RI Energetics Remedial

Name: ………………………………………. ( ) Class: ……………. Date: ……………. Raffles Institution Year 5 H2 Chemistry 2024 ChemFocus T3W6 – Energetics Part 1 Notes: ⎯ Pre-ChemFocus Activity ⎯ • Familiarise yourself with the definitions of all enthalpies by reading through Pages 7 – 17 of your notes. • Watch the online Info-pack video IVY > C2025 – H2 CHEMISTRY > Pages > [ChemFocus] T3W6 Energetics > First video only • • •

Self-Check Questions 1. Complete the following table. Enthalpy change Equations Hcombustion (CH4(g)) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Hcombustion (C(s)) Hformation (CO2(g)) Hatomisation (Na(s)) Hhydration (Na+(g)) Hsolution (NaCl(s)) Hformation (NaCl(s)) Lattice energy of NaCl(s) 1st ionisation energy of Na 1st electron affinity of Cl Bond energy of Cl−Cl Hatomisation (Cl2(g)) Hatomisation (CH4(g)) 2. When 30 cm3 of 1.0 mol dm–3 CH3NH2, is added to 25 cm 3 of 1.0 mol dm–3 hydrochloric acid in a beaker, the temperature rose by 5.2 °C. The process efficiency is 90%. Calculate the standard enthalpy change of the reaction, assuming the specific heat capacity of water to be 4.18 J K–1 g–1. [2] Check your answers at IVY > C2025 – H2 CHEMISTRY > Pages > [ChemFocus] T3W6 Energetics > Remaining videos

3. Use the following data to c alculate the standard enthalpy change of solution of ammonium chloride. Lattice energy of ammonium chloride –705 kJ mol–1 Enthalpy change of hydration of NH4+ –307 kJ mol–1 Enthalpy change of hydration of Cl– –381 kJ mol–1 4. Use the following data to construct an energy cycle and hence calculate the enthalpy change for the hydrogenation of ethene gas, CH2=CH2(g), to ethane gas, CH3CH3(g). CH2=CH2(g) + H2(g) ⎯→ CH3CH3(g) Enthalpy change of combustion of CH3CH3(g) –1560 kJ mol–1 Enthalpy change of combustion of H2(g) –286 kJ mol–1 Enthalpy change of combustion of CH2=CH2(g) –1411 kJ mol–1

Tutorial Discussion Questions (to complete before Chemfocus session) 1a A student carried out an experiment to determine the enthalpy change of combustion of ethanol, CH3CH2OH, using the apparatus shown in the diagram. The results that the student obtained are shown in the table below. Mass of water / g 100 Initial temperature of water / °C 25.0 Final temperature of water / oC 31.5 Initial mass of burner and ethanol / g 62.57 Final mass of burner and ethanol / g 62.46 (i) Using the above experimental results, determine the experimental enthalpy change of combustion for ethanol. [3] (ii) Use the data below to calculate the standard enthalpy change of combustion of ethanol. [2] compound C2H5OH(l) CO2(g) H2O(l) Hf / kJ mol−1 –277.7 –394 –286 (iii) Suggest two reasons for the difference in the two values calculated in (a) and (b). ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… …………………………………………………………………………………………...[2] spirit burner containing ethanol tripod stand wire gauze water metal calorimeter thermometer xxxxxxxxxxxx

2a The standard enthalpy cha nges of formation of ammonia, hydrogen chloride and ammonium chloride are given below. Hf[NH3(g)] –46.0 kJ mol–1 Hf[HCl(g)] –91.0 kJ mol–1 Hf[NH4Cl(s)] –314.4 kJ mol–1 Gaseous ammonia reacts with hydrogen chloride to form solid ammonium chloride. NH3(g) + HCl(g) ⎯→ NH4Cl(s), H Using standard enthalpy changes given above, construct an energy cycle and show by calculations that the standard enthalpy change of reaction, H, is –177.4 kJ mol–1. [3]

