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but no matter can enter or leave.","correct":true,"explanation":"A closed system can exchange energy with its surroundings, for example as heat, while preventing reactants and products from entering or escaping."},{"text":"A closed system allows matter to be transferred, but no energy can enter or leave.","correct":false,"explanation":"This reverses the definitions of closed and isolated systems. A closed system allows energy transfer, but matter cannot enter or leave the system."},{"text":"A closed system allows particles to be transferred, but no heat can enter or leave.","correct":false,"explanation":"Particles are matter, so transferring them would make the system open. A closed system can transfer energy, including heat, but cannot exchange matter."},{"text":"A closed system allows products to be transferred, but no reactants can enter or leave.","correct":false,"explanation":"A closed system prevents both reactants and products from crossing its boundary. Only energy may be transferred between the system and its surroundings."}]},{"id":"3bf770a4-a657-49a9-9fb3-37604faa6561","cardType":"basic","points":2,"front":"Why must an equilibrium mixture be in a closed system?","back":"Reactants and products remain available to react both ways.","mcqOptions":[{"text":"Reactants and products remain available to react both ways.","correct":true,"explanation":"Preventing reactants and products from escaping means both forward and reverse reactions can continue, allowing their rates eventually to become equal."},{"text":"Reactants and products remain at equal concentrations throughout.","correct":false,"explanation":"Equilibrium concentrations are constant but are not necessarily equal. The system must be closed so reactants and products remain available for both reaction directions."},{"text":"Reactants and products stop colliding after equilibrium.","correct":false,"explanation":"Dynamic equilibrium does not stop particle collisions or reactions. A closed system retains reactants and products so forward and reverse reactions can continue."},{"text":"Reactants and products are continually replaced from outside.","correct":false,"explanation":"Adding substances from outside contradicts the definition of a closed system. Equilibrium requires reactants and products to remain available inside and react both ways."}]},{"id":"5b009101-ab05-4433-816b-e7f41cafd7a7","cardType":"basic","points":1,"front":"In a reversible reaction, what can products do?","back":"Products can reform reactants under the same conditions.","mcqOptions":[{"text":"Products can reform reactants under the same conditions.","correct":true,"explanation":"A reversible reaction proceeds in both directions under the same conditions, so product particles can collide successfully and form the original reactants again."},{"text":"Products can form more products under the same conditions.","correct":false,"explanation":"This confuses continuation of the forward reaction with reversibility. In the reverse reaction, products reform reactants under the same conditions."},{"text":"Products can stop all reactions under the same conditions.","correct":false,"explanation":"Products do not automatically halt a reversible reaction. They can undergo successful collisions that reform reactants while the forward reaction also continues."},{"text":"Products can reform catalysts under the same conditions.","correct":false,"explanation":"A catalyst is regenerated during a catalytic mechanism, but that does not define reversibility. In a reversible reaction, products can reform the original reactants."}]},{"id":"d3f812b0-da35-47e2-8fb8-0628cc690fdd","cardType":"basic","points":2,"front":"Starting with only reactants, why is the forward reaction initially fast?","back":"Reactant concentration is high.","mcqOptions":[{"text":"Reactant concentration is high.","correct":true,"explanation":"Initially, many reactant particles occupy a given volume, producing frequent successful collisions and therefore a high rate for the forward reaction."},{"text":"Product concentration is high.","correct":false,"explanation":"Starting with only reactants means product concentration is initially zero. The forward reaction is fast because the reactant concentration, and hence collision frequency, is high."