Question

Calculate the enthalpy of the reaction 2NO(g) + O2(g)2NO2(g) Hess's law states that "the heat released or absorbed in a chemical process is the same whether the process takes place in one or in several steps." It is important to recall the following rules: given the following reactions and enthalpies of formation: 1. ¢ N2(g) + O2(g)+NO2(g), AH; = 33.2 kJ 2. N2(g) + O2(g) NO(g), AHB = 90.2 kJ 1. When two reactions are added, their enthalpy values are added. 2. When a reaction is reversed, the sign of its enthalpy value changes. 3. When the coefficients of a reaction are multiplied by a factor, the enthalpy value is multiplied by that same factor. Express your answer numerically in kilojoules. View Available Hint(s) AO + O 2 ? AH- Submit Part B Calculate the enthalpy of the reaction 4B(s) + 302(g) 2B,0:(s) given the following pertinent information: 1. B,0,(s) + 3H2O(g) +302(g) + B.H.(g), AH = +2035 kJ 2.2B (s) + 31128-B2016 (8), AND - +36 kJ 3. H2(g) + O2(g) →H2O(l), AH = -285 kJ 4. H2O(1) H2O(g), AHD = +44 kJ Express your answer numerically in kilojoules. View Available Hint(s) 0 ANO O ? Allº-

Answer 1

S. 2NO(g) + O₂(g) - 2 NO₂C9 AH = ? 1. 1 2 N29) + O2(g) NO₂0 2. 2 N₂ (g) + O2(0) - NOO AAA = 33.2 kJ Ato = 90.2 kJ Revorse the equation 2 and add with 1. fait fr ) - 10₂ @) NO(g) 18 & 20. 8) NO(g) +202 (9) NO₂ (9) AHF AHA + (-48) A H= 133.2-90.2) kJ A = -57KJ Multiply net equation by 2. 2 Nog + O2 (9) 2 NO2 AH = 2x Ate AH=2xt-57) kJ = -114 kJ @ Port B 4B15) + 302(g) → 2 B₂O₃ (5) AH=? Reverse 30₂ (9) + By H6 (2) B₂O3 8) + 3H2019) write some 2.816) + 3H₂(g) - By H6 (8) Remorse and relations 3 Holl) -> 3H2 (3) + 30.09 Reverse and multiply 3 H₂O (9) - 3H₂010 - 15 02162B3) > B20 multiply net equation by 2. AH=2x -AHO + AM3 -3AH8 - 3448} A H=2x L-2035 + 36 - 3xl-285) - 34447 LAH=- 2552 kJ SO2 + by 3 of equation 4

Part A

∆H = -114 kj

Part B

∆H = -2552 kj