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IB Chemistry R1.1: Measuring enthalpy changes worked solutions

Worked solutions to every practice question in R1.1: Measuring enthalpy changes of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question Q01

R1.1 graded and connected practice[1 mark]

A solid melts at constant temperature in a warmer bath. Which description is correct? A. Exothermic; the solid loses heat. B. Exothermic; the solid gains heat. C. Neither, because temperature is constant. D. Endothermic; the solid gains heat.

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D. Melting takes energy from the warmer bath. During the plateau, this energy changes the intermolecular arrangement rather than increasing the average kinetic-energy scale. A zero temperature change does not require a zero transfer.

Final answer

Question Q02

R1.1 graded and connected practice[1 mark]

A sensor measures a sealed vessel’s temperature. Which statement distinguishes temperature from heat? A. Temperature measures total vessel energy. B. Heat is transfer due to a temperature difference; temperature relates to average particle kinetic energy. C. They are the same quantity in different units. D. Temperature is proportional only to particle number.

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B. Heat measures energy crossing a boundary due to a temperature difference. Temperature describes an average kinetic-energy scale. A larger sample at the same temperature contains more particles, so particle number alone cannot determine its temperature.

Final answer

Question Q03

R1.1 graded and connected practice[1 mark]

Products lie below reactants on a common potential-energy scale. Which pair is correct? A. More stable products; \Delta H<0. B. More stable products; \Delta H>0. C. Less stable products; \Delta H<0. D. Less stable products; \Delta H>0.

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A. The products have lower potential energy on the shared reference scale and are more energetically stable. Subtracting initial from final energy gives a negative enthalpy change. This inference needs no statement about reaction speed.

Final answer

Question Q04

R1.1 graded and connected practice[1 mark]

Quartz of mass 45.0\,\mathrm g warms by 8.00\,\mathrm K, with c=0.730\,\mathrm{J\,g^{-1}K^{-1}}. Find its heat gain. A. 32.9\,\mathrm J B. 263\,\mathrm J C. 360\,\mathrm J D. 9240\,\mathrm J

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B. Use Q=mc\Delta T=45.0(0.730)(8.00)=262.8\,\mathrm J, which rounds to 263\,\mathrm J. The interval is already in kelvin; no temperature offset is added.

Final answer

B; 263\,\mathrm J

Question Q05

R1.1 graded and connected practice[1 mark]

Surroundings lose 0.456\,\mathrm{kJ} as 0.0120\,\mathrm{mol} of reaction occurs. With no other heat exchange, find the molar reaction enthalpy. A. -38.0\,\mathrm{kJ\,mol^{-1}} B. +0.456\,\mathrm{kJ\,mol^{-1}} C. +38.0\,\mathrm{kJ\,mol^{-1}} D. +3.80\times10^4\,\mathrm{kJ\,mol^{-1}}

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C. The surroundings have Q_{\rm sur}=-0.456\,\mathrm{kJ}, so conservation gives Q_{\rm sys}=+0.456\,\mathrm{kJ}. Thus \Delta_{\rm r}H=+0.456/0.0120=+38.0\,\mathrm{kJ\,mol^{-1}}.

Final answer

C; +38.0\,\mathrm{kJ\,mol^{-1}}

Question Q06

R1.1 graded and connected practice[1 mark]

The curve represents one reaction pathway. Energy values use a common reference. Find the enthalpy change and the barrier for the reverse reaction.

Profile data: R=60, P=18, peak=130, all in \mathrm{kJ\,mol^{-1}}.

A \Delta H=-42\,\mathrm{kJ\,mol^{-1}}; E_{\mathrm{a,reverse}}=70\,\mathrm{kJ\,mol^{-1}}.

B \Delta H=-42\,\mathrm{kJ\,mol^{-1}}; E_{\mathrm{a,reverse}}=112\,\mathrm{kJ\,mol^{-1}}.

C \Delta H=+42\,\mathrm{kJ\,mol^{-1}}; E_{\mathrm{a,reverse}}=112\,\mathrm{kJ\,mol^{-1}}.

