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is a pair of opposing arrows, showing that the reaction can proceed in both directions. A one-way arrow would represent an irreversible reaction under the stated conditions."},{"text":"A reversible reaction is shown with $\\rightarrow$, not a two-way arrow.","correct":false,"explanation":"A single forward arrow shows only one direction, so it is the wrong symbol for a reversible reaction. The correct notation is opposing arrows, $\\rightleftharpoons$, not a one-way arrow."},{"text":"A reversible reaction is shown with $=$, not a one-way arrow.","correct":false,"explanation":"An equals sign does not show chemical change in two directions and is not used for reversible reactions. The correct notation is $\\rightleftharpoons$, with opposing arrows, not a one-way arrow."},{"text":"A reversible reaction is shown with $\\rightarrow$, not a one-way arrow.","correct":false,"explanation":"A two-way arrow is the correct idea for reversibility, so calling it the symbol used instead of a one-way arrow makes this completion wrong. The correct symbol is $\\rightleftharpoons$."}]},{"id":"6a158a39-30bc-4643-9d2b-d2741b04ac31","cardType":"basic","points":1,"front":"What does the $\\rightleftharpoons$ symbol not automatically mean?","back":"It does not mean the two sides are equal amounts.","mcqOptions":[{"text":"It does not mean the two sides are equal amounts.","correct":true,"explanation":"The opposing arrows show that both forward and reverse reactions can occur, but they give no information about the relative amounts of reactants and products. Those amounts depend on the equilibrium position."},{"text":"It does mean the two sides are equal amounts.","correct":false,"explanation":"Equilibrium does not require equal amounts of reactants and products, but this option reverses the statement. The symbol does not automatically mean the two sides are equal amounts."},{"text":"It does not mean both reactions can occur.","correct":false,"explanation":"The opposing arrows indicate that forward and reverse reactions can occur, so claiming they cannot is a misconception. The symbol does not automatically mean the two sides are equal amounts."},{"text":"It does mean products are always favoured.","correct":false,"explanation":"A reversible reaction may have more reactants or more products at equilibrium; the symbol does not guarantee that products are favoured. It does not automatically mean the two sides are equal amounts."}]},{"id":"4ac1bd69-5daa-4f80-969c-064b1ef28a02","cardType":"basic","points":1,"front":"In $A + B \\rightleftharpoons C + D$, what is the forward reaction?","back":"$A + B \\rightarrow C + D$.","mcqOptions":[{"text":"$A + B \\rightarrow C + D$.","correct":true,"explanation":"The forward reaction is defined as the reaction written from left to right, so reactants $A$ and $B$ form products $C$ and $D$."},{"text":"$C + D \\rightarrow A + B$.","correct":false,"explanation":"This reverses the direction of the reaction shown in the equation. The forward reaction goes from left to right: $A + B \\rightarrow C + D$."},{"text":"$A + C \\rightarrow B + D$.","correct":false,"explanation":"The reactants and products have been paired incorrectly rather than following the equation as written. The forward reaction is $A + B \\rightarrow C + D$."},{"text":"$A + B \\rightleftharpoons C + D$.","correct":false,"explanation":"This uses an equilibrium arrow instead of showing the forward direction only. The forward reaction is written with a single arrow: $A + B \\rightarrow C + D$."}]},{"id":"409fa4ec-a3ad-4761-890f-bf2f2c4e8f12","cardType":"basic","points":1,"front":"When $NH_4Cl(s)$ is heated, which gases form?","back":"$NH_3(g)$ and $HCl(g)$.","mcqOptions":[{"text":"$NH_3(g)$ and $HCl(g)$.","correct":true,"explanation":"Heating ammonium chloride causes it to dissociate into ammonia and hydrogen chloride gases: $NH_4Cl(s) \\rightleftharpoons NH_3(g) + HCl(g)$."},{"text":"$N_2(g)$ and $H_2(g)$.","correct":false,"explanation":"Nitrogen and hydrogen are not the products of heating ammonium chloride in this reversible dissociation. The gases formed are ammonia, $NH_3(g)$, and hydrogen chloride, $HCl(g)$."},{"text":"$NH_3(g)$ and $Cl_2(g)$.","correct":false,"explanation":"Chlorine gas is not released because the chloride remains in hydrogen chloride. Heating produces $NH_3(g)$ and $HCl(g)$, not ammonia and chlorine."