This confuses thermodynamic feasibility with kinetics. A negative total entropy change makes the reaction infeasible, whereas a feasible reaction may not occur because of high activation energy."}]},{"id":"b0c7001f-553e-43da-a2f2-2886b2f28646","cardType":"cloze","points":null,"front":"The Gibbs energy equation is $$\\text{[...]}$$.","back":"The Gibbs energy equation is $$\\boxed{\\Delta G = \\Delta H - T\\Delta S_{\\text{system}}}$$.","mcqOptions":[{"text":"$$$\\Delta G = \\Delta H - T\\Delta S_{\\text{system}}$$","correct":true,"explanation":"Gibbs energy change equals enthalpy change minus absolute temperature multiplied by the system’s entropy change. This equation predicts whether a process is thermodynamically feasible under the stated conditions."},{"text":"$$$\\Delta G = \\Delta H + T\\Delta S_{\\text{system}}$$","correct":false,"explanation":"The entropy term is subtracted, not added. The correct equation is $\\Delta G = \\Delta H - T\\Delta S_{\\text{system}}$, so increasing entropy makes $\\Delta G$ more negative."},{"text":"$$$\\Delta G = T\\Delta H - \\Delta S_{\\text{system}}$$","correct":false,"explanation":"Temperature multiplies the entropy change, not the enthalpy change. The correct equation is $\\Delta G = \\Delta H - T\\Delta S_{\\text{system}}$, which also gives consistent energy units."},{"text":"$$$\\Delta G = \\Delta H - \\frac{\\Delta S_{\\text{system}}}{T}$$","correct":false,"explanation":"The entropy change must be multiplied by temperature rather than divided by it. The correct relationship is $\\Delta G = \\Delta H - T\\Delta S_{\\text{system}}$."}]},{"id":"8c41d96e-33b2-4ad0-be65-1a88f87cb8b6","cardType":"basic","points":1,"front":"State the condition for feasibility in terms of Gibbs energy change ($\\Delta G$).","back":"$$$\\Delta G < 0$$ (or negative)","mcqOptions":[{"text":"$$$\\Delta G < 0$$ (or negative)","correct":true,"explanation":"A negative Gibbs energy change means the forward process is thermodynamically feasible under the stated conditions, although it does not indicate how quickly the reaction occurs."},{"text":"$$$\\Delta G > 0$$ (or positive)","correct":false,"explanation":"A positive Gibbs energy change means the forward reaction is not feasible under the stated conditions. Feasibility requires $\\Delta G < 0$, meaning Gibbs energy decreases."},{"text":"$$$\\Delta G = 0$$ (or zero)","correct":false,"explanation":"A zero Gibbs energy change describes equilibrium, where there is no net thermodynamic tendency in either direction. Forward feasibility requires $\\Delta G < 0$."},{"text":"$$$\\Delta G \\leq 0$$ (or non-positive)","correct":false,"explanation":"Including zero incorrectly treats equilibrium as a feasible net forward change. The stated condition for forward feasibility is strictly $\\Delta G < 0$."}]},{"id":"62e51c81-ff92-4829-9c22-772ecb884180","cardType":"cloze","points":null,"front":"When calculating Gibbs energy, you must ensure units match; usually by converting $\\Delta S$ from [ ] to kJ.","back":"When calculating Gibbs energy, you must ensure units match; usually by converting $\\Delta S$ from J to kJ.","mcqOptions":[{"text":"J","correct":true,"explanation":"Entropy change is commonly supplied in $\\text{J mol}^{-1}\\text{ K}^{-1}$, while enthalpy change is commonly in kJ mol⁻¹, so the entropy value is converted from J to kJ."},{"text":"kJ","correct":false,"explanation":"This reverses the required conversion: entropy change is usually given in J rather than kJ. Convert it from J to kJ so its energy unit matches $\\Delta H$."},{"text":"K","correct":false,"explanation":"Kelvin measures temperature, not the energy component of entropy units. Entropy change is normally given in J mol⁻¹ K⁻¹, so J must be converted to kJ."},{"text":"mol","correct":false,"explanation":"Moles describe amount of substance and are not the energy unit being converted. The J in the entropy unit must be converted to kJ to match enthalpy units."}]},{"id":"59adfd5b-c6f6-4c89-9950-e63396e263cc","cardType":"cloze","points":null,"front":"When calculating Gibbs energy, you must ensure units match; usually by converting $\\Delta S$ from J to [ ].","back":"When calculating Gibbs energy, you must ensure units match; usually by converting $\\Delta S$ from J to kJ.","mcqOptions":[{"text":"kJ","correct":true,"explanation":"The Gibbs equation combines ΔH with TΔS, so their energy units must match. Converting entropy from joules to kilojoules makes TΔS compatible with ΔH when values are expressed in kJ mol⁻¹."},{"text":"MJ","correct":false,"explanation":"Converting to megajoules would not match the usual kJ units used for ΔH and would introduce an extra factor of 1000. The required conversion is from joules to kilojoules."},{"text":"cal","correct":false,"explanation":"Calories are not the energy units normally paired with ΔH in Gibbs calculations, and this conversion would still require matching ΔH separately. The required unit is kilojoules."},{"text":"eV","correct":false,"explanation":"Entropy is an energy change per temperature, so converting it to electronvolts is inappropriate for the usual chemistry data and would not match ΔH. The required unit is kilojoules."}]},{"id":"2cd6575c-ce7f-417d-90aa-80acf2c3246e","cardType":"basic","points":2,"front":"How is the threshold temperature for feasibility calculated?","back":"$$$T = \\frac{\\Delta H}{\\Delta S}$$ (the temperature where $\\Delta G = 0$)","mcqOptions":[{"text":"$$$T = \\frac{\\Delta H}{\\Delta S}$$ (the temperature where $\\Delta G = 0$)","correct":true,"explanation":"At the feasibility threshold, $\\Delta G=0$. Substitution into $\\Delta G=\\Delta H-T\\Delta S$ and rearrangement gives $T=\\Delta H/\\Delta S$, provided the units are consistent."},{"text":"$$$T = -\\frac{\\Delta H}{\\Delta S}$$ (the temperature where $\\Delta G = 0$)","correct":false,"explanation":"The extra negative sign comes from rearranging the Gibbs equation incorrectly. Setting $0=\\Delta H-T\\Delta S$ gives $T\\Delta S=\\Delta H$, so $T=\\Delta H/\\Delta S$."},{"text":"$$$T = \\frac{\\Delta S}{\\Delta H}$$ (the temperature where $\\Delta G = 0$)","correct":false,"explanation":"This inverts the ratio and does not produce units of temperature. Rearranging $0=\\Delta H-T\\Delta S$ gives the threshold temperature as $T=\\Delta H/\\Delta S$."},{"text":"$$$T = \\Delta H\\Delta S$$ (the temperature where $\\Delta G = 0$)","correct":false,"explanation":"Multiplying enthalpy and entropy does not follow from rearranging the Gibbs equation and gives incorrect units. At $\\Delta G=0$, the correct result is $T=\\Delta H/\\Delta S$."}]},{"id":"60c6a3b0-9d09-4fd4-a98e-770d31515b0c","cardType":"basic","points":2,"front":"If $\\Delta H$ is positive and $\\Delta S$ is positive, how does temperature affect feasibility?","back":"The reaction becomes feasible at high temperatures.","mcqOptions":[{"text":"The reaction becomes feasible at high temperatures.","correct":true,"explanation":"With positive $\\Delta H$ and $\\Delta S$, raising temperature makes the negative $-T\\Delta S$ term larger. Above the threshold temperature, it outweighs $\\Delta H$ and makes $\\Delta G$ negative."},{"text":"The reaction becomes feasible at low temperatures.","correct":false,"explanation":"At low temperatures, the positive enthalpy term dominates, leaving $\\Delta G$ positive. The reaction becomes feasible only at high temperatures, when $-T\\Delta S$ outweighs $\\Delta H$."},{"text":"The reaction is feasible at all temperatures.","correct":false,"explanation":"Positive entropy does favour feasibility, but positive enthalpy opposes it. The reaction is not feasible at all temperatures; it becomes feasible only above the threshold temperature."},{"text":"The reaction is feasible at no temperatures.","correct":false,"explanation":"This overlooks the growing contribution of the negative $-T\\Delta S$ term. Because $\\Delta S$ is positive, sufficiently high temperature makes $\\Delta G$ negative and the reaction feasible."}]},{"id":"f1258d77-c4f8-499c-b2b7-fcc026dc7706","cardType":"basic","points":2,"front":"Under what sign conditions for $\\Delta H$ and $\\Delta S$ is a reaction never feasible?","back":"