},{"text":"Reverse activation energy is low.","correct":false,"explanation":"The initial forward rate depends primarily on the high reactant concentration, not the reverse activation energy. Initially there are no products to undergo the reverse reaction."},{"text":"Equilibrium has already been reached.","correct":false,"explanation":"An initial mixture containing only reactants is not normally at equilibrium. Its high reactant concentration causes frequent collisions and a fast forward reaction."}]},{"id":"73f326f8-10b9-4b80-b939-6797203f937b","cardType":"basic","points":1,"front":"At dynamic equilibrium, how do the forward and reverse reaction rates compare?","back":"They are equal.","mcqOptions":[{"text":"They are equal.","correct":true,"explanation":"Dynamic equilibrium is reached when forward and reverse reactions occur at identical rates, so there is no overall change in reactant or product concentrations."},{"text":"They are both zero.","correct":false,"explanation":"This confuses dynamic equilibrium with reactions stopping. Both reactions continue, but their rates are equal, producing no net concentration change."},{"text":"They are constantly increasing.","correct":false,"explanation":"Reaction rates do not need to increase at equilibrium. The defining condition is that the forward and reverse rates are equal and remain balanced."},{"text":"They are equal to concentrations.","correct":false,"explanation":"Rates and concentrations are different quantities with different units. At dynamic equilibrium the forward and reverse reaction rates are equal, while concentrations remain constant."}]},{"id":"746c33cc-3447-44f8-baca-33a409028699","cardType":"cloze","points":null,"front":"At dynamic equilibrium, concentrations are [ ], but particles still react in [ ].","back":"At dynamic equilibrium, concentrations are constant, but particles still react in both directions.","mcqOptions":[{"text":"At dynamic equilibrium, concentrations are constant, but particles still react in both directions.","correct":true,"explanation":"Equal forward and reverse rates cause no net concentration change, although individual particles continue converting between reactants and products in both directions."},{"text":"At dynamic equilibrium, concentrations are equal, but particles still react in both directions.","correct":false,"explanation":"Equilibrium concentrations do not have to be equal; their values depend on the equilibrium position. They are constant because both reaction directions have equal rates."},{"text":"At dynamic equilibrium, concentrations are constant, but particles still react in one direction.","correct":false,"explanation":"Constant concentrations cannot be maintained by a reaction proceeding in only one direction. At dynamic equilibrium, particles react in both directions at equal rates."},{"text":"At dynamic equilibrium, concentrations are decreasing, but particles still react in opposite directions.","correct":false,"explanation":"Equal opposing reaction rates prevent any overall concentration decrease. Concentrations remain constant while particles continue reacting in both the forward and reverse directions."}]},{"id":"613d2f2d-d2a6-4234-b7c9-63e9d585af48","cardType":"basic","points":1,"front":"If the forward rate is greater than the reverse rate, what happens to product concentration overall?","back":"It increases.","mcqOptions":[{"text":"It increases.","correct":true,"explanation":"When the forward reaction is faster, products are formed faster than they are consumed by the reverse reaction, so their concentration increases overall."},{"text":"It decreases.","correct":false,"explanation":"A decrease would occur if the reverse rate exceeded the forward rate. Here products form faster than they are consumed, so their concentration increases."},{"text":"It remains constant.","correct":false,"explanation":"Constant product concentration requires equal forward and reverse rates. A greater forward rate creates a net production of products, increasing their concentration."},{"text":"It falls to zero.","correct":false,"explanation":"This wrongly treats the faster forward reaction as product removal. The forward reaction produces products, so a greater forward rate makes product concentration increase."}]},{"id":"1372fb58-832f-4a06-92f0-5541f2c53419","cardType":"cloze","points":null,"front":"Le Chatelier’s principle: an equilibrium shifts to [ ].","back":"Le Chatelier’s principle: an equilibrium shifts to oppose the change.","mcqOptions":[{"text":"oppose the change","correct":true,"explanation":"An equilibrium responds to a changed condition by shifting in the direction that reduces the effect of that change and establishes a new equilibrium."