},{"text":"$NH_4(g)$ and $HCl(g)$.","correct":false,"explanation":"$$NH_4(g)$ is not the gaseous product in this reaction; ammonium chloride dissociates into ammonia and hydrogen chloride. The correct gases are $NH_3(g)$ and $HCl(g)$."}]},{"id":"b178902c-2b70-45e5-a257-e73e8e0dfac8","cardType":"basic","points":1,"front":"At the cooler end after heating ammonium chloride, what reforms?","back":"White $NH_4Cl(s)$.","mcqOptions":[{"text":"White $NH_4Cl(s)$.","correct":true,"explanation":"At the cooler end, ammonia and hydrogen chloride recombine to form solid ammonium chloride, seen as a white deposit: $NH_3(g) + HCl(g) \\rightarrow NH_4Cl(s)$."},{"text":"Colourless $NH_3(g)$.","correct":false,"explanation":"Ammonia remains a colourless gas and does not form the solid deposit at the cooler end. The gases recombine to produce white solid $NH_4Cl(s)$."},{"text":"White $CuSO_4(s)$.","correct":false,"explanation":"Copper(II) sulfate is unrelated to the ammonium chloride equilibrium. Cooling allows ammonia and hydrogen chloride to recombine, forming white $NH_4Cl(s)$."},{"text":"Blue $NH_4Cl(s)$.","correct":false,"explanation":"The solid deposit is ammonium chloride, whose colour is white; it is not blue. The cooler end reforms white $NH_4Cl(s)$ when the two gases recombine."}]},{"id":"ac83d309-3f7c-48ff-9357-e5af13149f11","cardType":"cloze","points":null,"front":"Hydrated copper(II) sulfate is [ ]; anhydrous copper(II) sulfate is [ ].","back":"Hydrated copper(II) sulfate is blue; anhydrous copper(II) sulfate is white.","mcqOptions":[{"text":"Hydrated copper(II) sulfate is blue; anhydrous copper(II) sulfate is white.","correct":true,"explanation":"Hydrated copper(II) sulfate contains water of crystallisation and is blue. Removing that water by heating produces anhydrous copper(II) sulfate, which is white."},{"text":"Hydrated copper(II) sulfate is white; anhydrous copper(II) sulfate is blue.","correct":false,"explanation":"The colours have been reversed. Hydrated copper(II) sulfate is blue because of its water of crystallisation, whereas anhydrous copper(II) sulfate is white."},{"text":"Hydrated copper(II) sulfate is blue; anhydrous copper(II) sulfate is blue.","correct":false,"explanation":"Anhydrous copper(II) sulfate is white, not blue. The blue colour belongs to the hydrated compound, so the correct order is blue for hydrated and white for anhydrous."},{"text":"Hydrated copper(II) sulfate is white; anhydrous copper(II) sulfate is white.","correct":false,"explanation":"Removing water makes copper(II) sulfate white, so hydrated and anhydrous samples cannot both be white in this test. The hydrated form is blue and the anhydrous form is white."}]},{"id":"08604e1f-2896-4497-a567-aced2b9037ff","cardType":"basic","points":1,"front":"What does heating hydrated copper(II) sulfate drive off?","back":"Water from the crystals.","mcqOptions":[{"text":"Water from the crystals.","correct":true,"explanation":"Heating hydrated copper(II) sulfate removes its water of crystallisation. The remaining anhydrous copper(II) sulfate changes from blue to white."},{"text":"Oxygen from the crystals.","correct":false,"explanation":"Oxygen is not the substance responsible for hydration. Heating drives off water of crystallisation, leaving white anhydrous copper(II) sulfate."},{"text":"Copper from the crystals.","correct":false,"explanation":"Heating does not remove copper from the compound; copper remains in the anhydrous copper(II) sulfate. The substance driven off is water from the crystals."},{"text":"Hydrogen from the crystals.","correct":false,"explanation":"Hydrogen is not released as the crystals lose their water. The complete water molecules of crystallisation are driven off, leaving anhydrous copper(II) sulfate."}]},{"id":"24cd840c-ece5-46c1-aa05-5bcd526a5875","cardType":"basic","points":1,"front":"What happens when water is added to white anhydrous copper(II) sulfate?","back":"It becomes blue hydrated copper(II) sulfate.","mcqOptions":[{"text":"It becomes blue hydrated copper(II) sulfate.","correct":true,"explanation":"Water restores the water of crystallisation in anhydrous copper(II) sulfate. The solid therefore becomes blue hydrated copper(II) sulfate."},{"text":"It becomes white anhydrous copper(II) sulfate.","correct":false,"explanation":"Adding water does not leave the compound anhydrous; it hydrates the salt and produces its blue form. The correct result is blue hydrated copper(II) sulfate."},{"text":"It becomes black copper(II) oxide.","correct":false,"explanation":"Water is not incorporated to produce copper(II) oxide in this test. It restores hydrated copper(II) sulfate, which is blue."