D \Delta H=-112\,\mathrm{kJ\,mol^{-1}}; E_{\mathrm{a,reverse}}=42\,\mathrm{kJ\,mol^{-1}}.

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B. Read the product, reactant and peak levels from the common scale. The endpoint change is 18-60=-42\,\mathrm{kJ\,mol^{-1}}. For the reverse pathway the starting level is 18, so its barrier is 130-18=112\,\mathrm{kJ\,mol^{-1}}. The forward barrier is 130-60=70\,\mathrm{kJ\,mol^{-1}}.

Final answer

B; -42, 112\,\mathrm{kJ\,mol^{-1}}

Question Q07

R1.1 graded and connected practice[1 mark]

A gas is in its standard state at 298.15\,\mathrm K. Which statement applies? A. The pure gas is at 100\,\mathrm{kPa}. B. The gas must occupy 22.7\,\mathrm{dm^3}. C. The gas must be at 273.15\,\mathrm K. D. The gas is mixed with air at total pressure 100\,\mathrm{kPa}.

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A. A gaseous standard state uses the pure gas at 100\,\mathrm{kPa} and the stated temperature. The stated 298.15\,\mathrm K is not replaced by 273.15\,\mathrm K. A standard state imposes no particular sample volume.

Final answer

Question Q08

R1.1 graded and connected practice[1 mark]

A rapid insulated reaction raises temperature by 9.60\,\mathrm K. Repeat with the same released heat and c, but triple the heated mass with inert solvent. Neglect cup heating. Find the rise. A. 28.8\,\mathrm K B. 9.60\,\mathrm K C. 0.312\,\mathrm K D. 3.20\,\mathrm K

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D. With captured Q and c fixed, \Delta T=Q/(mc). Tripling m therefore gives 9.60/3=3.20\,\mathrm K. The chemical conversion releases the same heat for the same reacting amount; adding inert solvent does not change that molar enthalpy under the given model.

Final answer

D; 3.20\,\mathrm K

Question Q09

R1.1 graded and connected practice[1 mark]

A burner loses 0.800\,\mathrm g, including 0.100\,\mathrm g evaporated unburned. Water heat is measured correctly. How does using total mass loss as burned fuel affect the calculated combustion enthalpy? A. Too negative: evaporation creates heat. B. Less negative: the fuel amount is too large. C. Unchanged: the balance measures mass correctly. D. The sign becomes positive.

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B. The reacted mass is 0.800-0.100=0.700\,\mathrm g. Using 0.800\,\mathrm g makes the amount assigned to combustion too large. Dividing the unchanged negative reaction heat by that amount gives a value with smaller magnitude. Its magnitude is 0.700/0.800=0.875 of the correct result, a 12.5\% underestimate.

Final answer

Question Q10

R1.1 graded and connected practice[1 mark]

Complete combustion of 0.400\,\mathrm g fuel (M=80.0\,\mathrm{g\,mol^{-1}}) releases 18.0\,\mathrm{kJ} in total. Find its molar enthalpy. A. -45.0\,\mathrm{kJ\,mol^{-1}} B. +3600\,\mathrm{kJ\,mol^{-1}} C. -3600\,\mathrm{kJ\,mol^{-1}} D. -0.0900\,\mathrm{kJ\,mol^{-1}}

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C. The reacted amount is n=0.400/80.0=0.00500\,\mathrm{mol}. The fuel releases 18.0\,\mathrm{kJ}, so \Delta_{\rm r}H=-18.0/0.00500=-3.60\times10^3\,\mathrm{kJ\,mol^{-1}}. Dividing only by grams would instead give 45.0\,\mathrm{kJ\,g^{-1}}.

