R1.1 Measuring enthalpy changes

1. The form of energy that flows from a hot object to a cold one — A student who treats heat and temperature as the same thing picks this. What flows from a hot object to a cold one is heat, energy transferred because of the temperature difference; temperature measures the average kinetic energy of the particles.
2. A measure of the total kinetic energy of all the particles it has — A student who thinks temperature depends on the amount of substance picks this. A total would be larger for a larger sample, but a bucket and a cup of water can be at the same temperature. Temperature measures the average kinetic energy of the particles.
3. A quantity that has to rise whenever heat is transferred into it — A student who equates being heated with getting hotter picks this. During a change of state, such as boiling, heat is transferred into the substance while its temperature stays constant. Temperature measures the average kinetic energy of the particles.
4. A measure of the average kinetic energy of its particles — Temperature is an average over the particles: it measures how fast, on average, they are moving. It does not depend on how many particles there are, and it is not itself a form of energy that is transferred.

Syllabus statement R1.1.1 · Read this in Learn

1. Energy is transferred from the water to the dissolving solid, so the process is endothermic. — The dissolving solid is the system and the water is part of the surroundings. The system takes in energy from the water, so the water's temperature falls: the process is endothermic, with a positive ΔH.
2. The solution is the system; it loses energy, so its temperature falls and the process is exothermic. — A student who treats the solution as the system reads the fall in temperature as the system losing energy. The thermometer measures the water, which is part of the surroundings; the water has lost energy to the system, so the process is endothermic.
3. Cold is released by the dissolving solid and spreads out through all of the water in the polystyrene cup. — A student who thinks of cold as something that can be given out picks this. Only energy is transferred: the water becomes colder because energy passes from it into the dissolving solid.
4. Energy is used up as the solid dissolves, so the total energy of everything decreases. — A student who thinks processes use energy up picks this. Energy is conserved: the energy that leaves the water is taken in by the dissolving solid, so the total energy is unchanged.

Syllabus statement R1.1.2 · Read this in Learn

1. The products are more stable than the reactants, as they have more energy. — A student who thinks more energy means more stable picks this. The products do have more potential energy, which makes them less stable than the reactants: lower potential energy means greater stability.
2. It shows an energy barrier, which the profile of an exothermic reaction lacks. — A student who thinks exothermic reactions need no energy to start picks this. Both profiles show an energy barrier (the activation energy). The difference is where the products end: above the reactants for endothermic, below for exothermic.
3. The products are at a higher potential energy than the reactants. — In an endothermic reaction the system gains energy from the surroundings, so the products end at a higher potential energy than the reactants. The difference between the two levels is ΔH, which is positive.
4. The distance along the x-axis shows how long the reaction takes. — A student who reads the x-axis as time picks this. The reaction coordinate shows the progress of the particles from reactants to products; the profile gives no information about how long the reaction takes.

Syllabus statement R1.1.3 · Read this in Learn

1. It is the heat transferred at constant pressure with every substance in its standard state. — This is the definition: ΔH⦵ is the heat transferred at constant pressure, under standard conditions (100 kPa), with each substance in its standard state, for the amounts in the equation, in kJ mol⁻¹.
2. Its value is the same whether a product such as water forms as a liquid or as a gas. — A student who thinks physical state does not affect ΔH picks this. For H₂ + ½O₂ → H₂O, ΔH⦵ is −286 kJ mol⁻¹ for H₂O(l) but −242 kJ mol⁻¹ for H₂O(g), because condensing steam releases energy. ΔH⦵ applies to the states shown.
3. It is the total heat transferred in an experiment, whatever amount of each reactant is used. — A student who equates the heat measured with ΔH picks this. The heat measured depends on how much reacts. ΔH⦵ refers to the amounts in the equation and is quoted per mole, in kJ mol⁻¹.
4. It is positive when heat passes from the system into the surroundings at 100 kPa. — A student who thinks energy given out is a positive quantity picks this. ΔH describes the system. When heat passes from the system to the surroundings, the system loses energy and ΔH⦵ is negative.