},{"text":"It becomes blue anhydrous copper(II) sulfate.","correct":false,"explanation":"The anhydrous salt is white before water is added, but hydration changes it to blue. It does not become a blue anhydrous compound; it becomes blue hydrated copper(II) sulfate."}]},{"id":"89e415f9-6b52-4ae2-bbe9-cd41c1bb1be4","cardType":"basic","points":2,"front":"Why is heating hydrated copper(II) sulfate in an open dish not dynamic equilibrium?","back":"Water vapour escapes, so the system is not sealed.","mcqOptions":[{"text":"Water vapour escapes, so the system is not sealed.","correct":true,"explanation":"Dynamic equilibrium requires a closed system. In an open dish, water vapour leaves, so it is unavailable to react with anhydrous copper(II) sulfate in the reverse reaction."},{"text":"Heat energy escapes, so the system is not sealed.","correct":false,"explanation":"A closed system may exchange heat with its surroundings, so energy transfer does not prevent dynamic equilibrium. The problem is that water vapour escapes from the open dish."},{"text":"Copper sulfate escapes, so the system is not sealed.","correct":false,"explanation":"Copper(II) sulfate remains in the dish because it is a non-volatile solid. Dynamic equilibrium is not established because water vapour escapes and cannot take part in the reverse reaction."},{"text":"Oxygen gas enters, so the system is not sealed.","correct":false,"explanation":"Oxygen entering is not the relevant change in this reversible hydration reaction. The system fails to reach dynamic equilibrium because water vapour escapes from the open dish."}]},{"id":"f85ef3fa-41c2-46d3-a10c-d8d24c01e274","cardType":"cloze","points":null,"front":"A sealed container is a [ ]: substances cannot [ ].","back":"A sealed container is a closed system: substances cannot enter or leave.","mcqOptions":[{"text":"A sealed container is a closed system: substances cannot enter or leave.","correct":true,"explanation":"A sealed container is a closed system because matter cannot cross its boundary. Reactants and products remain inside, although energy may still be transferred to or from the surroundings."},{"text":"A sealed container is a open system: substances cannot enter or leave.","correct":false,"explanation":"An open system allows substances to enter or leave, whereas sealing prevents matter transfer. A sealed container is therefore a closed system in which substances cannot enter or leave."},{"text":"A sealed container is a closed system: substances cannot exchange energy.","correct":false,"explanation":"A closed system can exchange energy with its surroundings while preventing matter transfer. A sealed container stops substances from entering or leaving, not necessarily energy from being exchanged."},{"text":"A sealed container is a isolated system: substances cannot gain or lose energy.","correct":false,"explanation":"An isolated system exchanges neither matter nor energy, which is a stronger condition than simply being sealed. A sealed container is a closed system, and substances cannot enter or leave."}]},{"id":"c61db5cb-77ba-4bae-ab66-f256964c3e5b","cardType":"basic","points":2,"front":"Why must gas or vapour stay inside for dynamic equilibrium?","back":"Particles remain available for the reverse reaction.","mcqOptions":[{"text":"Particles remain available for the reverse reaction.","correct":true,"explanation":"Keeping gases and vapours inside preserves product particles that can collide and reform the reactants. Their continued availability allows the reverse reaction to occur alongside the forward reaction."},{"text":"Particles remain available for the forward reaction.","correct":false,"explanation":"The forward reaction can proceed from the original reactants, but equilibrium also requires products to react backwards. Gas or vapour must remain available specifically for the reverse reaction."},{"text":"Particles remain available for the catalyst.","correct":false,"explanation":"A catalyst is not consumed and does not need escaping particles to remain available. Gas or vapour must stay inside so its particles can participate in the reverse reaction."},{"text":"Particles remain available for the stationary state.","correct":false,"explanation":"Equilibrium depends on particles continuing to collide, not becoming stationary. Retaining gas or vapour keeps particles available for the reverse reaction that reforms the reactants."