Question

3.(15pts) In each case, find the missing AH. (kJ/mole) (a) If 2 Al(s) + Fe2O3(s) ----> Al2O3(s) + 2 Fe(s) AH,xº = -851.5kJ/mol then 1/8 Al2O3(s) + 1/4 Fe(s) --> 1/4 Al(s) + 1/8 Fe2O3(s) AHX = ? (b) If CaO(8) + 3 C(s) --> CaC2(s) + CO(g) AH,” = 464,8 kJ/mol then 6 CaO(s) + 18 C(s) ---> 6 CaCz(s) +6 CO(g) AHX = ? (c) If N2(g) + 2 O2(g) ---> N204(g) and 2 NO2(g) ----> N2(g) + 2 02(g) AHex = 9.2 kJ/mol AH X = -33.2 kJ/mol then 3N204(g) ----> 6 NO2(g) AHX = ? 4. (14 pts) Calculate AH xº (kJ/mole) for the reaction C2H4(g) + 6 F2(g) ---> 2 CF.(g) + 4 HF(g) AHım (kJ/mole) = ? Using only the following data H2(g) + F2(g) ---> 2 HF(g) C(s) + 2 F2(g) -----> CF4(g) 2 C(s) + 2 H2(g) -----> CH.(g) AH(kJ/mole) -537.0 - 680.0 + 52.3

Answer 1

3(a) 2 Al (s) + Fez Oz (s) → A1,03(S) + 2 Fers) SM-85).Sks)mot on reversing the equ we get A1,03(s) + zte(s) - 2Al(s) + Fez 03 (5) shio son = 851.5kJ/mol Dividing the above equ by 8 we get $4130318) + Fe(s) 32415) + Fe2O3(s) stigma = 851,5 kJ/mol SHoxn = 106.4 kJ/mol (6) C9015) +36(S) → Cacz(s) + cocg) shormon = 464.8 kJ /mol ми/hairy w love as a b w 6 laols) + 180 (8) 6 Calz(s) + 6 cocg) shpix = 6* 464.885 /mol = 2788.8 kJ/mol

(C) N2 (9) + 202 (9) ► N204 CG) A Momen = 9.2 kJ/mol Reversing and multiplying the above equ by 3 we get 31,0469) - 3N2 CG) + 60,69) i) SHE = -3x9.2KJ/mol = - 27.6kJ/mo). 2002 (9) Nq Cq) + 202 (9) Ahryn = -33.2 KJ/mal Reversing and multiplying the above equ by 3 we get 3N₂ (9) + 602 (9) 6 Noz (9) - (H) Sh ran = 33.2 ks 1 mol X 3 Adding eqn (i) and til we = 99.6KJ/mol get: 3 N, 04 (9) 6N02 (9) Ahram = -27.6 kJ/mol + 99.6kJ/mol suram = 72.0 kJ/mol

4) 2015) +2H2 (9) → (zHy (9) om = 52.3 kJ/mol Reversing the above equ we get ... C2 Hy (9) 22(5) + 2H₂ C9) DH8 xn=-52.3 KJ/mol - (i) C(s) +26219) multiplying the 2065) + 462 (g) = (fy (9) Stox = -680.0 kJ/mol above equ by 2 we get → 26f429) Steam = -2x680.0kJ/mol = -1360.0 KJmol - cii) H2 (9) + f(g) -> 2HF(g shisn = -537.0 XJ/mol multiplying the above equ by 2 we get 2 H2(g) + 27269) ► 4Hf (g) shºorn = -2% 537.0kJ/mol 10) = -1074.0 kJ/mol Adding eq" (i), cii) and cin) we get Czty (g) + 672 (9) ► 2014 (9) + 4 +fcg) A Horyn = -52.3 kJ/mod -1360.0 kJ/mol-167400 ki/mol shorxh = -2486.3 kJ/mol