Problem: Calculate the energy involved in the oxidation of elemental sulfur to sulfur trioxide from reactions: 1) S (s) + Ô2 (g) SO, (g) H = -296.0 kJ 2) 2 SO, (g) + O, (g) 2 SO3 (g) H, = -198.2 k.J 3) S (s) + 3/2 O2 (g) SO, (g) H3 = ?
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- Consider the problem below: (Equation 1) H2g) + 1/2 O2(g) H2O) AH = -285.8 kJ/mol ---> (Equation 2) C(s) + O2(g) CO2(g) AH = -393.5 kJ/mol --- (Equation 3) 2 C(s) + H2(g) ---> C2H2(g) AH = 226.7 kJ/mol %3D (Equation 4) 2 C2H2(g) + 5 O2e) ---> 4 CO2(e) + 2 H20) AH = ??????? ---> 4 CO2(g) %3D To find AH for equation 4 you must: Screen Reader Version Consider the problem below: (Equation 1) H2(g) + 1/2 O2(g) arrow H201) AH = minus 285.8 kJ/mol (Equation 2) C(s) + O2(g) arrow CO2(g) AH = minus 393.5 kJ/mol 2 C(s) + H2(g) arrow C2H2(g) AH = 226.7 kJ/mol (Equation 4) 2 C2H2(g) + 5 O2(g) arrow 4 CO2(g) + 2 H2OU) AH = ??????? %3D To find AH for equation 4 you must:6. (a) How much does the free energy (kJ/mole) of 15 g of pure Cu change by adding 5 g of pure Zn at a temperature of 1100 degrees C? Assume an ideal solution, and that Go(Cu) = -1200 kJ/mole, and Go(Zn) = -2000 kJ/mole at this temperature. A(Cu) = 63.55, and A(Zn) = 65.39. b) By how much more does the system free energy change by adding 5 g of Ga to the mixture in part a)? A(Ga) = 69.7, and Go(Ga) = - 3000 kJ/mole. %3DPROBLEM: A key step in the production of sulfuric acid is the oxidation of SO₂(g) to SO3(g): 2SO₂(g) + O₂(g) → 2SO3(g) At 298 K, AG = -141.6 kJ; AH = -198.4 kJ; and AS = -187.9 J/K (a) Use the data to decide if this reaction is spontaneous at 25°C, and predict how AG will change with increasing T. (b) Assuming AH and AS are constant with increasing T, is the reaction spontaneous at 900.° C? PROBLEM: ΔΗ = 58.1 kJ At 25°C (298 K), the reduction of copper(1) oxide is nonspontaneous AG = 8.9 kJ). Calculate the temperature at which the reaction becomes spontaneous. AS = 0.165 kJ/K
- [4] rxn 2. Given the following information, calculate AH for the reaction. 2C3Hs(g) +1002(g) → 6CO2(g) + 8H₂O(1) | AH¦ (C³H8(g)) = -103.9 kJ/mol; AH; (CO2(g)) = -393.5 kJ/mol, AH; (H₂O(1) = -285.5 kJ/molCH,(g) + 2 0,(g) AH = -809.0 kJ/mol rxn co,(g) + 2 H,0(g) AH cond = -81.3 kJ 2 H,0(1) Given the energy diagram above, what is AH when 0.5 mol of methane is combusted? -802.3 kJ +802.3 kJ -1604.6 kJ +401.15 kJ +1604.6 kJ -401.15 kJCalculate AH for the reaction: 2NH3 (g) + O2(g) → N2H4 (1) + H2O(1) given the following data: 2NH3 (g) + 3N20(g) → 4N2(g) + 3H2O(1) AH = –1010. kJ N20(g) + 3H2 (9) → N¿H4(1)+H2O(1) AH = -317 kJ N2H4 (1) + O2 (g) → N2(g)+ 2H2O(1) AH = -623 kJ H2 (9) + 02 (9) → H2O(1) AH = -286 kJ ΔΗ - kJ
- (a) Fe(OH);(s) – FeO(OH)(s) + H;0() AH =-24.3 kJ/mol AH - 35.6 kJ/mol 298 (b) Fe;O(s) + 3H:O(() 2Fe(OH)3(s) (c) 2FEO(OH)(s) Fe,Os(s) + H20() !! 298 AG98 for process (a) is -31.3 kJ / mol and AG for process (c) is 1.7 kJ /mol. Calculate G2 O A) 179.0 kJ/mol O B)-372.6 kJ/mol OC) -490.6 kJ/mol OD) 231.2 kJ/mol Q 5: for FeO(OH)(s)! 298Given the following enthalpy change values for the reactions below: A + 2B – C; A,H = 350.1 kJ mol-1 E + F - 2C + D;A,H = 476.2 kJ mol-1 ½ E - G + 2B ; A;H = -229.6 kJ mol-1 Calculate A,H for the following reaction (in kJ mol-1) using Hess's Law: 2A + D - F + 2GCHALLENGE PROBLEM (If time permits) One reaction involving the conversion of iron ore to iron metal is Fe0 (s) + CO (g) → Fe (s) + CO, (g) Calculate AH for the reaction given the following data. 4. 1. 3Fe,0, (s) + CO (g) →2Fe,0, (s) + CO, (g) AH = -47.0 kJ 2. Fe,0; (s) + 3CO (g) →2F (s) +3 CO, (g) AH = -25.0 kJ 3. Fe,0, (s) + CO (g) →3FE0 (s) + CO, (g) AH = +19.0 kJ
- PLease just explain steps to me: What quantity, in moles, of hydrogen is consumed when 151.1 kJ of energy is evolved from the combustion of a mixture of H2O (g) and O2 (g)? H2 (g) + 1/2 O2 (g) --> H2O (l); delta rH= -285.8 kJ/mol-rxn6. Given that the AHf for CaO, H3PO4, Ca3(PO4)2 and H2O, are, respectively, -635, -1267, -4126, and - 286 kJ/mole, (a) calculate the AH in kJ for the reaction: 3CaO + 2H3PO4 → Ca3(PO4)2 + 3H2O (b) calculate how much heat is released or absorbed when 15g of CaO reacts with excess phosphoric acidCalculate AH for the reaction NH3 (g) + CH4 (g) → HCN (g) + 3 H2 (g) , from the following data. N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔΗ- – 91. 8 kJ /mol C (s, graphite) + 2 H2 (g) → CH4 (g) - 74. 9 kJ / mol AH = 2 C (s, graphite) + H2 (g) + N2 (g) → 2 HCN (g) AH = ΔΗ: 270. 3 kJ / mol