}]},{"id":"5d176c70-9a4d-463f-8587-5e5f8fd9d9cf","cardType":"cloze","points":null,"front":"At dynamic equilibrium, forward and reverse reaction rates are [ ], so concentrations remain [ ].","back":"At dynamic equilibrium, forward and reverse reaction rates are equal, so concentrations remain constant.","mcqOptions":[{"text":"At dynamic equilibrium, forward and reverse reaction rates are equal, so concentrations remain constant.","correct":true,"explanation":"Equal forward and reverse reaction rates mean reactants and products are formed as quickly as they are consumed. Their concentrations therefore remain constant, although reactions continue microscopically."},{"text":"At dynamic equilibrium, forward and reverse reaction rates are unequal, so concentrations remain constant.","correct":false,"explanation":"Unequal reaction rates would produce a net change in composition, so concentrations could not remain constant. At dynamic equilibrium, the two rates are equal and concentrations are constant."},{"text":"At dynamic equilibrium, forward and reverse reaction rates are equal, so concentrations remain equal.","correct":false,"explanation":"Equal reaction rates keep concentrations constant but do not make reactant and product concentrations equal. The equilibrium mixture can contain very different amounts of reactants and products."},{"text":"At dynamic equilibrium, forward and reverse reaction rates are zero, so concentrations remain constant.","correct":false,"explanation":"Calling both rates zero confuses dynamic equilibrium with a reaction that has stopped. Both reactions continue at equal non-zero rates, causing concentrations to remain constant."}]},{"id":"c24735f5-d8dd-4a7c-b2a9-ff667981dbab","cardType":"basic","points":1,"front":"At equilibrium, must reactant and product concentrations be equal?","back":"No - concentrations are constant, not necessarily equal.","mcqOptions":[{"text":"No - concentrations are constant, not necessarily equal.","correct":true,"explanation":"Equilibrium requires the forward and reverse rates to be equal, which keeps each concentration constant. It does not require reactants and products to have identical concentrations."},{"text":"Yes - concentrations are equal and remain constant.","correct":false,"explanation":"This confuses equal reaction rates with equal concentrations. At equilibrium, concentrations remain constant, but their actual values depend on the equilibrium position and need not be equal."},{"text":"No - concentrations are changing, although rates are equal.","correct":false,"explanation":"Concentrations stop changing once equilibrium has been reached because the forward and reverse rates are equal. They are constant over time, though not necessarily equal to one another."},{"text":"Yes - concentrations are constant, so they must be equal.","correct":false,"explanation":"Constant concentrations do not imply equal concentrations; each substance simply maintains its own unchanged value. At equilibrium, reactant and product concentrations can differ substantially."}]},{"id":"a3855b74-45ce-4cc1-b2da-4e86ee17ef08","cardType":"basic","points":1,"front":"If the position of equilibrium lies to the right, which side is favoured?","back":"Products are favoured.","mcqOptions":[{"text":"Products are favoured.","correct":true,"explanation":"An equilibrium position to the right means the equilibrium mixture contains a greater proportion of products than reactants. The products are therefore described as favoured."},{"text":"Reactants are favoured.","correct":false,"explanation":"Reactants are favoured when equilibrium lies to the left, not the right. A position to the right indicates that the equilibrium mixture favours products."},{"text":"Both sides are favoured.","correct":false,"explanation":"Equal forward and reverse rates occur at every dynamic equilibrium, but this does not mean both sides are equally favoured. A right-hand position specifically favours products."},{"text":"Neither side is favoured.","correct":false,"explanation":"An equilibrium position describes which side has the greater equilibrium proportion, so one side can be favoured. If it lies to the right, products are favoured."}]},{"id":"49d7e098-57d9-45e0-9851-f1ddf2685522","cardType":"basic","points":1,"front":"In a reversible reaction, how does a catalyst affect the forward and reverse reactions?","back":"It speeds up both reactions.","mcqOptions":[{"text":"It speeds up both reactions.","correct":true,"explanation":"A catalyst provides an alternative pathway with lower activation energy for both directions. It therefore accelerates both reactions equally and allows equilibrium to be reached sooner."