Final answer

C; -3.60\times10^3\,\mathrm{kJ\,mol^{-1}}

Tutorial practice

Question Q11

R1.1 graded and connected practice[5 marks]

For a solubility investigation, 2.24\,\mathrm{g} of potassium chloride dissolves completely in 80.00\,\mathrm{g} of water. The final solution cools from 23.40\,{}^{\circ}\mathrm{C} to 21.85\,{}^{\circ}\mathrm{C}. Use M(\mathrm{KCl})=74.55\,\mathrm{g\,mol^{-1}} and c_{\mathrm{solution}}=4.18\,\mathrm{J\,g^{-1}\,K^{-1}}. Heat transfer to the cup and room is negligible.

(a) Calculate the heat change of the solution, with its sign.

(b) Determine the molar dissolution enthalpy.

(c) Explain why using only the original water mass would bias the result.

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(a) Include the dissolved salt: m=80.00+2.24=82.24\,\mathrm g. The signed interval is 21.85-23.40=-1.55\,\mathrm K. Thus Q_{\rm sol}=82.24(4.18)(-1.55)=-532.83296\,\mathrm J=-0.533\,\mathrm{kJ} to three significant figures. (b) The amount is n=2.24/74.55=0.030046948\ldots\,\mathrm{mol}. The dissolution absorbs the solution's lost sensible heat, giving \Delta_{\rm sol}H=+0.53283296/n=+17.73334695\,\mathrm{kJ\,mol^{-1}}, or +17.7\,\mathrm{kJ\,mol^{-1}}. (c) A water-only mass makes the heat-loss magnitude too small, so it understates the positive dissolution enthalpy.

Final answer

-0.533\,\mathrm{kJ}; +17.7\,\mathrm{kJ\,mol^{-1}}; water-only mass understates the positive result.

Question Q12

R1.1 graded and connected practice[6 marks]

An isomerization has a forward activation energy of 64\,\mathrm{kJ\,mol^{-1}} and a molar enthalpy change of +38\,\mathrm{kJ\,mol^{-1}}.

(a) Sketch its potential-energy profile. Label both axes, reactants, products, the forward activation energy and the enthalpy change.

(b) Add a possible catalysed pathway to your sketch and state how the catalyst affects the enthalpy change.

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(a) Use reaction coordinate horizontally and relative potential energy vertically. Place reactants at an arbitrary zero, products at +38\,\mathrm{kJ\,mol^{-1}} and the peak at +64\,\mathrm{kJ\,mol^{-1}}. Draw a forward barrier from 0 to 64 and an upward endpoint arrow from 0 to 38. (b) Add a lower peak with the same endpoints. A possible catalysed peak is 52, which lies above products. The enthalpy change remains +38\,\mathrm{kJ\,mol^{-1}}. The uncatalysed and catalysed reverse barriers are 26 and 14\,\mathrm{kJ\,mol^{-1}}.

Final answer

Products 38 above reactants; peak 64 above reactants; lower catalysed peak, same endpoints.

Question Q13

R1.1 graded and connected practice[1 mark]

The following reaction warms 72.0\,\mathrm{g} of solution by 4.00\,\mathrm{K}.

\mathrm{Ba(OH)_2(aq)}+2\,\mathrm{HBr(aq)}\rightarrow\mathrm{BaBr_2(aq)}+2\,\mathrm{H_2O(l)}

Initially there are 0.0300\,\mathrm{mol} of \mathrm{Ba(OH)_2} and 0.0400\,\mathrm{mol} of \mathrm{HBr}. Use c=4.18\,\mathrm{J\,g^{-1}\,K^{-1}} and neglect heat loss. Determine \Delta H per mole of reaction as written, in \mathrm{kJ\,mol^{-1}}.

A -30.1

B -60.2

C -40.1

D +60.2

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B. Compare available moles of reaction: 0.0300/1=0.0300 for the base and 0.0400/2=0.0200 for the acid. Acid limits, so n_{\rm reaction}=0.0200\,\mathrm{mol}. The solution gains Q=72.0(4.18)(4.00)=1203.84\,\mathrm J. Thus \Delta_{\rm r}H=-1.20384/0.0200=-60.192\,\mathrm{kJ\,mol^{-1}}, or -60.2\,\mathrm{kJ\,mol^{-1}}.

