Question

Part 11: Calculating the enthalpy of formation of MgO(s) Fill in the missing value (4) Ha(g) + 1/2O2(g) → H2O(1) AH = -286 kJ/mol (5) Mg(s) + 2 HCl(aq) → MgCl(aq) + Halg) Hºex (5)= -554 kJ/mol (6) MgO(s) + 2 HCl(aq) → MgCl(aq) + H2O(1) Hev(6) = -193 kJ/mol Calculate AHx(7) for the goal reaction: Mg(s) + 1/2 0.(g) → MgO(s) and show your work. AH'expe() = kJ/mol Calculate the percent error in your calculated value for the above reaction compared with the theoretical value of a no(7) = -608.1 kJ/mol. % error

Answer 1

Ons i to find there for reachón Mg(s) + LO2(g) - MOOCs) - @ Curien mol H2(g) + 102(g) H2O(l) -0 DH'= -286 kJ/mol Mals) + 2HQ (aq) → Mgu, (aq) + H₂19) - Hexp = -55 4 kJ/ Mgo(s) + 2H610q) Mg C2 (aq) + H2O(l) - Shexp =- 193 kJ/mol To obtain the required reaction @ rouerse reaction ③ and add it to and ③ we get H₂(g) + 10₂ (9) Hote) 1H=-286 kJ/mol reg(s) + 2HClaa) rafhlaq) + H2(g) AH = -193 kJ/mol rgth(aq) + H011) angols) +246 laq) AH = 193 kJ/mol . 554 Mg(s) + 102(g) -> Mgo(s) exphº (7) = -286-554+193 = - 647 kJ/mol 1969

Error= | Aritmeo - A Hexpl = 1.608.1 +6471 = 38.9 kJ/mol percentange error: lettoreX 60 error: Error 16 = 38.9 x Ico 608.1 percentage error = 6.4%
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