},{"text":"It speeds up only the forward reaction.","correct":false,"explanation":"A catalyst does not selectively accelerate product formation. It lowers the activation energy for both the forward and reverse pathways, so both reactions speed up."},{"text":"It speeds up only the reverse reaction.","correct":false,"explanation":"A catalyst affects both directions because the same alternative pathway can be followed forwards and backwards. It speeds up both reactions, rather than only the reverse reaction."},{"text":"It slows down both reactions.","correct":false,"explanation":"Catalysts lower activation energy and increase reaction rates; they do not slow reactions. In a reversible reaction, both the forward and reverse reactions are accelerated."}]},{"id":"513a808b-ac7d-4e16-b10d-75de2dc10e12","cardType":"basic","points":1,"front":"What happens to the position of equilibrium when a catalyst is added?","back":"The position of equilibrium is unchanged.","mcqOptions":[{"text":"The position of equilibrium is unchanged.","correct":true,"explanation":"A catalyst increases the forward and reverse rates equally, so it does not change the equilibrium composition. It only reduces the time needed to reach equilibrium."},{"text":"The position of equilibrium shifts right.","correct":false,"explanation":"A catalyst does not favour product formation because it accelerates both directions equally. The position of equilibrium remains unchanged, although equilibrium is reached more quickly."},{"text":"The position of equilibrium shifts left.","correct":false,"explanation":"A catalyst does not favour the reactants or shift equilibrium left. By increasing both reaction rates equally, it leaves the position of equilibrium unchanged."},{"text":"The position of equilibrium keeps changing.","correct":false,"explanation":"A catalyst helps the system reach equilibrium faster rather than preventing equilibrium. The equilibrium position remains unchanged because both forward and reverse reactions are accelerated equally."}]},{"id":"0dee6c2a-7dbe-45a5-9856-8346ad5048f6","cardType":"cloze","points":null,"front":"A catalyst provides a [ ] pathway and is not [ ].","back":"A catalyst provides a lower activation energy pathway and is not used up.","mcqOptions":[{"text":"A catalyst provides a lower activation energy pathway and is not used up.","correct":true,"explanation":"A catalyst offers an alternative reaction pathway with lower activation energy, increasing the reaction rate. It remains chemically unchanged overall, so it is not used up."},{"text":"A catalyst provides a higher activation energy pathway and is not used up.","correct":false,"explanation":"Catalysts do not increase activation energy; they provide a pathway with lower activation energy. They are not used up, so only the first part of this statement is wrong."},{"text":"A catalyst provides a lower activation energy pathway and is not changed into products.","correct":false,"explanation":"The activation-energy description is correct, but catalysts are not defined by avoiding conversion into products. The required statement is that a catalyst is not used up during the reaction."},{"text":"A catalyst provides a higher activation energy pathway and is not changed into products.","correct":false,"explanation":"Both parts are incorrect: a catalyst lowers activation energy, and the standard description is that it is not used up. These properties explain how it speeds reactions without being consumed."}]},{"id":"47d63593-9340-4e76-a3d6-9298917ecacc","cardType":"cloze","points":null,"front":"An endothermic reaction [ ]; an exothermic reaction [ ].","back":"An endothermic reaction takes in heat; an exothermic reaction releases heat.","mcqOptions":[{"text":"An endothermic reaction takes in heat; an exothermic reaction releases heat.","correct":true,"explanation":"Endothermic reactions absorb heat energy from the surroundings, whereas exothermic reactions transfer heat energy to the surroundings."},{"text":"An endothermic reaction releases heat; an exothermic reaction takes in heat.","correct":false,"explanation":"The two definitions have been reversed. Endothermic reactions take in heat, while exothermic reactions release heat to the surroundings."},{"text":"An endothermic reaction takes in light; an exothermic reaction releases heat.","correct":false,"explanation":"The exothermic description is correct, but endothermic refers to taking in heat, not specifically light. Light may provide energy in some reactions but is not the definition."