Question

RULES 1. If reverse equation, change sign of AH value. 2. If double equation, double AH value. 3. If halve equation, halve AH value. 4. NOTE that as states (solid, liquid, or gas) change, energy is absorbed or released so AH values likewise change. 5. AH' - O for all elements in their most stable form at 1 atm. pressure and 25°C. 6. AH is calculated as ZAHproducts. ZAHreactants. Rearrange the thermochemical equations given to determine the AH value for the equation of interest. Determine: Ca(OH)2(s) + H2S040 - CaSO4(s) + 2 H2O() Given Enthalpy (k/mol) -72.48 1. CaO(s) + H2O) = Ca(OH)2(s) 2. Cao(s) + H2SO4(0) – CaSO4(s) + H2O(0) -299.59 kl mol Question Help: Gwritten Example

Answer 1

® The given reactions are s- 17 Cao (s) + H₂011) An (a lon) (s) - 72.48 2. Ca o(s) + H₂SO4 l) ~ Casoy (5) kJ/mol + H₂olu) -29959 ki/mal We need to find all value for the reactions - Ca 1042 (s) + H₂SO4 (1) lasoy (8) + 2H₂oll) We need to reverse reaction (1) and then add it to reaction (2) to get the desired reaction. On Reversing reaction (1) the sign of all changes and on becomes equal to & 72.48 kJ/mol do Ca 10W₂ (s) cao (s) tl oll) AH, = +72.48 kJ/mol Ca0(s) + H₂SO4 (l) Casoy (s) + H₂0 (1) os Adding these two we get kJ/mol Ca (OH (s) + H₂SO4(l) CaSO4(s) + 2H₂011) én Ihreaction - All + AH2 = (72148-299.59) kJ/mol Alreation =-227.11 kJ/mol Alg = -299.59 . All value for reaction Calon), (3) + H₂SO4 ( → Casoy (s) + 2 H₂0 (1) comes out to be - 227: 11 kJ/mol

The given reactions are 3 - 1.) 2Fe(s) + 30219) 2-) C(s) + 1 ₂19) 3) Colg) + 1 O2 (g) We need to find all reaction Fe2O315) colg Colg) for Enthalpy (Ah) ,-833.6 kJ/mol - 119.6 kJ/mol - 23116 kJ/mol 2 fez 03 (s) + 36(S) — 360_19) +ufe (s) Multiply equation 2 and 3 by (3) and add them we get in 306) + 3 0219) -> 3 lotgi olla = 3x (-119,6) - = - 358 kJ/mol 3 colg) +3 0₂ (9) 310_19) Allz= 3x1-231 6] 1 = – 694.8kJ) 4.) 3 C(s) + 30, 19) 3 W219) Oly=? on ally = alla + Alz = (-358 - 694.8) kJ/mol ally = -10 52.8 kJ/mol e mol

Now multiply reaction (1) multiply reaction (1) by 2 and reverse it and then ad it to reaction 4.) we will get our desired reaction afe2O₃ (5) 30to) + 4 fe(s) All, = (2x 83. = 166712 ki/mal 3 C(s) + 30279). - 3 co2(g) oly= - 1052.8 kJ/mol rection 2 fez O₂ (s) + 3 C(s) 4 fels) + 3 o219) Oltration = ? лимоном - ди, + вич - (1667.2 - 105218) = 614.4ki/mal o. 1 Al reaction = 614,4 kJ/mol o. All value for reaction afe & O₂ (s) +36(s) 3 0₂ (9) tyfels) comes out to be 614.4 kJ/mol