Final answer

B; -60.2\,\mathrm{kJ\,mol^{-1}}

Question Q16

R1.1 graded and connected practice[3 marks]

Two insulated samples receive 900\,\mathrm J each. A has mass 120\,\mathrm g and c=2.50\,\mathrm{J\,g^{-1}K^{-1}}; B has mass 80.0\,\mathrm g and c=4.00\,\mathrm{J\,g^{-1}K^{-1}}. Neither changes phase. (a) Calculate each temperature rise. (b) Explain why the smaller sample need not have the larger rise.

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The heat capacities are 120(2.50)=300\,\mathrm{J\,K^{-1}} and 80.0(4.00)=320\,\mathrm{J\,K^{-1}}. Using \Delta T=Q/(mc) gives 900/300=3.00\,\mathrm K and 900/320=2.8125\,\mathrm K, or 2.81\,\mathrm K. B has the larger total heat capacity despite its smaller mass. Mass alone is insufficient when the specific heat capacities differ.

Final answer

A: 3.00\,\mathrm K; B: 2.81\,\mathrm K; compare mc.

Question Q17

R1.1 graded and connected practice[4 marks]

A supplied reaction 2\mathrm A\rightarrow\mathrm B has \Delta_{\rm r}H=-72.0\,\mathrm{kJ\,mol^{-1}} for the equation as written. (a) Find the system heat when 0.150\,\mathrm{mol} of A is consumed. (b) Write the reversed equation with all coefficients doubled and give its reaction enthalpy.

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The reaction amount is 0.150/2=0.0750\,\mathrm{mol}, so Q_{\rm sys}=0.0750(-72.0)=-5.40\,\mathrm{kJ}. Reverse the equation to change the sign, then double its coefficients to double the molar enthalpy attached to that equation. Thus 2\mathrm B\rightarrow4\mathrm A has \Delta_{\rm r}H=+144\,\mathrm{kJ\,mol^{-1}}.

Final answer

-5.40\,\mathrm{kJ}; 2\mathrm B\rightarrow4\mathrm A, +144\,\mathrm{kJ\,mol^{-1}}.

Question Q18

R1.1 graded and connected practice[4 marks]

Mix 30.0\,\mathrm{cm^3} of 0.600\,\mathrm{mol\,dm^{-3}} HCl with 45.0\,\mathrm{cm^3} of 0.300\,\mathrm{mol\,dm^{-3}} NaOH at the same initial temperature. The rise is 2.25\,\mathrm K. Use density 1.00\,\mathrm{g\,cm^{-3}}, additive volumes and c=4.18\,\mathrm{J\,g^{-1}K^{-1}}; neglect apparatus and room heat exchange. (a) Identify the limiting reagent and reacting amount. (b) Calculate the solution heat gain. (c) Find the enthalpy per mole of water formed.

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The acid supplies 0.0300(0.600)=0.0180\,\mathrm{mol} and the base supplies 0.0450(0.300)=0.0135\,\mathrm{mol}. The one-to-one neutralisation consumes all the NaOH and forms 0.0135\,\mathrm{mol} water. Both liquids warm, so m=75.0\,\mathrm g and Q_{\rm sol}=75.0(4.18)(2.25)=705.375\,\mathrm J. The reacting system loses that heat: \Delta H=-0.705375/0.0135=-52.25\,\mathrm{kJ\,mol^{-1}}, or -52.3\,\mathrm{kJ\,mol^{-1}} to three significant figures.

Final answer

NaOH; 0.0135\,\mathrm{mol}; 705\,\mathrm J; -52.3\,\mathrm{kJ\,mol^{-1}}.

Question Q19

R1.1 graded and connected practice[4 marks]

Burning 0.650\,\mathrm g of a dry food warms 100.0\,\mathrm g of water by 18.0\,\mathrm K. Take c=4.18\,\mathrm{J\,g^{-1}K^{-1}}. The stated food energy is 18.0\,\mathrm{kJ\,g^{-1}}. (a) Find the energy captured per gram of food. (b) Calculate the percentage of the stated energy captured. (c) Explain one physical reason for incomplete capture.

