5 Hess’ law

Hess’s Law Hess’s law states that total enthalpy change for a reaction is independent of the route by which the chemical change takes place Hess’s law is a version of the first law of thermodynamics, which is that energy is always conserved. 2H (g) + 2Cl(g) H2 + Cl2 2HCl (g) a b ΔH On an energy level diagram the directions of the arrows can show the different routes a reaction can proceed by. In this example one route is arrow ‘a’ The second route is shown by arrows ΔH plus arrow ‘b’ So a = ΔH + b And rearranged ΔH = a – b H+ (g) + Br – (g) H+ (aq) + Br – (aq) H (g) + Br (g) HBr (g) a c d ΔH Interconnecting reactions can also be shown diagrammatically. In this example one route is arrow ‘a’ plus ΔH The second route is shown by arrows ‘c’ plus arrow ‘d’ So a+ ΔH = c + d And rearranged ΔH = c + d – a CuSO4 (aq) CuSO4 (s) + 5H2O (l) CuSO4 .5H2O (s) + 11kJmol-1 = -66.1 kJmol-1 ∆H reaction + aq + aq ∆H reaction +11kJmol-1 -66.1 kJmol-1 ∆H reaction = -66.1 – 11 = -77.1 kJmol-1 This Hess’s law is used to work out the enthalpy change to form a hydrated salt from an anhydrous salt. This cannot be done experimentally because it is impossible to add the exact amount of water and it is not easy to measure the temperature change of a solid. Often Hess’s law cycles are used to measure the enthalpy change for a reaction that cannot be measured directly by experiments. Instead alternative reactions are carried out that can be measured experimentally. Instead both salts are dissolved in excess water to form a solution of copper sulphate. The temperature changes can be measured for these reactions. Using Hess’s law to determine enthalpy changes from enthalpy changes of formation. ∆H reaction = Σ ∆fH products – Σ ∆fH reactants Example 6. What is the enthalpy change for this reaction ? Al2O3 + 3 Mg
3 MgO + 2 Al ∆H = Σ∆fH products – Σ ∆fH reactants ∆H = 3 x ∆fH (MgO) – ∆fH (Al2O3 ) ∆H = (3 x –601.7) – –1675.7 = -129.4 kJ mol-1 Remember elements have ∆Hf = 0 ∆fH(MgO)= -601.7 kJ mol-1 ∆fH(Al2O3 ) = -1675.7 kJ mol-1 Example 7. Using the following data to calculate the heat of combustion of propene ∆Hf C3H6 (g) = +20 kJ mol-1 ∆Hf CO2 (g)= –394 kJ mol-1 ∆Hf H2O(g)= –242 kJ mol-1 C3H6 + 4.5 O2
3CO2 + 3H2O ∆cH = Σ ∆fH products – Σ ∆fH reactants ∆cH = [3 x ∆fH (CO2 ) + 3 x ∆fH (H2O)] – ∆fH (C3H6 ) ∆cH = [(3 x –394) + (3 x –242)] – 20 = -1928 kJ mol-1 Using Hess’s law to determine enthalpy changes from enthalpy changes of combustion. ∆H reaction = Σ ∆cH reactants – Σ ∆cH products Example 8. Using the following combustion data to calculate the heat of reaction CO (g) + 2H2 (g)
CH3OH (g) ∆cH CO(g) = -283 kJ mol-1 ∆cH H2 (g)= –286 kJ mol-1 ∆cH CH3OH(g)= –671 kJ mol-1 ∆H reaction = Σ ∆cH reactants – Σ ∆cH products ∆H = ∆cH (CO) + 2 x ∆cH (H2 ) – ∆cH (CH3OH) ∆H = -283+ 2x –286 – -671 = -184 kJ mol-1 Elements in standard states Reactants Products Σ ∆ Σ ∆fH products fH reactants ∆H reaction 2Al (s) + 3 Mg (s) + 1.5O2 (g) Al2O3 (s) + 3 Mg (s) 3 MgO (s) + 2 Al (s) ∆ 3 x ∆fH (MgO) fH(Al2O3 ) ∆H reaction 3C (s) + 3 H2 (g) + 4.5 O2 (g) C3H6 (g) + 4.5 O2 (g) 3 CO2 (g) + 3 H2O (g) 3 x ∆fH (CO2 ) ∆fH(C3H6 ) ∆cH 3 x ∆fH (H2O) Combustion Products Reactants Products Σ ∆cH reactants Σ ∆cH products ∆H reaction CO2 (g) + 2 H2O (l) CO (g) + 2H2 (g) CH3OH(g) ∆cH(CO) + ∆cH(CH3OH) Example 9. Using the following combustion data to calculate the heat of formation of propene 3C (s) + 3H2 (g)
C3H6 (g) ∆cH C (s) = -393kJ mol-1 ∆cH H2 (g)= –286 kJ mol-1 ∆cH C3H6 (g)= –-2058 kJ mol-1 ∆H = Σ ∆cH reactants – Σ ∆cH products ∆fH = 3 x ∆cH (C) + 3 x ∆cH (H2 ) – ∆cH (C3H6 ) ∆fH = 3x -393+ 3x –286 – -2058 = +21 kJ mol-1 3 CO2 (g) + 3 H2O (l) 3C (s) + 3H2 (g) C3H6 (g) ∆cH(C3H6 3 x ∆cH (C) + )