Syllabus statement R1.1.4 · Read this in Learn

1. Energy is created by the reaction, so the total energy of the system and the surroundings increases. — A student who thinks reactions produce energy picks this. Energy is never created: the energy that warms the water comes from the chemicals, whose energy falls by the same amount, so the total is unchanged.
2. The reacting chemicals gain energy, and the rise in the temperature of the solution shows this. — A student who treats the solution as the system picks this. The thermometer is in the water, which is part of the surroundings. The water's temperature rises because it gains energy from the chemicals, which therefore lose energy.
3. The energy of the reacting chemicals falls by exactly the same amount as the energy of the surroundings rises. — The reacting chemicals (the system) lose energy, and exactly that amount is transferred to the surroundings, mainly the water, whose temperature rises. Total energy is conserved: nothing is created or destroyed.
4. Temperature is transferred from the reacting chemicals to the water, which is why the water warms up. — A student who treats temperature as the thing that flows picks this. What is transferred is energy, as heat; the water's temperature rises as a result of gaining that energy.

Syllabus statement R1.1.1 · Read this in Learn

1. Mg(s) added to dilute HCl(aq): the temperature rises by 11.5 °C — A student who treats the solution as the system reads the temperature rise as the chemicals gaining energy. The water is part of the surroundings: it has gained energy from the reacting chemicals, so this reaction is exothermic.
2. NaHCO₃(s) added to citric acid solution: the temperature falls by 4.5 °C — The temperature of the solution, which is part of the surroundings, falls, so energy has been transferred from the surroundings to the reacting chemicals. This is an endothermic reaction.
3. Fe(s) and S(s) heated briefly: the glow continues after the heating stops — A student who thinks any reaction that needs heating to start is endothermic picks this. The brief heating only supplies the activation energy; the reaction then continues on its own and releases energy as light and heat, so it is exothermic.
4. Hydrogen burned in oxygen to form liquid water: ΔH⦵ = −286 kJ mol⁻¹ — A student who thinks a negative ΔH means energy is taken in picks this. A negative ΔH means the system loses energy to the surroundings: the formation of water from hydrogen and oxygen is exothermic.

Syllabus statement R1.1.2 · Read this in Learn

1. −110 kJ mol⁻¹ — ΔH = potential energy of products − potential energy of reactants = 40 − 150 = −110 kJ mol⁻¹. The products lie below the reactants, so the reaction is exothermic and ΔH is negative.
2. +110 kJ mol⁻¹ — A student who thinks energy given out is a positive quantity reads the fall of 110 kJ mol⁻¹ correctly but gives it a plus sign. ΔH describes the system, which loses energy, so ΔH = 40 − 150 = −110 kJ mol⁻¹.
3. +130 kJ mol⁻¹ — A student who confuses ΔH with the barrier height takes 280 − 150 = +130 kJ mol⁻¹. That is the activation energy, Ea. ΔH is products minus reactants: 40 − 150 = −110 kJ mol⁻¹.
4. −240 kJ mol⁻¹ — A student who counts only the fall from the top of the barrier to the products takes 40 − 280 = −240 kJ mol⁻¹. That ignores the 130 kJ mol⁻¹ taken in to reach the top. The net change is 40 − 150 = −110 kJ mol⁻¹.

Working ΔH = potential energy of products − potential energy of reactants = 40 kJ mol⁻¹ − 150 kJ mol⁻¹ = −110 kJ mol⁻¹. (The barrier height above the reactants, 280 − 150 = 130 kJ mol⁻¹, is the activation energy and does not enter ΔH.)

Syllabus statement R1.1.3 · Read this in Learn

1. −6.75 × 10⁵ kJ mol⁻¹ — A student who does not convert joules to kilojoules divides 13 480 J by 0.01997 mol and quotes −675 000 as kJ mol⁻¹. Q must be divided by 1000 first: ΔH = −13.48 kJ / 0.01997 mol = −675 kJ mol⁻¹.
2. −6.75 × 10² kJ mol⁻¹ — Q = mcΔT = 150.0 × 4.18 × 21.5 = 13 480 J = 13.48 kJ. n(C₂H₅OH) = 0.920 / 46.08 = 0.01997 mol. ΔH = −Q/n = −13.48 / 0.01997 = −675 kJ mol⁻¹. (This is far less exothermic than the accepted value, −1367 kJ mol⁻¹, largely because much of the energy is transferred to the air and the can rather than to the water.)
3. −9.25 × 10³ kJ mol⁻¹ — A student who adds 273 to the temperature change uses ΔT = 294.5 K, giving Q = 184.7 kJ and ΔH = −9250 kJ mol⁻¹. A temperature change has the same value in K as in °C: ΔT = 21.5 K.
4. −1.35 × 10¹ kJ mol⁻¹ — A student who takes the heat measured as ΔH gives it the negative sign of an exothermic change but does not divide by the amount burned, writing ΔH = −13.5 kJ mol⁻¹. ΔH is per mole, so Q must be divided by the amount of ethanol burned, 0.01997 mol, giving −675 kJ mol⁻¹.