},{"text":"reinforce the change","correct":false,"explanation":"This reverses Le Chatelier’s principle. The equilibrium shifts in the direction that opposes the imposed change rather than making its effect greater."},{"text":"stop all reactions","correct":false,"explanation":"A disturbance does not stop the forward and reverse reactions. The equilibrium shifts to oppose the change until equal reaction rates are re-established."},{"text":"favour the products","correct":false,"explanation":"The shift is not always towards products; its direction depends on the disturbance. Le Chatelier’s principle states that equilibrium shifts to oppose the change."}]},{"id":"2bb7e3e1-1af6-4761-a014-f60bd8e37503","cardType":"basic","points":1,"front":"If more of a reactant is added, which way does equilibrium shift?","back":"Towards products.","mcqOptions":[{"text":"Towards products.","correct":true,"explanation":"Adding a reactant increases its concentration, so equilibrium shifts towards the products to consume some of the added reactant and oppose the concentration change."},{"text":"Towards reactants.","correct":false,"explanation":"Shifting towards reactants would produce more of the substance that was added. Equilibrium instead shifts towards products, consuming some of the additional reactant."},{"text":"Towards neither side.","correct":false,"explanation":"Changing a reactant concentration disturbs equilibrium, so a shift occurs. The system moves towards products to reduce the concentration of the added reactant."},{"text":"Towards the endothermic side.","correct":false,"explanation":"The endothermic direction determines the response to temperature, not concentration. Adding a reactant shifts equilibrium towards products so that some of it is consumed."}]},{"id":"c799ed9d-3d89-4f13-8690-e1f44fa4c66c","cardType":"basic","points":1,"front":"Increasing pressure in a gaseous equilibrium shifts equilibrium to which side?","back":"The side with fewer moles of gas.","mcqOptions":[{"text":"The side with fewer moles of gas.","correct":true,"explanation":"Increasing pressure favours the side containing fewer moles of gas because shifting to that side reduces the total number of gaseous particles and opposes the pressure increase."},{"text":"The side with more moles of gas.","correct":false,"explanation":"The side with more gas moles is favoured when pressure decreases. Increasing pressure shifts equilibrium towards fewer moles of gas to reduce the number of gaseous particles."},{"text":"The side with lower gas concentration.","correct":false,"explanation":"Pressure effects are predicted by comparing stoichiometric moles of gas, not individual gas concentrations. Increased pressure favours the side with fewer total moles of gas."},{"text":"The endothermic side of the equilibrium.","correct":false,"explanation":"The endothermic side is relevant to a temperature increase, not a pressure increase. Higher pressure favours whichever side has fewer moles of gas."}]},{"id":"5b51fc84-433b-40e6-8488-3a2d13b32e62","cardType":"cloze","points":null,"front":"If both sides of a gaseous equilibrium have the same total gas moles, pressure has [ ] on equilibrium position.","back":"If both sides of a gaseous equilibrium have the same total gas moles, pressure has no effect on equilibrium position.","mcqOptions":[{"text":"no effect","correct":true,"explanation":"When both sides contain the same total moles of gas, neither direction reduces the number of gaseous particles, so changing pressure does not shift equilibrium."},{"text":"a large effect","correct":false,"explanation":"A pressure change only shifts equilibrium when the two sides have different total gas moles. With equal gas moles, pressure has no effect on equilibrium position."},{"text":"an increasing effect","correct":false,"explanation":"Higher pressure does not automatically favour products. If both sides have equal total moles of gas, neither side is favoured and pressure has no effect."},{"text":"a decreasing effect","correct":false,"explanation":"Lowering or raising pressure cannot favour either side when their total gas moles are equal. Pressure therefore has no effect on the equilibrium position."