Question

6. Using the enthalpy of formation values in Appendix 4 of this lab manual, calculate AH' for the following reaction. 2 Na(g) + 2 HyOz - 2 NaOH(aq) + H2 7. The following reaction has a AH of -1196 kJ. Use this value and the information in Appendix 4 to calculate the AH of CIF3(Q). 2 CIF3() + 2NH3() - N2 + 6 HF() + Cl2 AH'= -1196 kJ

Answer 1

O Enthalpy of Reaction from Enthalpies of Formation: The enthalpy of reaction can be calculated by Elements in Standard States of AH;° = { vB AH{°, products – ve AHt°, reactants 2HI(g) ve is the stoichiometric coefficient Heat released AH, = 52.5 kJ mol-' 1 Reactants AH, H2(g) +12(9) above equation holds good for any reaction as the same reference state is used for reactants and products (shown in figure). Products Enthalpy of Reaction

2 Nals) + 2 H₂Old) 2 NaOH(aq) + H₂ (3) How = 2xore ( Ng Noor, aq) + 1x AHG (M2, 3, OMEO (N9,5) + 2x DHE" (4,0,) from Appendix SHE (Nas) =0 AH ( H, 3) = 0 Ohp (H₂0, d) = -285.8 kg/mol BHpº (Nach, ag) = -469.45 kJ/mol 10 DHMW = 2x- 469.15 + 1X0 - 2X0 -2*6-9858) Shox = -366.7 KI So, CIF3 (91 + 2 Nike 19) — N219) + 6HF(9) + 213) Othew = -1196 kg are no = 1x One (N219) + 6x One (HF, g) + 1x0Mp 1 1419) — Hp (C1f310) - 9x OHF° (N73')

AHp (N2, 91=0 AHE (Claigl=0 DHp (HF1975 -273. 3 kg/mol Heº (n + 1) = -4 10 13/- -1196 = 140+ 6x-273.3 +1% 0 - OHpo (C1F3131 - 2x - 45.90 *AHE (CIF3, 9) = 362.0 Kolonel