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The water gains 100.0(4.18)(18.0)=7524\,\mathrm J=7.524\,\mathrm{kJ}. Dividing by 0.650\,\mathrm g gives 11.5753846\ldots\,\mathrm{kJ\,g^{-1}}, or 11.6\,\mathrm{kJ\,g^{-1}}. Compare unrounded captured energy with 0.650(18.0)=11.70\,\mathrm{kJ} supplied: capture is 100(7.524/11.70)=64.3077\ldots\%=64.3\%. Hot gases and the apparatus receive energy that is absent from the water-only calculation; incomplete burning can further reduce the release.

Final answer

11.6\,\mathrm{kJ\,g^{-1}}; 64.3\%; energy also heats the air or apparatus.

Question Q20

R1.1 graded and connected practice[4 marks]

You must compare the dissolution enthalpies of two safe, soluble salts in an insulated cup. Both dissolve rapidly and completely. (a) Describe the measurements and sequence needed for a signed molar result. (b) Give two controls or corrections, each with its purpose.

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Weigh the dry salt and water separately. Record the initial water temperature, add the salt, stir and record temperature against time through dissolution and the later drift. Use the corrected signed interval and the combined solution mass in Q=mc\Delta T, then divide -Q by salt moles from its mass and molar mass. Use matched starting temperatures so each comparison refers to the same thermal conditions. Calibrate cup heat capacity to include energy stored in the apparatus, or extrapolate a justified post-reaction trace to estimate room exchange. These corrections address different terms and must not be counted twice. A lid limits evaporation and external exchange; repeated runs estimate spread.

Final answer

Measure salt/water masses and temperature trace; use solution heat and salt moles. Control starting temperature and correct cup/room heat exchange.

Advanced practice

Question Q14

R1.1 graded and connected practice[9 marks]

A cup contains 80.00\,\mathrm{g} of water. At t=60\,\mathrm{s}, 2.220\,\mathrm{g} of anhydrous calcium chloride is added and dissolves completely. Temperature is recorded as shown. The cup has heat capacity 18.0\,\mathrm{J\,K^{-1}}; use c_{\mathrm{solution}}=4.18\,\mathrm{J\,g^{-1}\,K^{-1}} and M(\mathrm{CaCl_2})=110.98\,\mathrm{g\,mol^{-1}}.

Synthetic trace (t/\mathrm s,T/^{\circ}\mathrm C): (0,20.40),(30,20.40),(60,20.40) just before addition; (90,24.70),(120,24.40),(150,24.10),(180,23.80) after reaction. Addition occurs at 60\,\mathrm s.

(a) Extrapolate the straight cooling section back to the addition time and determine the corrected temperature rise.

(b) Calculate the corrected heat gained by the solution and cup together.

(c) Determine the molar dissolution enthalpy.

(d) State one assumption required for the extrapolation in (a).

(e) Treat the initial and extrapolated temperatures as each uncertain by \pm0.10\,\mathrm{K}. Calculate the maximum percentage uncertainty in the rise, and give one way to reduce it.

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(a) The post-reaction points have slope -0.0100\,\mathrm{K\,s^{-1}}. Their line is T=25.60-0.0100t for numerical t in seconds and T in degrees Celsius. At addition, T(60)=25.00\,{}^{\circ}\mathrm C and the corrected rise is 25.00-20.40=4.60\,\mathrm K. (b) The total heat capacity is (80.00+2.220)(4.18)+18.0=361.6796\,\mathrm{J\,K^{-1}}. Hence Q=361.6796(4.60)=1663.72616\,\mathrm J=1.66\,\mathrm{kJ}. (c) Use n=2.220/110.98=0.020003604\ldots\,\mathrm{mol}. Then \Delta_{\rm sol}H=-1.66372616/n=-83.1713\ldots\,\mathrm{kJ\,mol^{-1}}, or -83.2\,\mathrm{kJ\,mol^{-1}}. (d) Assume rapid dissolution and a later locally linear heat-transfer trend representative of the brief earlier interval. (e) The worst-case interval uncertainty is 0.10+0.10=0.20\,\mathrm K, giving 100(0.20/4.60)=4.3478\ldots\%=4.3\%. Use a probe with smaller temperature-reading uncertainty. A matched control does not replace the extrapolation correction.

