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Industrial Chemistry

Industrial Chemistry. Hess’s law. Index. Hess’s Law. Hess’s Law and its experimental verification. Hess’s Law calculations, 4 examples. Hess’s Law and calculations. Hess’s law states that “enthalpy change is independent of the route taken”. Route 2. NaOH (s). NaCl (aq).  H 2.

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Industrial Chemistry

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  1. Industrial Chemistry Hess’s law

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

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

  4. Route2 NaOH (s) NaCl (aq) H 2 H 3 NaOH (aq) H 1 Route1 NaOH (s) NaCl (aq) H 1 = H 2 + H 3 Verification of Hess’s Law H = enthalpy change 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.

  5. Route 2 H 2 + H 3 Route 1 H 1 50 ml 1mol l-1 HCl 50 ml HCl 50 ml H2O then H = c m H 1 = c m T T Verification of Hess’s Law 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. 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 H2 and H 3 H 2 Knowing the specific heat capacity for water, it is then possible to calculate the Enthaply change for this reaction. H 3 H 1 = H 2 + H 3 will verify Hess’s Law

  6. 3C (s) + 3H2 (g) C3H6 (g) Route 2a Route2b 3CO2 (g) + 3H2O (l) Hess’s Law Calculations Hess’s Law can be used to calculate enthalpy changes that cannot be directly measured by experiment. Route 1 Route 1cannot 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 bothcarbon and hydrogen and 3C (s) + 3O2 (g)  3CO2 (g) 3H2 (g) + 1½O2 (g)  3H2O (l) Route 2b involves the reverse combustion of propane 3CO2 (g) + 3H2O(l)  C3H6 (g) + 4½O2 (g)

  7. H 1 = H 2a + H 2b Route 1 3C (s) + 3H2 (g) Route 2a H c C= -394 kJ mol –1 H c H = -286 kJ mol –1 C3H6 (g) + H 2a = -( x 394) = -1182 kJ mol -1 -( x 286) = -858 kJ mol -1 H 2a= -2040kJ Route2a Route2b 3CO2 (g) + 3H2O(l) H c Propene = -2058.5 kJ mol –1 Route 2b H 2b = + 2058.5 kJ (note the reverse sign) = +18.5 kJ mol -1 + = ( 2058.5) -2040 kJ H1 Example 1 3 3 3 3 H 1

  8. Alternative approach for example 1 C(s) + O2 (g)  CO2(g) ΔHo298 = -394 kJ mol-1 H2(g) + ½O2(g)  H2O(g) ΔHo298 = -286 kJ mol-1 C3H6(g) + 4½O2(g)  3H2O(g) + 3CO2(g)ΔHo298 = -2058.5 kJ mol-1 3C(s) + 3H2 (g)  C3H6(g) ΔHf = ? 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. ΔHc = 3 x -394 kJ 3C(s) + 3O2 (g)  3CO2(g) 3H2(g) + 1½O2(g)  3H2O(g) ΔHc = 3 x -286 kJ 3H2O(g) + 3CO2(g)  C3H6(g) + 4½O2(g) ΔHc = +2058.5 kJ Equation has been reversed; (enthalpy now has opposite sign)

  9. 3C(graphite) + 3O2 (g)  3CO2(l) ΔHc = 3 x -394 3H2(g) + 1½O2(g)  3H2O(l) ΔHc = 3 x -286 3H2O(g) + 3CO2(g)  C3H6(g) + 4½O2(l) ΔHc = +2058.5 Now add the equations and also the corresponding enthalpy values 3C(s) + 3H2(g)  C3H6(g) ΔHf = (3 x -394) + (3 x -286) + (+2058.5) ΔHf = +18.5 kJ mol-1

  10. Route 1 ? C6H6(l) + 3H2 (g)  C6H12(l) Route 2a Route 2b 6H2O(l) + 6CO2(g) Route 2ainvolves the combustion of bothbenzene and hydrogen H c benzene = -3268 kJ mol –1 C6H6(g) + 7½O2(g)  3H2O(l) + 6CO2(g) H2(g) + ½O2(g)  H2O(l) H c hydrogen = -286 kJ mol –1 Route 2binvolves the reverse combustion of cyclohexane 6CO2 (g) + 6H2O(l) => C6H12 (g) + 7½O2 (g) H c cyclohexane = -3924 kJ mol –1 H 1 = H 2a +H 2b = ( -3268 + ( x - 286)) + 3924 = 202 kJ mol -1 Calculate the enthalpy change for the reaction: Example 2 3 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). 3

  11. Alternative approach for example 2 C6H12(l) + 9O2 (g)  6H2O(l) + 6CO2(g) ΔH = -3924 kJmol-1 H2(g) + ½O2(g)  H2O(l) ΔH = -286 kJmol-1C6H6(l) + 7½O2(g)  3H2O(l) + 6CO2(g) ΔH = -3268 kJmol-1 C6H6(l) + 3H2(g)  C6H12(l) ΔHf = ? 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. C6H6(l) + 7½O2 (g)  6CO2(g) + 3H2O(g) ΔHc = -3268 kJ 3H2(g) + 1½O2(g)  3H2O(l) ΔHc = 3 x -286 kJ 6H2O(g) + 6CO2(g)  C6H12(g) + 9O2(g) ΔHc = +3924 kJ Equation has been reversed; (enthalpy now has opposite sign)

  12. C6H6(l) + 7½O2 (g)  6CO2(g) + 3H2O(g) ΔHc = -3268 3H2(g) + 1½O2(g)  3H2O(g) ΔHc = 3 x -286 6H2O(g) + 6CO2(g)  C6H12(l) + 9O2(g) ΔHc = +3924 Now add the equations and also the corresponding enthalpy values C6H6(l) + 3H2(g)  C6H12(l) ΔH f = -3268 + (3 x -286) + 3924 ΔH f = -202 kJ mol-1

  13. 3. Use the enthalpy changes of combustion shown in the table to work out the enthalpy change of formation of ethyne, C2H2. “Using the Second method” “Required” equation, 2C(s) + H2(g)  C2H2(g) ΔHf = ? • 2C(s) + 2O2(g)  2CO2(g)ΔHc = 2 x -394 kJ mol-1 • H2(g) + ½O2(g)  H2O(g)ΔHc = -286 kJ mol-1 C2H2(g) + 2½O2(g)  2CO2(g) + H2O(g)ΔHc = -1299 kJ mol-1 • 2CO2(g) + H2O(g) C2H2(g) + 2½O2(g)ΔHc = +1299 kJ mol-1 Adding “bulleted” equations gives us 2C(graphite) + H2(g)  C2H2(g) ΔHf = (2 x -394) + (-286) + 1299 ΔHf = +225 kJ mol-1

  14. C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) ΔHc = ? 4. Using the following standard enthalpy changes of formation, ΔHof / kJmol-1 : CO2(g), -394; H2O(g), -286; C2H5OH(l), -278 calculate the standard enthalpy of combustion of ethanol i.e. the enthalpy change for the reaction C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) “Using the Second method” 2C(s) + 3H2(g) + ½O2(g)  C2H5OH(l) ΔHf = -278 kJ ● 2C(s) + 2O2(g)  2CO2(g) ΔHf = 2 x -394 kJ ● 3H2(g) + 1½O2(g)  3H2O(g) ΔHf = 3 x -286 kJ ● C2H5OH(l) 2C(s) + 3H2(g) + ½O2(g) ΔHf = +278 kJ Add bulleted equations C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) Solve equation for ΔHc ΔHc = + 278 + (2 x -394) + (3 x -286) ΔHc = -1368 kJ mol-1

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