},{"text":"An endothermic reaction releases heat; an exothermic reaction takes in light.","correct":false,"explanation":"Both descriptions are wrong for the required definitions. Endothermic reactions take in heat, and exothermic reactions release heat, regardless of whether light is involved."}]},{"id":"14ec3dc0-5408-466b-b9c6-7d4dd64777c3","cardType":"cloze","points":null,"front":"Increasing temperature favours the [ ] direction; decreasing temperature favours the [ ] direction.","back":"Increasing temperature favours the endothermic direction; decreasing temperature favours the exothermic direction.","mcqOptions":[{"text":"Increasing temperature favours the endothermic direction; decreasing temperature favours the exothermic direction.","correct":true,"explanation":"Adding heat by increasing temperature favours the endothermic direction, which absorbs heat. Removing heat by decreasing temperature favours the exothermic direction, which releases heat."},{"text":"Increasing temperature favours the exothermic direction; decreasing temperature favours the endothermic direction.","correct":false,"explanation":"The temperature effects have been reversed. An increase favours the endothermic direction because it absorbs added heat, while a decrease favours the exothermic direction."},{"text":"Increasing temperature favours the endothermic direction; decreasing temperature favours the endothermic direction.","correct":false,"explanation":"Only the first part is correct. Increasing temperature favours the endothermic direction, but decreasing temperature removes heat and therefore favours the exothermic direction."},{"text":"Increasing temperature favours the exothermic direction; decreasing temperature favours the exothermic direction.","correct":false,"explanation":"Only the second part has the correct direction. Increasing temperature favours the endothermic direction, whereas decreasing temperature favours the exothermic direction."}]},{"id":"2f5a6c14-5ca0-47b7-99b8-6464f22e5cb9","cardType":"basic","points":2,"front":"For $CuSO_4 \\cdot 5H_2O(s) \\rightleftharpoons CuSO_4(s) + 5H_2O(g)$, what does increasing temperature do?","back":"It shifts right, forming more $CuSO_4(s)$ and $H_2O(g)$.","mcqOptions":[{"text":"It shifts right, forming more $CuSO_4(s)$ and $H_2O(g)$.","correct":true,"explanation":"The forward reaction is endothermic because heating hydrated copper sulfate removes water. Increasing temperature therefore shifts equilibrium right, producing more anhydrous $CuSO_4$ and water vapour."},{"text":"It shifts left, forming more $CuSO_4 \\cdot 5H_2O(s)$.","correct":false,"explanation":"Heating favours the endothermic forward direction, which removes water from hydrated copper sulfate. The equilibrium therefore shifts right, not left, and forms more $CuSO_4(s)$ and $H_2O(g)$."},{"text":"It shifts right, forming more $CuSO_4 \\cdot 5H_2O(s)$.","correct":false,"explanation":"The direction is correct, but the products are not. A rightward shift forms anhydrous $CuSO_4(s)$ and $H_2O(g)$, rather than reforming hydrated copper sulfate."},{"text":"It shifts left, forming more $CuSO_4(s)$ and $H_2O(g)$.","correct":false,"explanation":"The products shown are formed by a rightward shift, not a leftward shift. Increasing temperature favours the endothermic forward reaction, so the equilibrium shifts right."}]},{"id":"a2422b58-411a-4430-93c5-610c926ff6f1","cardType":"basic","points":1,"front":"When predicting pressure effects, which states should be counted?","back":"Gases only.","mcqOptions":[{"text":"Gases only.","correct":true,"explanation":"Pressure changes affect the equilibrium position only when gases are involved. Solids and liquids have negligible volume changes and are not counted when comparing gas moles."},{"text":"Solids only.","correct":false,"explanation":"Solids are not counted for pressure effects because their volumes change negligibly. Count the gaseous species only, since pressure changes influence the number of gas moles."},{"text":"Liquids and gases.","correct":false,"explanation":"Liquids, like solids, are not counted when predicting pressure effects because their volumes are nearly unchanged. Only gases contribute to the relevant comparison of moles."},{"text":"All substances.","correct":false,"explanation":"Counting every substance incorrectly includes solids and liquids. Pressure effects are predicted by comparing the number of gaseous moles on each side only."