Working ΔT = 40.5 °C − 19.0 °C = 21.5 °C = 21.5 K. Q = mcΔT = 150.0 g × 4.18 J g⁻¹ K⁻¹ × 21.5 K = 13 480.5 J = 13.48 kJ. M(C₂H₅OH) = 2(12.01) + 6(1.01) + 16.00 = 46.08 g mol⁻¹; n = 0.920 g / 46.08 g mol⁻¹ = 0.019965 mol. ΔH = −Q/n = −13.48 kJ / 0.019965 mol = −675 kJ mol⁻¹ = −6.75 × 10² kJ mol⁻¹ (3 s.f.).

Syllabus statement R1.1.4 · Read this in Learn

1. P has the greater potential energy, so it is the more stable of the two. — A student who thinks more energy means more stable picks this. P does have more potential energy, but that makes it the less stable substance; Q, with the lower potential energy, is the more stable.
2. Q has more potential energy than P, so P is the more stable substance. — A student who reads a negative ΔH as energy taken in places Q above P. A negative ΔH means the system loses energy, so Q lies below P and Q is the more stable.
3. Because ΔH is negative, the reaction's energy profile has no energy barrier. — A student who thinks exothermic reactions need no energy to start picks this. An exothermic reaction still has an energy barrier, the activation energy; the negative ΔH only means that the products lie below the reactants.
4. P has more potential energy than Q, so Q is the more stable of the two substances. — ΔH is negative, so the system loses energy as P turns into Q: Q lies 60 kJ mol⁻¹ below P. The substance with the lower potential energy is the more stable, so Q is more stable than P.

Syllabus statement R1.1.3 · Read this in Learn

1. −28.2 kJ mol⁻¹ — A student who divides by the amount of HCl uses n = 0.0500 mol, giving −1.411 / 0.0500 = −28.2 kJ mol⁻¹. HCl is in excess; only 0.0250 mol of NaOH is available, so only 0.0250 mol of reaction occurs. ΔH = −56.4 kJ mol⁻¹.
2. −37.6 kJ mol⁻¹ — A student who uses only the mass of the acid already in the cup takes m = 50.0 g, forgetting that the 25.0 g of NaOH(aq) added is also warmed. That gives Q = 940.5 J and ΔH = −37.6 kJ mol⁻¹. Both solutions change temperature, so m = 75.0 g, giving −56.4 kJ mol⁻¹.
3. −1.41 kJ mol⁻¹ — A student who takes the heat measured as ΔH gives it the negative sign of an exothermic change but does not divide by the amount that reacted, writing ΔH = −1.41 kJ mol⁻¹. ΔH is per mole, so Q must be divided by the amount of the limiting reactant, 0.0250 mol of NaOH, giving −56.4 kJ mol⁻¹.
4. −56.4 kJ mol⁻¹ — Q = mcΔT = 75.0 × 4.18 × 4.50 = 1411 J = 1.411 kJ, using the mass of the whole mixture (50.0 + 25.0 = 75.0 g). NaOH is limiting: n = 0.0250 dm³ × 1.00 mol dm⁻³ = 0.0250 mol (HCl, 0.0500 mol, is in excess). ΔH = −1.411 / 0.0250 = −56.4 kJ mol⁻¹.

Working m(solution) = (50.0 + 25.0) cm³ × 1.00 g cm⁻³ = 75.0 g. Q = mcΔT = 75.0 g × 4.18 J g⁻¹ K⁻¹ × 4.50 K = 1410.75 J = 1.411 kJ. n(HCl) = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol; n(NaOH) = 0.0250 dm³ × 1.00 mol dm⁻³ = 0.0250 mol. The mole ratio is 1:1, so NaOH is limiting and 0.0250 mol of reaction occurs. ΔH = −Q/n = −1.411 kJ / 0.0250 mol = −56.4 kJ mol⁻¹.