}]},{"id":"b2e9e03c-1fd0-4f2d-a7e6-b09f6bab6f9a","cardType":"basic","points":1,"front":"For an exothermic forward reaction, what does increasing temperature favour?","back":"The reverse endothermic direction.","mcqOptions":[{"text":"The reverse endothermic direction.","correct":true,"explanation":"Increasing temperature adds heat, so equilibrium shifts in the endothermic direction to absorb it. If the forward reaction is exothermic, the reverse reaction is endothermic."},{"text":"The forward exothermic direction.","correct":false,"explanation":"This treats heating as favouring heat production. Increasing temperature instead favours the reverse endothermic direction, which absorbs some of the added thermal energy."},{"text":"The reverse exothermic direction.","correct":false,"explanation":"If the forward reaction is exothermic, the reverse reaction must be endothermic, not exothermic. Heating favours this reverse endothermic direction."},{"text":"The forward endothermic direction.","correct":false,"explanation":"The forward reaction is specified as exothermic, so it cannot also be endothermic. Increasing temperature favours the reverse direction because that direction is endothermic."}]},{"id":"b2427c8c-d362-4f67-8d9d-400898c82c71","cardType":"basic","points":1,"front":"For an endothermic forward reaction, which direction does increasing temperature favour?","back":"The forward direction.","mcqOptions":[{"text":"The forward direction.","correct":true,"explanation":"Increasing temperature favours the endothermic direction because it absorbs added heat. As the forward reaction is endothermic, equilibrium shifts forwards."},{"text":"The reverse direction.","correct":false,"explanation":"The reverse direction is exothermic when the forward direction is endothermic. Increasing temperature instead favours the forward endothermic direction, which absorbs added heat."},{"text":"The exothermic direction.","correct":false,"explanation":"This reverses the temperature rule. Added heat favours the endothermic direction, so an endothermic forward reaction is favoured in the forward direction."},{"text":"The lower-pressure direction.","correct":false,"explanation":"Pressure and temperature changes follow different rules. Increasing temperature favours the endothermic direction, which here is the forward reaction, regardless of pressure considerations."}]},{"id":"58e4d677-73a3-4c01-b4b3-8f0a3b5bcaa7","cardType":"basic","points":2,"front":"Why does a catalyst not change equilibrium position?","back":"It speeds forward and reverse reactions by the same factor.","mcqOptions":[{"text":"It speeds forward and reverse reactions by the same factor.","correct":true,"explanation":"A catalyst provides an alternative pathway for both directions, increasing both rates by the same factor. Equilibrium is reached faster, but its position remains unchanged."},{"text":"It speeds only the forward reaction until equilibrium is reached.","correct":false,"explanation":"A catalyst does not selectively accelerate the forward reaction. It speeds both forward and reverse reactions by the same factor, so equilibrium position remains unchanged."},{"text":"It slows forward and reverse reactions by the same factor.","correct":false,"explanation":"A catalyst increases reaction rates rather than slowing them. It speeds both directions by the same factor, allowing equilibrium to be reached faster without shifting it."},{"text":"It changes forward and reverse reactions by different factors.","correct":false,"explanation":"Different rate changes would alter the balance and shift equilibrium. A catalyst speeds forward and reverse reactions by the same factor, leaving equilibrium position unchanged."}]},{"id":"ffb59d25-b776-4de5-b92b-c0525d4828a7","cardType":"cloze","points":null,"front":"In a homogeneous equilibrium, all species are in the [ ].","back":"In a homogeneous equilibrium, all species are in the same physical state.","mcqOptions":[{"text":"same physical state","correct":true,"explanation":"A homogeneous equilibrium contains all reactants and products in the same physical state, such as an entirely gaseous mixture or an entirely aqueous mixture."},{"text":"different physical states","correct":false,"explanation":"Species in different physical states form a heterogeneous equilibrium. In a homogeneous equilibrium, every reactant and product is in the same physical state."