Final answer

4.60\,\mathrm K; 1.66\,\mathrm{kJ}; -83.2\,\mathrm{kJ\,mol^{-1}}; rapid reaction; 4.3\%, smaller reading uncertainty.

Question Q15

R1.1 graded and connected practice[7 marks]

Two aqueous solutions, 30.0 and 70.0\,\mathrm g, start at 20.0 and 30.0\,{}^{\circ}\mathrm C, respectively. They react immediately on mixing through 0.0200\,\mathrm{mol} of reaction, reaching 26.2\,{}^{\circ}\mathrm C. Both have water's specific heat capacity. Neglect cup and room heat exchange; no phase change occurs.

(a) Show that the mixed temperature would have been 27.0\,{}^{\circ}\text{C} if no reaction occurred.

(b) Calculate the reaction enthalpy to two significant figures.

(c) Explain why comparing the final temperature only with the colder solution would misclassify the reaction.

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(a) Without reaction, conservation requires 30.0c(T-20.0)+70.0c(T-30.0)=0. Cancel the common c and solve: T=(30.0\times20.0+70.0\times30.0)/100.0=27.0\,{}^{\circ}\mathrm C. (b) The reaction lowers the final temperature below this baseline: \Delta T=26.2-27.0=-0.8\,\mathrm K. The liquid heat is Q_{\rm sur}=100.0(4.18)(-0.8)=-334.4\,\mathrm J. The reaction absorbs 0.3344\,\mathrm{kJ}, so \Delta_{\rm r}H=+0.3344/0.0200=+16.72\,\mathrm{kJ\,mol^{-1}}, reported as +17\,\mathrm{kJ\,mol^{-1}} to the requested two significant figures. (c) The hotter starting solution would warm the colder one without a reaction. Comparing only with 20.0\,{}^{\circ}\mathrm C would mistake that mixing transfer for reaction heat. Relative to the weighted baseline, the reaction is endothermic.

Final answer

27.0\,{}^{\circ}\mathrm C; +17\,\mathrm{kJ\,mol^{-1}}; compare with the weighted baseline.

Question Q21

R1.1 graded and connected practice[7 marks]

An enclosed fuel trial burns 0.00600\,\mathrm{mol} completely. Water of mass 180.0\,\mathrm g warms by 8.00\,\mathrm K; its specific heat capacity is 4.18\,\mathrm{J\,g^{-1}K^{-1}}. The apparatus, excluding water, has heat capacity 95.0\,\mathrm{J\,K^{-1}}. A calibration states that water plus apparatus captures 80.0\% of the fuel energy. (a) Find the energy captured by water plus apparatus. (b) Infer the total release and molar combustion enthalpy. (c) For the same capture fraction and setup, predict the rise when 0.500\,\mathrm g of fuel with M=50.0\,\mathrm{g\,mol^{-1}} burns. (d) Explain why dividing only the water heat by fuel moles gives a less negative value.

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The receiving heat capacity is 180.0(4.18)+95.0=847.4\,\mathrm{J\,K^{-1}}. Captured energy is 847.4(8.00)=6779.2\,\mathrm J. Since that is 80.0\% of the release, the full release is 6779.2/0.800=8474.0\,\mathrm J. Thus \Delta_{\rm c}H=-8.4740/0.00600=-1412.3333\ldots\,\mathrm{kJ\,mol^{-1}}, or -1.41\times10^3\,\mathrm{kJ\,mol^{-1}}. The new burned amount is 0.500/50.0=0.0100\,\mathrm{mol}. At unchanged capture fraction and heat capacity the rise scales with fuel amount: 8.00(0.0100/0.00600)=13.3333\ldots\,\mathrm K=13.3\,\mathrm K. Water-only energy excludes apparatus heating and the energy escaping the calibrated boundary, understating the release magnitude.

