Hess’ Law energy changes are state functions. The amount of energy depends only on the states of the reactants and products, but not on the intermediate steps. Energy (enthalpy) changes in chemical reactions are the same, regardless whether the reactions occur in one or several steps. The total energy change in a chemical reaction is the sum of the energy changes in its many steps leading to the overall reaction.
Practice Problem 6.45 ΔH = + 2940 kJ
Practice Problem 6.46 ΔH = - 537.5 kJ
PCl3 should be a product - first equation is good! Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: PCl5(g) → PCl3(g) + Cl2(g) P4(s) + 6 Cl2(g) → 4 PCl3(g) ΔH = - 2439 kJ P4(s) + 10 Cl2(g) → 4 PCl5(g) ΔH = - 3438 kJ Write the thermochemical equations given such that the reactant and products are in the proper places PCl3 should be a product - first equation is good! PCl5 should be a reactant - second equation needs to be ‘flipped’ products P4(s) + 6 Cl2(g) → 4 PCl3(g) ΔH = - 2439 kJ 4 PCl5(g) → P4(s) + 10 Cl2(g) ΔH = + 3438 kJ reactant intermediate Notice: when I ‘flipped’ the 2nd equation, I flipped the sign of ΔH
¼ ¼ P4(s) + 6 Cl2(g) → 4 PCl3(g) ΔH = - 2439 kJ 4 PCl5(g) → P4(s) + 10 Cl2(g) ΔH = + 3438 kJ ¼ ¼ Apply ‘multipliers’ to arrive at the desired final equation 1/4 P4(s) + 3/2 Cl2(g) → PCl3(g) ΔH = - 609.75 kJ PCl5(g) → 1/4 P4(s) + 5/2 Cl2(g) ΔH = + 859.5 kJ Cl2(g) ΔH = + 249.75 kJ PCl5(g) → PCl3(g) + Cl2(g)
Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: N2H4(l) + H2(g) → 2 NH3(g) N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3 H2(g) ΔH = - 37 kJ N2(g) + 3 H2(g) → 2 NH3(g) ΔH = - 46 kJ CH4O(l) → CH2O(g) + H2(g) ΔH = - 65 kJ Write the thermochemical equations given such that the reactant and products are in the proper places N2H4 should be a reactant - first equation is good! NH3 should be a product - second equation is good! H2 should be a reactant - third equation needs to be flipped!
N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3 H2 (g) ΔH = - 37 kJ N2(g) + 3 H2(g) → 2 NH3(g) ΔH = - 46 kJ CH2O(g) + H2(g) → CH4O(l) ΔH = + 65 kJ N2H4(l) + H2(g) → 2 NH3(g) ΔH = -18 kJ Don’t need no ‘multipliers’ intermediates → CH2O, N2 produced, then consumed catalyst → CH4O consumed, then produced
Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: 2 C2H4O(l) + 2 H2O(l) → 2 C2H6O(l) + O2(g) C2H6O(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH = - 685.5 kJ C2H4O(l) + 5/2 O2(g)→ 2 CO2(g) + 2 H2O(l) ΔH = - 583.5 kJ Write the thermochemical equations given such that the reactant and products are in the proper places C2H6O should be a product - first equation is needs to be flipped! C2H4O should be a reactant - second equation is good!
