Chapter 5 Thermochemistry Part B Thermochemical Equations

First Law of Thermodynamics The work, w, is positive if it is done on the system and negative if it is done by the system. Positive q is heat gained by the system, negative q is heat lost by the system. Change in Internal Energy U is due to work and heat U = Ufinal – Uinitial = w + q At constant volume and no other work w = o U = Ufinal – Uinitial = qv Heat at constant volume is equal to internal energy change, which depends only on the system initial and final states

Enthalpy (H) H = U + PV Similar to U, H is a state function If there is only expansion work: w=– P V Internal energy change at constant pressure U= (qP + w) Enthalpy change at constant pressure H = U + PV H = (qP + w ) -w = qP

Enthalpy Change (H) qp = H = Hfinal – Hinitial Heat at constant pressure is equal to enthalpy change, which depends only on the initial and final states qp = H = Hfinal – Hinitial qp = H = Hproducts – Hreactants Endothermic reaction. System absorbs energy from surroundings. H positive Exothermic reaction. System loses energy to surroundings. H negative

Thermochemical Equation Write H immediately after equation N2(g) + 3H2(g)  2NH3(g) H = –92.38 kJ Must give physical states of products and reactants H different for different states CH4(g) + 2O2(g)  CO2(g) + 2H2O(l ) H ° rxn = –890.5 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H ° rxn = –802.3 kJ Difference is equal to the energy to vaporize water

Multiple Paths; Same H ° Can often get from reactants to products by several different paths Intermediate A Products Reactants Intermediate B Should get same H ° Enthalpy is state function and enthalpy change is path independent

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

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 steps. Chief Use Calculation of H °rxn for 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

Hess’s Law of Heat Summation 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 For any reaction that can be written into steps, value of H °rxn for reactions = sum of H °rxn values of each individual step

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

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.

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

Example Calculate H °rxn for 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

2C(s, gr) + H2(g)  C2H2(g) 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

Standard Enthalpy of Formation, Hf° A standard enthalpy of formation is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions.

Standard State Most stable form and physical state of element at 1 atm and 25 °C (298 K) Element Standard state O O2(g) C C (s, gr) H H2(g) Al Al(s) Ne Ne(g) Note: All Hf° of elements in their standard states = 0

Standard Enthalpy of Formation, Hf° Hf° of C2H5OH(l ) 2C(s, gr) + 3H2(g) + ½O2(g)  C2H5OH(l ) Hf° = –277.03 kJ/mol Hf° of Fe2O3(s) 2Fe(s) + 3/2O2(g)  Fe2O3(s) Hf° = –822.2 kJ/mol

Using Hf° H°reaction = – Sum of all H°f of all of the products Sum of all H°f of all of the reactants

Calculate Horxn Using Hf° Calculate H°rxn using Hf° data for the reaction SO3(g)  SO2(g) + ½O2(g) H°rxn = 99 kJ

Standard Enthalpy of combustion, Hc° Standard enthalpy of combustion, Hc°, is the enthalpy change when 1 mole of a substance combines with oxygen under standard state conditions C2H5OH(l) + 3O2(g)⟶2CO2 (g) + 3H2O(l) Hc°= −1366.8 kJ/mol

Using Hc° H°reaction = – Sum of all H°c of all of the reactants Sum of all H°c of all of the products