Final answer

6.78\,\mathrm{kJ}; 8.47\,\mathrm{kJ} released; -1.41\times10^3\,\mathrm{kJ\,mol^{-1}}; 13.3\,\mathrm K.

Question Q22

R1.1 graded and connected practice[7 marks]

A thermometric titration adds HCl to 40.0\,\mathrm{cm^3} of NaOH. Fitted branches are T=20.00+0.160v before the thermal endpoint and T=26.00-0.0800v after it; v is numerical acid volume in \mathrm{cm^3} and T is in ^{\circ}\mathrm C. Acid concentration is 0.400\,\mathrm{mol\,dm^{-3}}. Reagents start at 20.00\,{}^{\circ}\mathrm C. Use additive volumes, density 1.00\,\mathrm{g\,cm^{-3}}, c=4.18\,\mathrm{J\,g^{-1}K^{-1}} and negligible cup heat. (a) Find the fitted endpoint volume and NaOH concentration. (b) Find the molar neutralisation enthalpy at the endpoint. (c) Each temperature used for the rise has stated uncertainty \pm0.05\,\mathrm K. Find the worst-case percentage uncertainty in the rise. (d) Explain why the largest sampled temperature need not locate the endpoint.

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At the intersection 20.00+0.160v=26.00-0.0800v, so 0.240v=6.00 and v=25.0\,\mathrm{cm^3}. Acid moles are 0.0250(0.400)=0.0100\,\mathrm{mol}, equal to the initial base moles. Base concentration is 0.0100/0.0400=0.250\,\mathrm{mol\,dm^{-3}}. The endpoint temperature is 24.00\,{}^{\circ}\mathrm C and the rise is 4.00\,\mathrm K. Both reagent portions warm: Q=65.0(4.18)(4.00)=1086.8\,\mathrm J. Therefore \Delta H=-1.0868/0.0100=-108.68\,\mathrm{kJ\,mol^{-1}}, or -109\,\mathrm{kJ\,mol^{-1}}. The supplied synthetic lines need not reproduce the accepted value for a strong acid and base. Absolute temperature uncertainties add: 0.05+0.05=0.10\,\mathrm K, giving 100(0.10/4.00)=2.5\%. A sampling grid can straddle the intersection; use fitted branches rather than the nearest recorded maximum.

Final answer

25.0\,\mathrm{cm^3}; 0.250\,\mathrm{mol\,dm^{-3}}; -109\,\mathrm{kJ\,mol^{-1}}; 2.5\%.

Connected questions

Question C01

R1.1 graded and connected practice[6 marks]

R1.1 + R2.1 + R3.1. A spill model mixes 20.0\,\mathrm{cm^3} of 0.500\,\mathrm{mol\,dm^{-3}} HCl with 30.0\,\mathrm{cm^3} of 0.200\,\mathrm{mol\,dm^{-3}} NaOH at 22.00\,{}^{\circ}\mathrm C. Use \Delta H_{\rm neut}=-56.0\,\mathrm{kJ\,mol^{-1}} per mole of water, density 1.00\,\mathrm{g\,cm^{-3}}, additive volumes and c=4.18\,\mathrm{J\,g^{-1}K^{-1}}. Ignore cup and room exchange. (a) Find the amount neutralised and the final temperature. (b) Find the concentration of excess hydrogen ions. (c) Calculate the final pH, assuming complete dissociation.

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Acid and base amounts are 0.0200(0.500)=0.0100 and 0.0300(0.200)=0.00600\,\mathrm{mol}. Base limits and 0.00600\,\mathrm{mol} water forms. Released heat is 0.00600(56.0)=0.336\,\mathrm{kJ}. The solution rise is 336/[50.0(4.18)]=1.6076555\ldots\,\mathrm K, so final temperature is 23.61\,{}^{\circ}\mathrm C to two decimal places. Excess hydrogen ions amount to 0.00400\,\mathrm{mol} in 0.0500\,\mathrm{dm^3}, giving 0.0800\,\mathrm{mol\,dm^{-3}}. Thus \mathrm{pH}=-\log_{10}(0.0800)=1.09691\ldots=1.10. The limiting amount determines heat; the excess amount determines pH.