3a Heating lithium in a stream of hydrogen gas produces white crystalline, ionic lithium hydride, LiH. Construct a clearly labelled energy level diagram to calculate the lattice energy of LiH, using the following data, together with relevant data from the Data Booklet. [4] Enthalpy change of formation of LiH(s) −90.5 kJ mol−1 Enthalpy change of atomisation of Li(s) +159.5 kJ mol−1 Electron affinity of hydrogen atoms −73.0 kJ mol−1

Tutorial Practice Questions (to be attempted IN CLASS) 1b A student used the apparatus below for the combustion of propan-1-ol. The enthalpy change of combustion of propan -1-ol, C3H7OH, is –2021 kJ mol−1. Given that the specific heat capacity of water is 4.18 J g−1 K−1 and assuming no heat loss, calculate the final temperature of the water. [2] Mr of propan-1-ol 60.0 Initial mass of burner and propan-1-ol / g 32.5 Final mass of burner and propan-1-ol / g 31.9 Initial temperature of water / oC 21.0 Mass of water heated / g 200 2b (i) Using the data below, construct an energy cycle and calculate the enthalpy change for the following reaction: CH4(g) + 2O2(g) ⎯→ 2H2O(l) + CO2(g) Equation Enthalpy change, ∆H / kJ mol⁻1 [3] C(s) + 2H2(g) ⎯→ CH4(g) −75.0 C(s) + O2(g) ⎯→ CO2(g) −393.5 H2(g) + ½O2(g) ⎯→ H2O(l) −285.9

(ii) The enthalpy change for the following reaction is −801.7 kJ mol−1. CH4(g) + 2O2(g) ⎯→ 2H2O(g) + CO2(g) Use this information and your answer to (i) to calculate the enthalpy change of vaporisation of water. [2] 3b Magnesium nitride, Mg3N2, is produced when magnesium metal is burnt in pure nitrogen atmosphere. The lattice energy of magnesium nitride can be calculated from a Born–Haber cycle using information from the Data Booklet and the table below. H / kJ mol−1 standard enthalpy change of atomisation of magnesium +148 sum of the first three electron affinities of nitrogen +146 standard enthalpy change of formation of magnesium nitride −461

Self-Check Questions 1. Complete the following table. Enthalpy change Equations Hcombustion (CH4(g)) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) Hcombustion (C(s)) C(s) + O2(g) → CO2(g) Hformation (CO2(g)) C(s) + O2(g) → CO2(g) Hatomisation (Na(s)) Na(s) → Na(g) Hhydration (Na+(g)) Na+(g) + aq → Na+(aq) Hsolution (NaCl(s)) NaCl(s) + aq → Na+(aq) + Cl−(aq) Hformation (NaCl(s)) Na(s) + ½Cl2(g) → NaCl(s) Lattice energy of NaCl(s) Na+(g) + Cl−(g)→ NaCl(s) 1st ionisation energy of Na Na(g) → Na+(g) + eˉ 1st electron affinity of Cl Cl(g) + eˉ → Clˉ(g) Bond energy of Cl−Cl Cl2(g) → 2Cl(g) Hatomisation (Cl2(g)) ½Cl2(g) → Cl(g) Hatomisation (CH4(g)) CH4(g) → C(g) + 4H(g) 2. CH3NH2 + HCl → CH3NH3+Cl– 𝑛CH3NH2 = 30 1000 × 1.0 = 0.0300 mol 𝑛HCl = 25 1000 × 1.0 = 0.0250 mol CH3NH2 is in excess, hence 𝑛reaction=𝑛HCl q = mcΔT q = (30 + 25) × 4.18 × (+5.2) × 100 90 = +1328 J ΔH = – 𝑞 𝑛 = –(1328 / 0.0250) = –53132 J mol–1 = –53.1 kJ mol–1 3. ΔHsol = ∑ΔHhyd – L.E. = –307 + (–381) – (–705) = +17 kJ mol–1

4. 7 2O2(g) + CH2=CH2(g) + H2(g) ⎯→ CH3CH3(g) + 7 2O2(g) 2CO2(g) + 3H2O(l) ΔH = –1411 + (–286) –(–1560) = –137 kJ mol–1 ΔH –1411 –286 –1560