Question

1.(3pts) Use the following data to determine the AH... of 2 CHO(1) + 21 Og) --> 14 COMg) + 16 HẠO(g) Substance AH(kJ/mol) Substance AH?(kJ/mol Data: C Hag) CH,OH(L) C.H.O(1) CH, O) H.O(g) CO(g) O.(g) 20.9 -239.0 -277.0 -316 0 -241.82 -110.53 C.H.(5) CH,OH(g) C3H20 (1) C-HBO(l) H.O(1) CO; N (8) -104,0 -201.0 -298.6 -340.0 -285.83 -393.51 2.(2pts) If you are given 338,0 g 0 () and a sufficient amount of the other reactant, calculate the heat associated with reaction in problem 1 and indicate if the heat is absorbed or evolved. Atomic masse: O = 16.00 3.(5pts) Consider the thermochemical reaction: 4 FeO (s) + 4 CO(g) ----> Fe(s) + 4 CO2(g) AH® - ? Use the following data to determine the value of the missing quantity in the above equation. Fe 0,(s) + 3 CO(g) ----> Fe(s) + 3 CO2(g) 3 Fe,0,() + CO(g) ---> 2 Fe3O4(s) + CO2(g) 3 FO(s) + CO2(g) ----> Fe3O4(s) + CO(g) AH AH AH = -23.0 kJ/mol = -39.0 kJ/mol = -18,0 kJ/mol

Answer 1

Answer 3

Answer 1 AH°rxn = [14(AH°FCO2) + 16(AH°FH20)] - [2AH°F(C7H160) - 21AH°FO2] = [14(-393.51)+16(-285.83)] - [2(-340.0) - 21(0)] = -9402.42 kJ Answer 2. Moles of O2 = 338 g/32 g/mol = 10.56 mol Moles of C7H160 required for 10.56 moles of O2 = (2/21)*10.56 = 1.008 mol From Answer 1, AHrxn = -9402.42 kJ This is heat released for two moles. Heat released for combustion of 1.008 mol of C7H160 = (-9402.42/2)*1.008 = - 4738.8 kJ Heat is evolved.

I Cany tqno ko (s) + 3 (а) 2 (c) + 3 o, ta) 46 3,041) + co (4) - 2k194 +,1) Eqn . 3rego + CO₂(g) reg On (s) + Co (g) Мару а*@ 44 Чko+ час, — ЧР 30u +4 o (1) Дн. Чx-s-o) . -24-о кә*/ Мана 4" 6 2 3, 0, + 2 0 — це 40 + , Анах (-34-0) - 26 ока -407 Muluply ear ① by 2.0 че +6 дн° 2xt-s-o) 2 -26 oles | ли (6) ч 0 2со, — Lean дн° , че +6 2 Pe, 0, subtract eqn +с со from 0, + 2CD ди” 2 -ачо -t-aь о) + 2-ota) 2 fe,

Add eqn 7 and 6, we get,

4FeO + 4CO -----> 4Fe + 4CO2

enthalpy change = -36.0 +2.0 = -34.0 kJ