Question

9. Given the following data: 2C1F(g) + O2(g) → Cl2O(g) + F20(g) 2C1F3(g) + 2O2(g) → Cl2O(g) + 3F20 2F2(g) + O2(g) → 2F20(g) AH = 167.4 kJ AH = 341.4 kJ AH = -43.4 kJ Calculate AH for the reaction CIF(g) + F2(g) → CIF3(g)

Answer 1

Hess's law Enthalpy is a state function. - depends only on the current state of the system. The enthalpy change (OH) for a reaction depends only on the initial and final conditions and not on the path taken. o Whether a given reaction occurs in one step or many steps, the enthalpy change remains constant. Hess's law : The enthalpy change associated with a reaction carried out in multiple steps is equal to the sum of the enthalpy changes for each individual step Oriven reaction ostep-1 : Manipulate given equations to most closely resemble equation of interest. & step-2o Perform same speration on est as an equation. If we flip equation, flip the sign on SH.

0 Step-3: If you multiply or divide equation multin multiply or divide same numbers, the chemical it by the Step 4. Add new equations together. 10 Step-5; 3 cancel out any compounds that are Same on both sides of the reaction arrow, Step 6: Write new equation and double check to o make m sure it matches the equation of interest. at for calmlate - Clf g) + F2 (g) clf3(g) Given step wise realtions, 2 clF61+ O2(g) 42 (9) + f2 (2) TH= 167.4 kJ 2 uf3 g + 202 (g) ant 20e (9) cho Chog 3 + 3 f2002, OH = 314.4 kJ 2 f2 (97+ (a1 2620 G, CH = -43.4 kJ ① is multiply by / Clf (g) + 1/2 0:2 (9) 1/2 cl2o (9) + 1/2 F20 (2) TH= 1674 kJ Equation o is flipped, multiply by 1/2 = 83.7ky 12 1₂ 0(2) + 3/F2 0 (8) CIE3 (9) + 02 (2) alt= - 314.4 ng Equation is multiplied by 12 =-157.2 kJ togt 1/2029 F20 (9) Olt - 43.4 kJ Add above equations = -217 kg elf g) +2028) +2012°(8)+ 3/259391, CI +1/24,0%) 4H = (837-1672) + F₂ W) +1/ 22 (3) -217 73 cancellation, : -95.2 kg - CIF (&) the (2) = CIF 3 (3) 41 - 95.2 kJ & elf3(8) + O₂

@ The balaced chemi cal equation given, 4 MHz (9) +502 (9) UNO(g) + 6H₂O & MOW, Enthalpy of reaction = Enthalpy (Ho) Enthalpy of formation of products Enthalpy formation of realtants E SHf (producet) - EoH f (reactants > 49 : 4 x 4 (1) + 6 x 4 (194)-( 4 ) ( NHỊ) The + 5 498 (02)) rumorx ( 90 kJ/m.1) +6*(-24243/]-1444-8067/mals +5 (0.00 13/png) - 360 KJ - 1452 kg) -T - 320 kJ+oky] 4H A AH = (3 , 0 - [4 52 + 32 0 + - ) k y oto = -772 kg . Ho for the reaction= 772 kg " is diartomic AHF for O2 so Stable molecule = 0.00 kJ/mol