Running thermochemical equations in reverse

1. Running thermochemical equations in reverse 7.8 Hess’s Law Consider CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H°reaction= – 802.3 kJ • Reverse thermochemical equation • Must change sign of H CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) H°reaction = 802.3 kJ

2. Reverse Thermochemical Equation, • Makes sense: • If we get energy out when forming products • Must put energy in to go back to reactants • Consequence of Law of Conservation of Energy • More like a mathematical equation • If you know H ° for reaction, you also know H ° for the reverse Changes sign of H

3. Multiple Paths; Same H ° • Can often get from reactants to products by several different paths Products Reactants Intermediate A Intermediate B • Should get same H ° • Enthalpy is state function and path independent • Let’s see if this is true

4. Ex. Multiple Paths; Same H ° Path a: Single step C(s) + O2(g)  CO2(g) H°rxn = –393.5 kJ Path b:Two step Step 1: C(s) + ½O2(g)  CO(g) H °rxn = –110.5 kJ Step 2: CO(g) + ½O2(g)  CO2(g) H °rxn = –283.0 kJ Net Rxn: C(s) + O2(g)  CO2(g) H °rxn = –393.5 kJ • Chemically and thermochemically, identical results • True for exothermic reaction or for endothermic reaction

5. Ex. 8 Multiple Paths; Same H °rxn Path a: N2(g) + 2O2(g)  2NO2(g) H°rxn= 68 kJ Path b: Step 1: N2(g) + O2(g)  2NO(g) H°rxn= 180. kJ Step 2: 2NO(g)+ O2(g)  2NO2(g) H°rxn= –112 kJ Net rxn: N2(g) + 2O2(g)  2NO2(g) H°rxn= 68 kJ Hess’s Law of Heat Summation • For any reaction that can be written into steps, value of H°rxnfor reactions = sum of H°rxnvalues of each individual step

6. Enthalpy Diagrams • Graphical description of Hess’ Law • Vertical axis = enthalpy scale • Horizontal line =various states of reactions • Higher up = larger enthalpy • Lower down = smaller enthalpy

7. Enthalpy Diagrams • Use to measure Hrxn • Arrow down Hrxn = negative • Arrow up Hrxn = positive • Calculate cycle • One step process = sum of two step process Ex. H2O2(l )  H2O(l ) + ½O2(g) –286 kJ = –188 kJ + Hrxn Hrxn= –286 kJ – (–188 kJ ) Hrxn= –98 kJ

8. Hess’s Law Hess’s Law of Heat Summation • Going from reactants to products • Enthalpy change is same whether reaction takes place in one step or many • Chief Use • Calculation of H °rxnfor reaction that can’t be measured directly • Thermochemical equations for individual steps of reaction sequence may be combined to obtain thermochemical equation of overall reaction

9. Hess’s Law • Values of H °rxn can be determined by the manipulation of any combination of thermochemical equations that add up to the final net equation.

10. Rules for Manipulating Thermochemical Equations • When equation is reversed, sign of H°rxn must also be reversed. • If all coefficients of equation are multiplied or divided by same factor, value of H°rxn must likewise be multiplied or divided by that factor • Formulas canceled from both sides of equation must be for substance in same physical states

11. Strategy for Adding Reactions Together: • Choose most complex compound in equation for one-step path • Choose equation in multi-step path that contains that compound • Write equation down so that compound • is on appropriate side of equation • has appropriate coefficient for our reaction • Repeat steps 1 – 3 for next most complex compound, etc.

12. Strategy for Adding Reactions (Cont.) • Choose equation that allows you to • cancel intermediates • multiply by appropriate coefficient • Add reactions together and cancel like terms • Add energies together, modifying enthalpy values in same way equation modified • If reversed equation, change sign on enthalpy • If doubled equation, double energy

13. Ex. Calculate H°rxn for C (s, graphite)  C(s, diamond) Given C(s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ C(s, dia) + O2(g)  CO2(g) H°rxn = –396 kJ • To get desired equation, must reverse second equation and add resulting equations C(s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ CO2(g)  C(s, dia) + O2(g) H°rxn = –(–396 kJ) C(s, gr) + O2(g) + CO2(g)  C(s, dia) + O2(g)+ CO2(g) H° = –394 kJ + 396 kJ = + 2 kJ ] –1[

14. Example Calculate H °rxnfor 2 C(s, gr) + H2(g)  C2H2(g) Given the following: • C2H2(g) + 5/2O2(g) 2CO2(g) + H2O(l ) H °rxn= –1299.6 kJ • C(s, gr) + O2(g)  CO2(g) H °rxn= –393.5 kJ • H2(g) + ½O2(g)  H2O(l ) H °rxn= –285.8 kJ

15. Calculate for 2C(s, gr) + H2(g)  C2H2(g) Example continue 2CO2(g) + H2O(l )  C2H2(g) + 5/2O2(g) H°rxn= –(–1299.6 kJ) = +1299.6 kJ 2C(s, gr) + 2O2(g) 2CO2(g) H°rxn=(2–393.5 kJ) = –787.0 kJ H2(g) + ½O2(g)  H2O(l ) H°rxn= –285.8 kJ 2CO2(g) + H2O(l ) + 2C(s, gr) + 2O2(g) + H2(g) + ½O2(g)  C2H2(g) + 5/2O2(g) + 2CO2(g) + H2O(l) 2C(s, gr) + H2(g)  C2H2(g) H°rxn = +226.8 kJ

16. Example! Which of the following is a statement of Hess's Law? • H for a reaction in the forward direction is equal to H for the reaction in the reverse direction. • Hfor a reaction depends on the physical states of the reactants and products. • If a reaction takes place in steps, Hfor the reaction will be the sum of Hs for the individual steps. • If you multiply a reaction by a number, you multiply H by the same number. • H for a reaction in the forward direction is equal in magnitude and opposite in sign to Hfor the reaction in the reverse direction.

17. Example! Given the following data: C2H2(g) + O2(g)  2CO2(g) + H2O(l ) Hrxn= –1300. kJ C(s) + O2(g)  CO2(g) Hrxn = –394 kJ H2(g) + O2(g)  H2O(l ) Hrxn = –286 kJ Calculate for the reaction 2C(s) + H2(g)  C2H2(g) • 226 kJ • –1980 kJ • –620 kJ • –226 kJ • 620 kJ Hrxn = +1300. kJ + 2(–394 kJ) + (–286 kJ)