$\\Delta H$ is positive (endothermic)
$\\Delta S$ is negative

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

$\\Delta H$ is positive (endothermic)
$\\Delta S$ is negative

","correct":true,"explanation":"Positive $\\Delta H$ raises $\\Delta G$, while negative $\\Delta S$ makes $-T\\Delta S$ positive. Both contributions therefore keep $\\Delta G$ positive at every temperature."},{"text":"

$\\Delta H$ is positive (endothermic)
$\\Delta S$ is positive

","correct":false,"explanation":"Although positive $\\Delta H$ opposes feasibility, positive $\\Delta S$ favours it increasingly as temperature rises. Such a reaction becomes feasible at sufficiently high temperatures."},{"text":"

$\\Delta H$ is negative (exothermic)
$\\Delta S$ is negative

","correct":false,"explanation":"Negative $\\Delta H$ can make the reaction feasible at low temperatures, before the positive contribution from $-T\\Delta S$ becomes dominant. It is therefore not never feasible."},{"text":"

$\\Delta H$ is negative (exothermic)
$\\Delta S$ is positive

","correct":false,"explanation":"Negative $\\Delta H$ lowers $\\Delta G$, and positive $\\Delta S$ makes $-T\\Delta S$ negative. Both effects favour feasibility, so the reaction is feasible at all temperatures."}]}]}],"examCountdown":{"examDate":"2027-05-11","paperLabel":"Paper 1"},"heading":"Entropy","title":"Entropy Flashcards","topbar":{"backHref":"/as-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"66701290-6b5f-57ca-9588-aa7740b88172","walled":"$L38"}],"$L39","$L3a"]