Question

The salt ammonium chloride dissolves in water according to the reaction: NH_CI) —NH4+ (aq) + Cl(aq) (a) Calculate the standard enthalpy change AH° for this reaction, using the following data: AH NHACI() = -314.4 kJ mol-1 AHNH4+(aq) = -132.5 kJ mol-1 AH; Cr(aq) = -167.2 kJ moll kJ (b) Calculate the temperature reached by the solution formed when 36.8 g of NH4Cl is dissolved in 0.186 L of water at 23.7 °C. Approximate the heat capacity of the solution by the heat capacity of 186 g of pure water (specific heat capacity = 4.18 J glºc-), ignoring the mass of the salt. °C (c) Heats of reaction find practical application in hot packs or cold packs. Would this dissolution reaction be appropriate for the preparation of a hot pack or a cold pack? Submit Answer 2 question attempts remaining

Answer 1

a) AH creaction) = Eard products) -EDHA (reactants) for the reaction NHull(s) - NHY (aq) + Celars AH reaction) = 1 DH f (NH 4 ) tarif (ce) - AHF (NHyce) =-132. 5 kJimal - 16702 kJ. mol-1-3 1404 kJ. mol) = 14.7 kJ molt Standard enthalpy change AH = 14.7 kJ)

b (0494) (CS) Specific beat capacity - Heat capacity (0) mas's (m) Heat capacity CZ CS X mass of pure water (m) - 4018 Jgt oct X 186 g = 777.48°3,47 Now, Heat capacity (c) = Heat involved (0) Change in temperature hi Q = C. AT Q = AH = 14.7 kJ per mol. AT = 5T, 7,-2.7°C AT = o 14.7 x 103 J x 1 C ou imol 777048 Joct. AT = 18.91 ic for I mol. Moles of NHyll used = mass - molar mass = 36.89 = 0.68 8 mol in 53.49 g/mol 0.186 L AT = 18.91°C 0. 688 mol - 13.01°C i mol AT = T2-T, = T2- 23.7° C = 13.01°C T = 36.7°C Temperature reached = 36.7°C This dissolution reaction is an endothermic eaction. We need to provide heat for this reaction. Hence this dissolution. would be appropriate for a cold pack Tas it would take heat from the and give a cooling effe from the outside