Lecture 6: Thermochemistry

5.60 Spring 2008 Lecture #6 page 1 Thermochemistry • Goal: To predict ΔH for every reaction, even if it cannot be carried out in the laboratory The heat of reaction ΔHrx is the ΔH for the isothermal reaction at constant pressure. Fe2O3(s,T,p) + 3H2(g,T,p) = 2Fe(s,T,p) + 3H2O(l,T,p) e.g. ΔHrx (T , p ) = 2HFe (T , p ) + 3HH O (T , p ) − 3HH (T , p ) − HFe O (T , p ) 2 2 2 3 [ΔHrx = H ( products ) − H ( reactants )] We cannot know H values because enthalpy, like energy, is not measured on an absolute scale. We can only measure differences in enthalpy. • Define a reference scale for enthalpy H (298.15K, 1 bar) ≡ 0 e.g. • For every element in its most stable form at 1 bar and 298.15K HH° (g) (298.15K ) = 0 2 HC (graphite) (298.15K ) = 0 ΔHf° (298.15K ) : ° The “°” means 1 bar We can now write reactions to form every compound from its constituent atoms. The heat of reaction is the heat of formation of 1 mole of that compound from the constituent elements in their most stable forms. 5.60 Spring 2008 Lecture #6 page 2 Example (let T = 298.15 K) ½ H2 (g,T,1 bar) + ½ Br2 (l,T,1 bar) = HBr (g,T,1 bar) ΔHrx = ΔHf° ( products ) − ΔHf° ( reactants ) = ΔHf°,HBr ( g,T ) − 1 1 ΔHf°,H2 ( g,T ) − ΔHf°,Br2 (l ,T ) = ΔHf°,HBr ( g,T 2 2  ) 0 - elements in most stable forms We can tabulate ΔHf° (298.15K ) values for all known compounds. We can calculate ΔHrx° (T ) for any reaction (T = 298.15K). e.g. CH4 (g,T,1 bar) + 2O2 (g,T,1 bar) = CO2 (g,T,1 bar) + 2H2O(l,T, 1 bar) • First decompose reactants into elements • Second put elements together to form products • Use Hess’s law [A statement of the fact that because H is a function of state, we can add ΔH’’s around paths.] CH4 (g,T,1 bar) = Cgraphite (s,T,1 bar) + 2H2(g,T,p) ΔHI ΔHII 2O2 (g,T,1 bar) = 2O2 (g,T,1 bar) Cgraphite (s,T,1 bar) + O2 (g,T,1 bar) = CO2 (g,T,1 bar) ΔHIII 2H2(g,T,p) + O2 (g,T,1 bar) = 2H2O(l,T, 1 bar) ΔHIV ____________________________________________ CH4 (g,T,1 bar) + 2O2 (g,T,1 bar) = CO2 (g,T,1 bar) + 2H2O(l,T, 1 bar) 5.60 Spring 2008 Lecture #6 page 3 ΔHrx = ΔHI + ΔHII + ΔHIII + ΔHIV ΔHI = HC + 2HH − HCH = −ΔHf°,CH 2 ΔHII = HO − HO = 0 2 4 4 2 ΔHIII = HCO − HC − HO = ΔHf°,CO 2 2 2 ΔHIV = 2HH O − 2HH − HO = 2ΔHf°,H O 2 ∴ 2 2 2 ΔHrx = 2ΔHf°,H O + ΔHf°,CO − ΔHf°,CH 2 2 4 In general, ΔHrx = ∑ν i ΔHf°,i (products ) − ∑ν i ΔHf°,i (reactants ) i i ν ≡ stoichiometric coefficient • ΔH at constant p and for reversible process is ΔH = q p ⇒ The heat of reaction is the heat flowing into the reaction from the surroundings If ΔHrx < 0, qp < 0 heat flows from the reaction to the surroundings (exothermic) If ΔHrx > 0, qp > 0 heat flows into the reaction from the surroundings (endothermic) 5.60 Spring 2008 Lecture #6 page 4 Temperature dependence of ΔHrx ⎛ ∂H ⎞ ⎜ ⎟ = Cp ⎝ ∂T ⎠ p Recall ∴ ⎛ ∂ΔH ⎞ ⎜ ⎟ = ΔC p = ∑ν i C p ,i (products ) − ∑ν i C p ,i (reactants ) ⎝ ∂T ⎠ p i i e.g. CH4 (g,T,1 bar) + 2O2 (g,T,1 bar) = CO2 (g,T,1 bar) + 2H2O(l,T, 1 bar) ΔC p = C p ,CO2 ( g, T , 1 bar ) + 2C p ,H2O (l , T , 1 bar ) − C p ,CH4 ( g, T , 1 bar ) − 2C p ,O2 ( g, T , 1 bar ) T2 ∫T 1 ⎛ ∂ΔH ⎞ ⎜ ⎟ dT = ΔH (T2 ) − ΔH (T1 ) ⎝ ∂T ⎠ p  ΔC p T2 ΔH (T2 ) = ΔH (T1 ) + ∫ ΔC pdT T1 Especially simple when ΔC p is T-independent