Final answer

0.00600\,\mathrm{mol}; 23.61\,{}^{\circ}\mathrm C; 0.0800\,\mathrm{mol\,dm^{-3}}; pH 1.10.

Question C02

R1.1 graded and connected practice[6 marks]

R1.1 + R2.2. Two trials use the same 0.0200\,\mathrm{mol} reaction and the same insulated receiver of heat capacity 250\,\mathrm{J\,K^{-1}}. Each shows a 3.20\,\mathrm K rise. A catalyst changes the completion time from 80.0 to 20.0\,\mathrm s. The uncatalysed forward barrier is 75.0\,\mathrm{kJ\,mol^{-1}}; the catalyst lowers it by 18.0\,\mathrm{kJ\,mol^{-1}}. (a) Find the reaction enthalpy and both reverse barriers. (b) Find the mean heat-transfer power in each trial. (c) Explain why a faster trial gives the same temperature rise.

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The receiver gains 250(3.20)=800\,\mathrm J, so \Delta H=-0.800/0.0200=-40.0\,\mathrm{kJ\,mol^{-1}}. The reverse barrier is E_{\rm a,f}-\Delta H: without catalyst it is 75.0-(-40.0)=115\,\mathrm{kJ\,mol^{-1}}; with catalyst it is (75.0-18.0)-(-40.0)=97.0\,\mathrm{kJ\,mol^{-1}}. Mean powers are 800/80.0=10.0\,\mathrm W and 800/20.0=40.0\,\mathrm W. A catalyst changes the route and rate but leaves the endpoints and enthalpy unchanged. Equal reacting amount therefore delivers equal heat and equal rise to the same insulated heat capacity.

Final answer

-40.0\,\mathrm{kJ\,mol^{-1}}; reverse barriers 115, 97.0\,\mathrm{kJ\,mol^{-1}}; 10.0, 40.0\,\mathrm W.

Question C03

R1.1 graded and connected practice[6 marks]

R1.1 + S1.4 + R1.3. A biofuel has formula \ce{C3H8O} and M=60.10\,\mathrm{g\,mol^{-1}}. Burning 0.601\,\mathrm g completely warms 150.0\,\mathrm g water by 10.0\,\mathrm K. Use c=4.18\,\mathrm{J\,g^{-1}K^{-1}} and supplied reference \Delta_{\rm c}H=-1800\,\mathrm{kJ\,mol^{-1}} for matching states. (a) Write the balanced complete-combustion equation. (b) Find the water-based experimental molar combustion enthalpy. (c) Find the captured fraction and the amount of carbon dioxide produced.

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Balance carbon and hydrogen first: 2\ce{C3H8O}+9\ce{O2}\rightarrow6\ce{CO2}+8\ce{H2O}. Fuel moles are 0.601/60.10=0.0100\,\mathrm{mol}. Water gains 150.0(4.18)(10.0)=6270\,\mathrm J, giving experimental \Delta_{\rm c}H=-6.270/0.0100=-627\,\mathrm{kJ\,mol^{-1}}. The reference release for this amount is 0.0100(1800)=18.0\,\mathrm{kJ}. Captured fraction is 6.270/18.0=0.348333\ldots, or 34.8\%. Each mole of fuel gives three moles of carbon dioxide, hence 0.0300\,\mathrm{mol}. The experimental value records water heating rather than the complete released energy.

Final answer

2\ce{C3H8O}+9\ce{O2}\rightarrow6\ce{CO2}+8\ce{H2O}; -627\,\mathrm{kJ\,mol^{-1}}; 34.8\%; 0.0300\,\mathrm{mol\ CO2}.

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

Measuring enthalpy changes from the core ideas to full exam questions, with question types, worked examples and graded practice.

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