}]},{"id":"3d0f25a8-593f-4ad6-b818-ae19f435984f","cardType":"cloze","points":null,"front":"Increasing pressure shifts equilibrium towards [ ]; decreasing pressure shifts it towards [ ].","back":"Increasing pressure shifts equilibrium towards fewer moles of gas; decreasing pressure shifts it towards more moles of gas.","mcqOptions":[{"text":"Increasing pressure shifts equilibrium towards fewer moles of gas; decreasing pressure shifts it towards more moles of gas.","correct":true,"explanation":"A pressure increase favours the side occupying less gas volume, meaning fewer gas moles. A pressure decrease favours the side with more gas moles."},{"text":"Increasing pressure shifts equilibrium towards more moles of gas; decreasing pressure shifts it towards fewer moles of gas.","correct":false,"explanation":"The pressure effects have been reversed. Increasing pressure favours fewer gas moles because this reduces volume, while decreasing pressure favours more gas moles."},{"text":"Increasing pressure shifts equilibrium towards fewer moles of gas; decreasing pressure shifts it towards fewer moles of gas.","correct":false,"explanation":"The first effect is correct, but decreasing pressure favours more gas moles, not fewer. The system responds to oppose the pressure change by increasing its gas volume."},{"text":"Increasing pressure shifts equilibrium towards more moles of gas; decreasing pressure shifts it towards more moles of gas.","correct":false,"explanation":"Only the decreasing-pressure part is correct. Increasing pressure favours fewer gas moles, whereas decreasing pressure favours more gas moles."}]},{"id":"9ea7d867-e862-478f-bb70-f9476c067293","cardType":"basic","points":2,"front":"For $N_2(g) + 3H_2(g) \\rightleftharpoons 2NH_3(g)$, what does increasing pressure do?","back":"It shifts right, towards fewer gas moles.","mcqOptions":[{"text":"It shifts right, towards fewer gas moles.","correct":true,"explanation":"The left side contains four moles of gas, while the right side contains two. Increasing pressure favours the side with fewer gas moles, so equilibrium shifts right towards ammonia."},{"text":"It shifts left, towards more gas moles.","correct":false,"explanation":"Increasing pressure does not favour the side with more gas moles. There are four gaseous moles on the left and two on the right, so equilibrium shifts right."},{"text":"It has no effect because both sides contain gases.","correct":false,"explanation":"The presence of gases on both sides does not make pressure irrelevant. The left has four gas moles and the right has two, so increasing pressure shifts equilibrium right."},{"text":"It shifts right, towards more gas moles.","correct":false,"explanation":"The direction is right, but the reason is incorrect. The right side contains fewer gas moles, two compared with four on the left, and increased pressure favours that side."}]},{"id":"9f98fca1-2f45-4406-bc3b-d255b206e96b","cardType":"basic","points":1,"front":"If both sides have the same number of gas moles, what is the effect of changing pressure?","back":"It has no effect on the equilibrium position.","mcqOptions":[{"text":"It has no effect on the equilibrium position.","correct":true,"explanation":"When both sides contain the same number of gas moles, a pressure change favours neither side. The equilibrium position therefore remains unchanged."},{"text":"It shifts equilibrium right, towards fewer gas moles.","correct":false,"explanation":"A pressure change shifts equilibrium only when the sides contain different numbers of gas moles. Equal gas-mole totals mean neither direction is favoured, so there is no effect."},{"text":"It shifts equilibrium left, towards fewer gas moles.","correct":false,"explanation":"There is no fewer-gas-mole side when both sides have equal totals. Consequently, changing pressure does not shift the equilibrium position in either direction."},{"text":"It shifts equilibrium right, towards more gas moles.","correct":false,"explanation":"Pressure changes do not favour more gas moles, and equal gas-mole totals provide no direction to favour. The equilibrium position therefore has no effect from the pressure change."}]}]}],"examCountdown":{"examDate":"2027-05-19","paperLabel":"Paper 1"},"heading":"Reversible reactions and equilibria","title":"Reversible reactions and equilibria Flashcards","topbar":{"backHref":"/igcse/chemistry/edexcel/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"c932388e-cc1d-5e1f-a104-01bdf1487234","walled":"$L37"}],"$L38","$L39"]