},{"text":"equal molar concentrations","correct":false,"explanation":"Homogeneous describes physical states, not concentration values. The equilibrium concentrations may differ, but all species must be in the same physical state."},{"text":"the solid state","correct":false,"explanation":"A homogeneous equilibrium is not necessarily solid and may be entirely gaseous or aqueous. Its defining feature is that all species share the same physical state."}]},{"id":"7dbb731b-c0e2-4611-9d34-9eab249192b5","cardType":"basic","points":1,"front":"For $a\\,\\text{A}+b\\,\\text{B}\\rightleftharpoons c\\,\\text{C}+d\\,\\text{D}$, what is $K_c$?","back":"$K_c=\\frac{[\\text{C}]^c[\\text{D}]^d}{[\\text{A}]^a[\\text{B}]^b}$","mcqOptions":[{"text":"$K_c=\\frac{[\\text{C}]^c[\\text{D}]^d}{[\\text{A}]^a[\\text{B}]^b}$","correct":true,"explanation":"The equilibrium concentrations of products are placed in the numerator and reactants in the denominator, with each concentration raised to its stoichiometric coefficient."},{"text":"$K_c=\\frac{[\\text{A}]^a[\\text{B}]^b}{[\\text{C}]^c[\\text{D}]^d}$","correct":false,"explanation":"This inverts the equilibrium expression by placing reactants above products. For the forward reaction, product concentrations belong in the numerator and reactants in the denominator."},{"text":"$K_c=\\frac{[\\text{C}][\\text{D}]}{[\\text{A}][\\text{B}]}$","correct":false,"explanation":"This omits the stoichiometric coefficients as powers. Each equilibrium concentration must be raised to its coefficient, giving powers $c$, $d$, $a$ and $b$ respectively."},{"text":"$K_c=\\frac{[\\text{C}]^a[\\text{D}]^b}{[\\text{A}]^c[\\text{B}]^d}$","correct":false,"explanation":"The powers have been transferred by position rather than attached to their species. Each concentration is raised to its own coefficient: C to $c$, D to $d$, A to $a$ and B to $b$."}]},{"id":"588438dd-ae6b-4396-af6d-6503837ac6a5","cardType":"cloze","points":null,"front":"In $K_c$ expressions, square brackets mean [ ], and powers come from the [ ].","back":"In $K_c$ expressions, square brackets mean equilibrium concentration, and powers come from the balanced equation.","mcqOptions":[{"text":"In $K_c$ expressions, square brackets mean equilibrium concentration, and powers come from the balanced equation.","correct":true,"explanation":"Square brackets denote concentrations measured at equilibrium, while each power is the corresponding species’ stoichiometric coefficient in the balanced chemical equation."},{"text":"In $K_c$ expressions, square brackets mean initial concentration, and powers come from the balanced equation.","correct":false,"explanation":"Using initial concentrations ignores the changes occurring as equilibrium is established. Square brackets represent equilibrium concentration, while powers come from the balanced equation."},{"text":"In $K_c$ expressions, square brackets mean equilibrium concentration, and powers come from the rate equation.","correct":false,"explanation":"Reaction orders determine powers in rate equations, not equilibrium expressions. In a $K_c$ expression, brackets mean equilibrium concentration and powers come from the balanced equation."},{"text":"In $K_c$ expressions, square brackets mean initial concentration, and powers come from the rate equation.","correct":false,"explanation":"This confuses equilibrium expressions with kinetic calculations. Square brackets mean equilibrium concentration, and the powers are fixed by stoichiometric coefficients in the balanced equation."}]},{"id":"b5f59cfb-9db8-4435-bc7e-cd040f344a43","cardType":"basic","points":1,"front":"Which concentrations must be substituted into a $K_c$ expression?","back":"Equilibrium concentrations, not starting concentrations.","mcqOptions":[{"text":"Equilibrium concentrations, not starting concentrations.","correct":true,"explanation":"The equilibrium constant describes the mixture once equilibrium has been established, so every concentration inserted into the expression must be an equilibrium concentration."},{"text":"Starting concentrations, not equilibrium concentrations.","correct":false,"explanation":"Starting concentrations change as the forward and reverse reactions proceed. $K_c$ must be calculated using equilibrium concentrations, not the values present initially."},{"text":"Concentrations at completion, not equilibrium concentrations.","correct":false,"explanation":"An equilibrium reaction does not normally proceed to completion, so completion values are unsuitable. The expression requires the concentrations present at equilibrium."