Syllabus statement R1.1.4 · Read this in Learn

1. −25.9 kJ mol⁻¹ — A student who drops the minus sign in ΔH = −Q/n gives ΔH the sign of the temperature change. The water lost energy to the dissolving solid, so the system gained energy and ΔH is positive: +25.9 kJ mol⁻¹.
2. +2.07 kJ mol⁻¹ — A student who uses the mass of the solid in Q = mcΔT takes m = 4.00 g, giving Q = −103.7 J and ΔH = +2.07 kJ mol⁻¹. The temperature change measured is that of the 50.0 g of water, so m = 50.0 g, giving +25.9 kJ mol⁻¹.
3. +25.9 kJ mol⁻¹ — Q = mcΔT = 50.0 × 4.18 × (15.40 − 21.60) = −1296 J = −1.296 kJ (the water loses energy). n(NH₄NO₃) = 4.00 / 80.06 = 0.0500 mol. ΔH = −Q/n = −(−1.296) / 0.0500 = +25.9 kJ mol⁻¹: positive, because the process is endothermic.
4. −1.30 kJ mol⁻¹ — A student who takes the heat measured as ΔH stops at Q = mcΔT = 50.0 × 4.18 × (−6.20) = −1296 J = −1.30 kJ. Q is the energy gained by the water, not ΔH: ΔH = −Q/n, and dividing by the amount dissolved, 0.0500 mol, gives +25.9 kJ mol⁻¹.

Working ΔT = 15.40 °C − 21.60 °C = −6.20 K. Q = mcΔT = 50.0 g × 4.18 J g⁻¹ K⁻¹ × (−6.20 K) = −1295.8 J = −1.296 kJ (energy gained by the water is negative: the water lost energy). M(NH₄NO₃) = 2(14.01) + 4(1.01) + 3(16.00) = 80.06 g mol⁻¹; n = 4.00 g / 80.06 g mol⁻¹ = 0.04996 mol. ΔH = −Q/n = −(−1.296 kJ) / 0.04996 mol = +25.9 kJ mol⁻¹.

Syllabus statement R1.1.4 · Read this in Learn

1. +100 kJ mol⁻¹ — The size is right, but the reverse reaction goes downhill from 150 to 50 kJ mol⁻¹ and gives out energy, so ΔH is negative: −100 kJ mol⁻¹. A positive ΔH means energy is taken in.
2. −100 kJ mol⁻¹ — Reversed, the reaction starts at the product level, 150 kJ mol⁻¹, and ends at the reactant level, 50 kJ mol⁻¹: ΔH = 50 − 150 = −100 kJ mol⁻¹. The reverse of an endothermic reaction is exothermic, with ΔH equal in size and opposite in sign.
3. +150 kJ mol⁻¹ — 300 − 150 = 150 kJ mol⁻¹ is the height of the energy barrier for the reverse reaction, measured from its starting level. ΔH compares only the start and end levels: 50 − 150 = −100 kJ mol⁻¹.
4. −250 kJ mol⁻¹ — 300 − 50 = 250 kJ mol⁻¹ is the fall from the top of the barrier to the final level. Reaching the peak first absorbed 150 kJ mol⁻¹, so the net energy given out is 100 kJ mol⁻¹: ΔH = −100 kJ mol⁻¹.

Working For the reverse reaction the start is the product level (150 kJ mol⁻¹) and the end is the reactant level (50 kJ mol⁻¹): ΔH = 50 − 150 = −100 kJ mol⁻¹, equal in size and opposite in sign to ΔH(forward) = +100 kJ mol⁻¹. The peak at 300 kJ mol⁻¹ does not enter ΔH.

Syllabus statement R1.1.3 · Read this in Learn

1. Arrow W — W is the height of the energy barrier above the reactants, the energy that must be supplied to start the reaction. ΔH is not the barrier height; it is the difference between the reactant and product levels, arrow Y.
2. Arrow X — X is the fall from the top of the barrier to the products. That energy is released only after the energy W was absorbed to reach the peak, so the net energy given out is X − W, which is the drop from reactants to products, arrow Y.
3. Arrow Y — ΔH is the difference in potential energy between the products and the reactants, which is the arrow joining the reactant level to the product level. Here the products are lower, so ΔH is negative and the reaction is exothermic.
4. Arrow Z — Z is the potential energy of the reactants above the axis. That total is not released, because the products still hold most of it; the zero of the axis is arbitrary. Only the difference between the reactant and product levels, arrow Y, is transferred to the surroundings.

