1. Industrial Chemistry Hess’s law

2. Index Hess’s Law and its experimental verification Hess’s Law calculations, 4 examples. Hess’s Law

3. Hess’s Law and calculations Hess’s law states that “enthalpy change is independent of the route taken”

4. Verification of Hess’s Law The conversion of solid NaOH to NaCl solution can be achieved by two possible routes. Route 1 is a single-step process, (adding HCl (aq) directly to the solid NaOH) and Route 2 is a two-step process (dissolve the solid NaOH in water, then adding the HCl(aq)) All steps are exothermic. If Hess’s Law applies, the enthalpy change for route 1 must be the same as the overall change for route 2.  H = enthalpy change  H 1 =  H 2  H 3 + NaOH (s) NaCl (aq) Route 1  H 1 NaOH (s) NaOH (aq) NaCl (aq) Route 2  H 2  H 3

5. Route 2  H 2 +  H 3 50 ml H 2 O 50 ml HCl then 2.50g of NaOH added to a dry, insulated beaker. Before adding the acid, its temperature is recorded. The final temperature after adding the acid is also recorded. Knowing the specific heat capacity for water, it is then possible to calculate the Enthaply change for this reaction. 1. 2.50g of NaOH added to a dry, insulated beaker. 2. Before adding the water, its temperature is recorded. The final temperature rise after adding the water is also recorded. 3. Now add the acid, again, recording the final temperature. Use the equation below to calculate  H 2 and  H 3  H 2  H 1 =  H 2 +  H 3 will verify Hess’s Law Route 1  H 1 50 ml 1mol l -1 HCl  H 1 = c m TT  H = c m TT  H 3 Verification of Hess’s Law

6. Hess’s Law Calculations Hess’s Law can be used to calculate enthalpy changes that cannot be directly measured by experiment. Route 1 Route 1 cannot be carried out in a lab, as carbon and hydrogen will not combine directly. The enthalpy of combustion reactions can act as a stepping stone which enables a link with carbon and hydrogen (the reactants) with propene (the product) Route 2a involves the combustion of both carbon and hydrogen 3C (s) + 3O 2 (g)  3CO 2 (g) Route 2b involves the reverse combustion of propane 3CO 2 (g) + 3H 2 O(l)  C 3 H 6 (g) + 4½O 2 (g) 3H 2 (g) + 1½O 2 (g)  3H 2 O (l) and 3C (s) + 3H 2 (g) C 3 H 6 (g) 3CO 2 (g) + 3H 2 O (l) Route 2aRoute 2b

7.  H 1 3C (s) + 3H 2 (g) C 3 H 6 (g) Route 1 3CO 2 (g) + 3H 2 O(l) Route 2a Route 2b  H 1 =  H 2a  H 2b + H1H1 Route 2a  H c C= -394 kJ mol –1  H c H = -286 kJ mol –1  H 2a = -2040 kJ  H 2a= -( x 394) = -1182 kJ mol -1 -( x 286) = -858 kJ mol -1 + 3 3 33 Route 2b  H 2b = + 2058.5 kJ (note the reverse sign)  H c Propene = -2058.5 kJ mol –1 = -2040 kJ + ( 2058.5) = + 18.5 kJ mol -1 Example 1

8. Alternative approach for example 1 3C(s) + 3H 2 (g)  C 3 H 6 (g) C(s) + O 2 (g)  CO 2 (g) ΔH o 298 = -394 kJ mol -1 H 2 (g) + ½O 2 (g)  H 2 O(g) ΔH o 298 = -286 kJ mol -1 C 3 H 6 (g) + 4½O 2 (g)  3H 2 O(g) + 3CO 2 (g)ΔH o 298 = -2058.5 kJ mol - 1 ΔH f = ? Re-write the equations so that the reactants and products are on the same side of the “arrow” as the equation you are interested in. Multiply each equation so that there are the same number of moles of each constituent also. 3C(s) + 3O 2 (g)  3CO 2 (g) ΔHc = 3 x -394 kJ 3H 2 (g) + 1½O 2 (g)  3H 2 O(g)ΔHc = 3 x -286 kJ 3H 2 O(g) + 3CO 2 (g)  C 3 H 6 (g) + 4½O 2 (g)ΔHc = +2058.5 kJ Equation has been reversed; (enthalpy now has opposite sign)

9. 3C(graphite) + 3O 2 (g)  3CO 2 (l)ΔHc = 3 x -394 3H 2 (g) + 1½O 2 (g)  3H 2 O(l)ΔHc = 3 x -286 3H 2 O(g) + 3CO 2 (g)  C 3 H 6 (g) + 4½O 2 (l)ΔHc = +2058.5 Now add the equations and also the corresponding enthalpy values 3C(s) + 3H 2 (g)  C 3 H 6 (g) ΔH f = (3 x -394) + (3 x -286) + (+2058.5) ΔH f = +18.5 kJ mol -1