},{"text":"Average concentrations, not equilibrium concentrations.","correct":false,"explanation":"Average concentrations do not describe the specific equilibrium composition and therefore cannot be used. All values substituted into a $K_c$ expression must be equilibrium concentrations."}]},{"id":"537f1c9b-eb88-4cbe-bf9e-44642c123c83","cardType":"basic","points":1,"front":"What formula converts amount in moles into concentration for a $K_c$ calculation?","back":"$c=\\frac{n}{V}$","mcqOptions":[{"text":"$c=\\frac{n}{V}$","correct":true,"explanation":"Concentration is amount per unit volume, so dividing the amount in moles by the volume in dm$^3$ gives concentration in mol dm$^{-3}$."},{"text":"$c=nV$","correct":false,"explanation":"Multiplying by volume reverses the required operation and gives unsuitable units. Concentration is moles divided by volume, so $c=\\frac{n}{V}$."},{"text":"$c=\\frac{V}{n}$","correct":false,"explanation":"This calculates volume per mole rather than moles per unit volume. The concentration needed for $K_c$ is calculated using $c=\\frac{n}{V}$."},{"text":"$c=\\frac{V}{n^2}$","correct":false,"explanation":"Dividing volume by concentration does not isolate the concentration and gives incorrect units. Amount must be divided by volume, giving $c=\\frac{n}{V}$."}]},{"id":"943e35f9-8836-403e-a09c-1b4a646fd139","cardType":"basic","points":1,"front":"What does a large $K_c$ show about equilibrium position?","back":"Products are favoured.","mcqOptions":[{"text":"Products are favoured.","correct":true,"explanation":"A large $K_c$ means the equilibrium expression has a relatively large numerator compared with its denominator, indicating that products predominate at equilibrium."},{"text":"Reactants are favoured.","correct":false,"explanation":"Reactants are favoured when $K_c$ is small because the equilibrium expression’s denominator then dominates. A large $K_c$ instead shows that products are favoured."},{"text":"Equal amounts are present.","correct":false,"explanation":"Equal equilibrium amounts are not implied by a large equilibrium constant and may not occur even when $K_c=1$. A large $K_c$ shows products are favoured."},{"text":"Equilibrium is reached quickly.","correct":false,"explanation":"The size of $K_c$ describes equilibrium position, not how rapidly equilibrium is reached. A large value indicates that products are favoured."}]},{"id":"2c680614-acf9-48bb-accd-8ae995fb2bc9","cardType":"cloze","points":null,"front":"Changing [ ] changes $K_c$; concentration, pressure and catalysts do not if temperature is constant.","back":"Changing temperature changes $K_c$; concentration, pressure and catalysts do not if temperature is constant.","mcqOptions":[{"text":"temperature","correct":true,"explanation":"Only temperature changes the value of an equilibrium constant because it changes the relative energetic favourability of the forward and reverse reactions."},{"text":"concentration","correct":false,"explanation":"Changing concentration shifts the equilibrium position, but the concentration ratio readjusts to the same $K_c$ at constant temperature. Only changing temperature alters $K_c$."},{"text":"pressure","correct":false,"explanation":"Changing pressure can shift a gaseous equilibrium toward the side with fewer or more gas molecules, but it does not alter $K_c$ at constant temperature."},{"text":"catalyst","correct":false,"explanation":"A catalyst accelerates the forward and reverse reactions equally, so equilibrium is reached sooner without changing $K_c$. Only temperature changes the constant."}]},{"id":"d3ffbe72-7938-443a-85f0-a584932641d2","cardType":"basic","points":1,"front":"For an exothermic forward reaction, what happens to $K_c$ when temperature increases?","back":"$$K_c$ decreases.","mcqOptions":[{"text":"$$K_c$ decreases.","correct":true,"explanation":"Heating favours the endothermic reverse reaction, shifting equilibrium toward reactants. The resulting lower product-to-reactant concentration ratio means $K_c$ decreases."},{"text":"$$K_c$ increases.","correct":false,"explanation":"An increase would occur for an endothermic forward reaction. Here, heating favours the endothermic reverse direction, so the equilibrium shifts toward reactants and $K_c$ decreases."