Syllabus statement R1.1.3 · Read this in Learn

1. +12.4 kJ mol⁻¹ — Using the lowest recorded temperature, 17.0 °C, gives ΔT = 5.0 K and ΔH = +12.4 kJ mol⁻¹. By the time that reading was taken the solution was already gaining energy from the air, so the minimum was never reached. Extrapolating the warming line back to 2 min gives 16.5 °C, ΔT = 5.5 K and ΔH = +13.7 kJ mol⁻¹.
2. +13.7 kJ mol⁻¹ — ΔT from the extrapolated minimum: 22.0 − 16.5 = 5.5 K. Q = 50.0 × 4.18 × 5.5 = 1149.5 J = 1.1495 kJ. n(NH₄Cl) = 4.50 / 53.50 = 0.08411 mol. The water loses this energy to the dissolving solid, so ΔH is positive: +1.1495 / 0.08411 = +13.7 kJ mol⁻¹.
3. +1.23 kJ mol⁻¹ — Using m = 4.50 g, the mass of the solid, gives Q = 4.50 × 4.18 × 5.5 = 103.5 J and ΔH = +1.23 kJ mol⁻¹. It is the 50.0 g of water whose temperature falls, so m = 50.0 g, Q = 1149.5 J and ΔH = +13.7 kJ mol⁻¹.
4. +1.15 kJ mol⁻¹ — 1.15 kJ is Q, the energy absorbed in this particular experiment. ΔH is per mole of NH₄Cl, so Q must be divided by n = 4.50 / 53.50 = 0.08411 mol, giving +13.7 kJ mol⁻¹.

Working From the graph, the initial temperature is 22.0 °C and the warming line extrapolated back to 2 min gives 16.5 °C, so ΔT = 22.0 − 16.5 = 5.5 K (the recorded minimum, 17.0 °C, is not used because the solution was already gaining energy from the air). Q = mcΔT = 50.0 g × 4.18 J g⁻¹ K⁻¹ × 5.5 K = 1149.5 J = 1.1495 kJ. The water is the surroundings and loses this energy, so for the surroundings Q = −1.1495 kJ. M(NH₄Cl) = 14.01 + 4(1.01) + 35.45 = 53.50 g mol⁻¹; n = 4.50 g / 53.50 g mol⁻¹ = 0.08411 mol. ΔH = −Q/n = −(−1.1495 kJ) / 0.08411 mol = +13.7 kJ mol⁻¹ (endothermic, positive).

Syllabus statement R1.1.4 · Read this in Learn

1. Product B is more stable than product C. — Both profiles start from A at the same level. B lies below A (reaction 1 is exothermic) and C lies above A (reaction 2 is endothermic), so B has the lower potential energy of the two products and is the more stable.
2. Product C is more stable than reactant A. — C lies above A on the potential energy axis, so it has more potential energy and is less stable than A. Higher potential energy means lower stability; energy must be supplied to convert A into C, which is why reaction 2 is endothermic.
3. Reactions 1 and 2 are both endothermic. — Both profiles show an energy barrier that must be crossed, but the barrier does not decide the sign of ΔH. Reaction 1 ends with B below A, so it releases energy overall and is exothermic; only reaction 2, ending with C above A, is endothermic.
4. Reaction 2 takes less time than reaction 1. — The x-axis is the reaction coordinate, the progress from reactants to products, not time, so where a curve levels off says nothing about how long the reaction takes. If anything, reaction 2 has the higher energy barrier (activation energy), so at the same temperature it would be expected to be the slower reaction.

Syllabus statement R1.1.3 · Read this in Learn

1. 7.0 K — 27.0 − 20.0 = 7.0 K uses the highest reading actually recorded. Energy was being lost to the surroundings even before 5 min, so the mixture never reached the temperature the reaction alone would have produced. Extrapolating the cooling line back to 3.0 min gives 28.0 °C and ΔT = 8.0 K.
2. 4.5 K — 24.5 − 20.0 = 4.5 K takes the last reading as the final temperature. The fall from 27.0 °C to 24.5 °C is cooling to the room after the reaction, not part of the reaction's temperature change. The final temperature is the extrapolated value at 3.0 min, 28.0 °C, giving ΔT = 8.0 K.
3. 281 K — 8.0 + 273 = 281 K converts a temperature change as if it were a temperature. A difference of 8.0 °C is a difference of 8.0 K, because the kelvin and Celsius scales have the same size of degree; adding 273 applies only to a single temperature, not to ΔT.
4. 8.0 K — The cooling line extrapolated back to 3.0 min reaches 28.0 °C, so ΔT = 28.0 − 20.0 = 8.0 K. This corrects for the energy lost to the surroundings while the mixture was still warming.

Working The mixture starts at 20.0 °C. After the highest reading (27.0 °C at 5 min) the temperature falls along a straight line because energy is lost to the surroundings; this cooling was already happening while the mixture was warming. Extrapolating the cooling line back to the moment of mixing, 3.0 min, gives 28.0 °C, the temperature the mixture would have reached with no energy loss. ΔT = 28.0 − 20.0 = 8.0 °C = 8.0 K (a temperature difference has the same value in °C and K).

Syllabus statement R1.1.4 · Read this in Learn