2 C2H4O(l) + 2 H2O(l) → 2 C2H6O(l) + O2(g) 2 CO2(g) + 3 H2O(l) → C2H6O(l) + 3 O2(g) ΔH = + 685.5 kJ C2H4O(l) + 5/2 O2(g) → 2 CO2(g) + 2 H2O(l) ΔH = - 583.5 kJ 2 2 ‘multipliers’ 4 CO2(g) + 6 H2O(l) → 2 C2H6O(l) + 6 O2(g) ΔH = + 1371 kJ 2 C2H4O(l) + 5 O2(g) → 4 CO2(g) + 4 H2O(l) ΔH = - 1167 kJ 2 H2O(l) O2(g) 2 C2H4O(l) + 2 H2O(l) → 2 C2H6O(l) + O2(g) ΔH = +204 kJ
Standard Enthalpies of Formation The standard enthalpy of formation of a substance, denoted DHfo, is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state. Note that the standard enthalpy of formation for a pure element in its standard state is zero. 4
A p p e n d i x A - 14 A p p e n d i x A - 5 Standard Enthalpies of Formation A p p e n d i x A - 14 Standard Enthalpies of Combustion A p p e n d i x A - 5
Standard Enthalpies of Formation Hess’s Law for standard enthalpies of formation 4
You record the values of DHfo under the formulas in the equation, multiplying them by the coefficients in the equation. 4(-46.11) 5(0) 4(+90.25) 6(-241.8) You then calculate DHo by subtracting the values for the reactants from the values for the products. [6(-241.8) + 4(+90.25)] – [4(-46.11) + 5(0)] [(-1089.8)] – [(-184.44)] ΔH = -905.36 kJ 4
∆G = ∆H° - T∆S° ∆G = (-399.71) – (673)(-0.2776) ΔG = -212.9 kJ 2(-238.7) 3(0) (-393.51) 2(-241.8) 2(0.1268) 3(0.205) (0.2136) 2(0.1887) [2(-241.8) + (-393.51)] – [2(-238.7) + 3(0)] ΔH° = -399.71 kJ [2(0.1887) + (0.2136)] – [2(0.1268) + 3(0.205)] ΔS° = -0.2776 kJ/K ∆G = ∆H° - T∆S° ∆G = (-399.71) – (673)(-0.2776) ΔG = -212.9 kJ spontaneous 4
C2H4 (g) + H2 (g) → C2H6 (g) ΔH = -137. kJ C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l) ΔH = -1411. kJ 2 CO2 (g) + 3 H2O (l) → C2H6 (g) + 3½ O2 (g) ΔH = +1560 kJ H2 (g) + ½ O2 (g) → H2O (l) ΔH = -285.8 kJ C2H4 (g) + H2 (g) → C2H6 (g) ΔH = -137. kJ
2 2 3 ΔH = -1628 kJ 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) N2 (g) + O2 (g) → 2 NO (g) ΔH = -180.5 kJ 2 2 NH3 (g) → N2 (g) + 3 H2 (g) ΔH = +91.8 kJ 2 2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = -483.6 kJ 3 2 N2 (g) + 2 O2 (g) → 4 NO (g) ΔH = -361 kJ 4 NH3 (g) → 2 N2 (g) + 6 H2 (g) ΔH = +183.6 kJ 6 H2 (g) + 3 O2 (g) → 6 H2O (g) ΔH = -1450.8 kJ 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) ΔH = -1628 kJ
2 2 2 H2(g) + 2 C(s) + O2(g) → C2H5OH(l) ΔH = -486. kJ 2 CO2 (g) + 2 H2O (l) → C2H5OH (l) + 2 O2 (g) ΔH = +875 kJ C (s) + O2 (g) → CO2 (g) ΔH = -394.51 kJ 2 H2 (g) + ½ O2 (g) → H2O (l) ΔH = -285.8 kJ 2 2 CO2 (g) + 2 H2O (l) → C2H5OH (l) + 2 O2 (g) ΔH = +875 kJ 2 C (s) + 2 O2 (g) → 2 CO2 (g) ΔH = -789.02 kJ 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = -571.6 kJ 2 H2(g) + 2 C(s) + O2(g) → C2H5OH(l) ΔH = -486. kJ
CH4 (g) + NH3 (g) →HCN (g) + 3 H2 (g) ΔH = +256.0 kJ x½ NH3 (g) →½ N2 (g) + 1½ H2 (g) ΔH = ½(+91.8 kJ) R CH4 (g) → C (s) + 2 H2 (g) ΔH = +74.9 kJ ½ H2 (g) + C (s) + ½ N2 (g) → HCN (g) ΔH = ½(+270.3 kJ) CH4 (g) + NH3 (g) →HCN (g) + 3 H2 (g) ΔH = +256.0 kJ 2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g) ΔH = -1049. kJ x6 6 HCl (g) → 6 HCl (aq) ΔH = 6 (-74.8 kJ) x3 3 H2 (g) + 3 Cl2 (g) → 6 HCl (g) ΔH = 3 (-1845. kJ) Rx2 2 AlCl3 (aq) → 2 AlCl3 (s) ΔH = 2 (+323. kJ) 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s) ΔH = -6387. kJ