10. Calculate the enthalpy change for the reaction: The products of combustion act as a stepping stone which enables a link to be made with benzene and hydrogen (the reactants) with cyclohexane (the product). C 6 H 6 (l) + 3H 2 (g)  C 6 H 12 (l) Route 1 ? 6H 2 O(l) + 6CO 2 (g ) Route 2aRoute 2b Route 2a involves the combustion of both benzene and hydrogen C 6 H 6 (g) + 7½O 2 (g)  3H 2 O(l) + 6CO 2 (g)  H c benzene = -3268 kJ mol –1 H 2 (g) + ½O 2 (g)  H 2 O(l)  H c hydrogen = -286 kJ mol –1 Route 2b involves the reverse combustion of cyclohexane 6CO 2 (g) + 6H 2 O(l) => C 6 H 12 (g) + 7½O 2 (g)  H c cyclohexane = -3924 kJ mol –1  H 1 =  H 2a +  H 2b = ( -3268 + ( x - 286)) + 3924 = 202 kJ mol -1 3 3 Example 2

11. C 6 H 6 (l) + 3H 2 (g)  C 6 H 12 (l) ΔH f = ? Re-write the equations so that the reactants and products are on the same side of the “arrow” as the equation you are interested in. Multiply each equation so that there are the same number of moles of each constituent also. C 6 H 6 (l) + 7½O 2 (g)  6CO 2 (g) + 3H 2 O(g) ΔHc = -3268 kJ 3H 2 (g) + 1½O 2 (g)  3H 2 O(l) ΔHc = 3 x -286 kJ 6H 2 O(g) + 6CO 2 (g)  C 6 H 12 (g) + 9O 2 (g) ΔHc = +3924 kJ Equation has been reversed; (enthalpy now has opposite sign) C 6 H 12 (l) + 9O 2 (g)  6H 2 O(l) + 6CO 2 (g) ΔH = -3924 kJmol -1 H 2 (g) + ½O 2 (g)  H 2 O(l) ΔH = -286 kJmol -1 C 6 H 6 (l) + 7½O 2 (g)  3H 2 O(l) + 6CO 2 (g) ΔH = -3268 kJmol -1 Alternative approach for example 2

12. Now add the equations and also the corresponding enthalpy values C 6 H 6 (l) + 3H 2 (g)  C 6 H 12 (l) ΔH f = -3268 + (3 x -286) + 3924 ΔH f = -202 kJ mol -1 C 6 H 6 (l) + 7½O 2 (g)  6CO 2 (g) + 3H 2 O(g)ΔHc = -3268 3H 2 (g) + 1½O 2 (g)  3H 2 O(g)ΔHc = 3 x -286 6H 2 O(g) + 6CO 2 (g)  C 6 H 12 (l) + 9O 2 (g) ΔHc = +3924

13. 3. Use the enthalpy changes of combustion shown in the table to work out the enthalpy change of formation of ethyne, C 2 H 2. SubstanceC(graphite)H 2 (g)C 2 H 2 (g) ΔH o (combustion)-394 kJmol -1 -286 kJmol -1 -1299 kJmol -1 “Using the Second method” “Required” equation, 2C(s) + H 2 (g)  C 2 H 2 (g) ΔH f = ? 2C(s) + 2O 2 (g)  2CO 2 (g) ΔH c = 2 x -394 kJ mol -1 H 2 (g) + ½O 2 (g)  H 2 O(g) ΔH c = -286 kJ mol -1 C 2 H 2 (g) + 2½O 2 (g)  2CO 2 (g) + H 2 O(g) ΔH c = -1299 kJ mol -1 2CO 2 (g) + H 2 O(g)  C 2 H 2 (g) + 2½O 2 (g) ΔH c = +1299 kJ mol -1 ΔH f = (2 x -394) + (-286) + 1299 ΔH f = +225 kJ mol -1 Adding “bulleted” equations gives us 2C(graphite) + H 2 (g)  C 2 H 2 (g)

14. 4. Using the following standard enthalpy changes of formation, ΔH o f / kJmol -1 : CO 2 (g), -394; H 2 O(g), -286; C 2 H 5 OH(l), -278 calculate the standard enthalpy of combustion of ethanol i.e. the enthalpy change for the reaction C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(l) C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(l) ΔH c = ? 2C(s) + 3H 2 (g) + ½O 2 (g)  C 2 H 5 OH(l)ΔH f = -278 kJ ● 2C(s) + 2O 2 (g)  2CO 2 (g) ΔH f = 2 x -394 kJ ● 3H 2 (g) + 1½O 2 (g)  3H 2 O (g)ΔH f = 3 x -286 kJ ● C 2 H 5 OH(l)  2C(s) + 3H 2 (g) + ½O 2 (g) ΔH f = +278 kJ Add bulleted equations C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(l) Solve equation for ΔH c ΔH c = + 278 + (2 x -394) + (3 x -286) ΔH c = -1368 kJ mol -1 “Using the Second method”