},{"text":"$$K_c$ remains constant.","correct":false,"explanation":"Equilibrium constants remain unchanged when concentration, pressure or catalysts change at constant temperature, but temperature does affect them. For an exothermic forward reaction, $K_c$ decreases."},{"text":"$$K_c$ becomes zero.","correct":false,"explanation":"Heating shifts this equilibrium toward reactants but does not eliminate products entirely, so the constant does not become zero. For an exothermic forward reaction, $K_c$ decreases."}]},{"id":"96145f4d-5849-4691-9ad9-9a76a7f3aaa4","cardType":"basic","points":1,"front":"For an endothermic forward reaction, what happens to $K_c$ when temperature increases?","back":"$$K_c$ increases.","mcqOptions":[{"text":"$$K_c$ increases.","correct":true,"explanation":"Heating favours the endothermic forward reaction, shifting equilibrium toward products. The resulting higher product-to-reactant concentration ratio means $K_c$ increases."},{"text":"$$K_c$ decreases.","correct":false,"explanation":"A decrease would occur for an exothermic forward reaction. Because the forward reaction is endothermic, heating favours products and therefore increases $K_c$."},{"text":"$$K_c$ remains constant.","correct":false,"explanation":"Temperature changes can alter equilibrium constants, unlike concentration, pressure and catalysts at constant temperature. Heating an endothermic forward reaction increases $K_c$."},{"text":"$$K_c$ becomes infinite.","correct":false,"explanation":"Heating favours the forward reaction but does not force complete conversion into products, so $K_c$ does not become infinite. It increases to a finite value."}]},{"id":"5cc63a2f-8f7b-4e65-830b-e529e6af9498","cardType":"basic","points":1,"front":"For $2\\text{SO}_2(g)+\\text{O}_2(g)\\rightleftharpoons2\\text{SO}_3(g)$, what is the $K_c$ expression?","back":"$K_c=\\frac{[\\text{SO}_3]^2}{[\\text{SO}_2]^2[\\text{O}_2]}$","mcqOptions":[{"text":"$K_c=\\frac{[\\text{SO}_3]^2}{[\\text{SO}_2]^2[\\text{O}_2]}$","correct":true,"explanation":"Products appear in the numerator and reactants in the denominator, with powers equal to their balanced coefficients: 2 for SO$_3$, 2 for SO$_2$ and 1 for O$_2$."},{"text":"$K_c=\\frac{[\\text{SO}_2]^2[\\text{O}_2]}{[\\text{SO}_3]^2}$","correct":false,"explanation":"This reverses the product and reactant positions. Products belong in the numerator, so the correct expression is $K_c=\\frac{[\\text{SO}_3]^2}{[\\text{SO}_2]^2[\\text{O}_2]}$."},{"text":"$K_c=\\frac{[\\text{SO}_3]}{[\\text{SO}_2]^2[\\text{O}_2]}$","correct":false,"explanation":"The coefficient of SO$_3$ is 2, so its concentration must be squared rather than left to the first power. All powers come from the balanced equation."},{"text":"$K_c=\\frac{[\\text{SO}_3]^2}{[\\text{SO}_2]^2[\\text{O}_2]^2}$","correct":false,"explanation":"The coefficient of O$_2$ is 1, so its concentration is not squared. Only SO$_2$ and SO$_3$ have coefficients of 2 in the equation."}]},{"id":"63481af4-b107-47ab-a561-cc918f5ec353","cardType":"basic","points":1,"front":"For $2\\text{SO}_2(g)+\\text{O}_2(g)\\rightleftharpoons2\\text{SO}_3(g)$, what are the units of $K_c$?","back":"$\\text{dm}^3\\text{ mol}^{-1}$","mcqOptions":[{"text":"$\\text{dm}^3\\text{ mol}^{-1}$","correct":true,"explanation":"The concentration powers give $(\\text{mol dm}^{-3})^{2-3}=(\\text{mol dm}^{-3})^{-1}$, so the units of $K_c$ are $\\text{dm}^3\\text{ mol}^{-1}$."},{"text":"$\\text{mol dm}^{-3}$","correct":false,"explanation":"These are the units of concentration and result from reversing the overall exponent. The product powers total 2 while the reactant powers total 3, giving $\\text{dm}^3\\text{ mol}^{-1}$."},{"text":"$\\text{dm}^6\\text{ mol}^{-2}$","correct":false,"explanation":"This results from treating the difference in concentration powers as 2 rather than 1. The net exponent is $2-3=-1$, so the units are $\\text{dm}^3\\text{ mol}^{-1}$."},{"text":"$\\text{no units}$","correct":false,"explanation":"$$K_c$ is dimensionless only when the total concentration powers are equal on both sides. Here they are 2 and 3, so the units are $\\text{dm}^3\\text{ mol}^{-1}$."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Equilibrium constants and Le Chatelier's principle","title":"Equilibrium constants and Le Chatelier's principle Flashcards","topbar":{"backHref":"/a-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"d27b0f05-08b7-5620-8f0d-4d90c859bf28","walled":"$L39"}],"$L3a","$L3b"]