Question

Hi there

Could you answer these four parts of a question please? it is for a homework and i need it asap.

thanks

a. Write out the balanced chemical equation of titanium (III) iodide decomposing to titanium and iodine gas. Balance with whole number ratios. b. Calculate the heat of formation (kJ/mol) of titanium (III) iodide when 27.0 g of titanium (III) iodide decomposed to titanium (Ti) and iodide gas (12), and changed the temperature of 318 g of water from 42.9°C to 15.8°C c. Based on the AH®rxn is the reaction endothermic or exothermic? d. How many liters of iodine gas reacted if Ahºrxn=279.7 kJ at 28.9'C and 5.89 atm ? Use equation from a) for mole ratios.

Answer 1

I @ 2 Ti Iz o 2 Tig + 3 Iz cg ( + since Temperature of inster droped from a 42.9°C to T = 15.8°C o - 318 g, water - 4.184 Igor 9 = incoT - ma (tz-T). e = 3188 4.184 x (15.8-429) - - 36056.88 J

here - sige show that heat lost from water Now this heat is taken for the decomposition purpose of 27.0g of T (11) iodide (MI; I = 428.580 gmot) moles T. (111) iodide = 27.0 = 0,062998 428.580 - moks T require 360 56.88 So, 0.062998 motes 36056.88 so I not require 0.062998 3 5572342.88 J santholoog 570342 88 ks So, heat of formation of Ti Iz is AH = -572.34 kJ mol cs Scanned with CamScanner

mor © Reaction is Endothermic ] - As oHorn = + 372.34 ks this Heat is taken form water so that water's temperature get reduces. 1 Ohoan -+279.7 65 Imol Ti Iz 572.34 ks fror 279.7 ks - 279.7 mores of Tiu decomposed. 572-34 » 2 ее т., — Imol Ti Iz зи , z mor Iz 279.7 572.34 mol Tidz Nw 2 x 279-7 mort . 572.34 0.733. un Iz Ics Scanned with CamScanner

po= nRT V = nRT where p= 5.89 atu T=28-9°c = (28.9 +273)k m20.733 mol n-atin R - 6. 082057 mork 0.733 x 0.082057 x (301.9) 5-89 3.083 So, the volume of Iz cq is Tv=3.083 L Scanned with CamScanner