© 2006 Brooks/Cole - Thomson Some Thermodynamic Terms Notice that the energy change in moving from the top to the bottom is independent of pathway but the work required may not be! Some examples of state functions are: –T (temperature), P (pressure), V (volume),  E (change in energy),  H (change in enthalpy – the transfer of heat), and S (entropy) Examples of non-state functions are: –n (moles), q (heat), w (work)  ∆H along one path =  ∆H along another path This equation is valid because ∆H is a STATE FUNCTION These depend only on the state of the system and not how it got there. V, T, P, energy — and your bank account! Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.

© 2006 Brooks/Cole - Thomson Some Thermodynamic Terms The properties of a system that depend only on the state of the system are called state functions. –State functions are always written using capital letters. The value of a state function is independent of pathway. An analog to a state function is the energy required to climb a mountain taking two different paths. –E 1 = energy at the bottom of the mountain –E 1 = mgh 1 –E 2 = energy at the top of the mountain –E 2 = mgh 2 –  E = E 2 -E 1 = mgh 2 – mgh 1 = mg(  h)

© 2006 Brooks/Cole - Thomson Standard States and Standard Enthalpy Changes Thermochemical standard state conditions –The thermochemical standard T = K. –The thermochemical standard P = atm. Be careful not to confuse these values with STP. Thermochemical standard states of matter –For pure substances in their liquid or solid phase the standard state is the pure liquid or solid. –For gases the standard state is the gas at 1.00 atm of pressure. For gaseous mixtures the partial pressure must be 1.00 atm. –For aqueous solutions the standard state is 1.00 M concentration. ∆H f o = standard molar enthalpy of formation the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 6.2 and Appendix L

ENTHALPY Most chemical reactions occur at constant P, so and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = H final - H initial Heat transferred at constant P = q p q p = ∆H where H = enthalpy

If H final < H initial then ∆H is negative Process is EXOTHERMIC If H final > H initial then ∆H is positive Process is ENDOTHERMIC ENTHALPY ∆H = H final - H initial

Consider the formation of water H 2 (g) + 1/2 O 2 (g) → H 2 O(g) kJ USING ENTHALPY Exothermic reaction — heat is a “product” and ∆H = – kJ

Making liquid H 2 O from H 2 + O 2 involves two exothermic steps. USING ENTHALPY H 2 + O 2 gas Liquid H 2 O H 2 O vapor Making H 2 O from H 2 involves two steps. H 2(g) + 1/2 O 2(g) → H 2 O (g) kJ H 2 O (g) → H 2 O (l) + 44 kJ H 2 O (g) → H 2 O (l) + 44 kJ H 2(g) + 1/2 O 2(g) → H 2 O (l) kJ Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.

Enthalpy Values H 2(g) + 1/2 O 2(g) → H 2 O (g) ∆H˚ = -242 kJ 2H 2(g) + O 2(g) → 2H 2 O (g) ∆H˚ = -484 kJ H 2 O (g) → H 2(g) + 1/2 O 2(g) ∆H˚ = +242 kJ H 2(g) + 1/2 O 2(g) → H 2 O (l) ∆H˚ = -286 kJ Depend on how the reaction is written and on phases of reactants and products

Hess’s Law & Energy Level Diagrams Forming H 2 O can occur in a single step or in a two steps. ∆H total is the same no matter which path is followed. Active Figure 6.18

© 2006 Brooks/Cole - Thomson Thermochemical Equations Thermochemical equations are a balanced chemical reaction plus the  H value for the reaction. –For example, this is a thermochemical equation. The stoichiometric coefficients in thermochemical equations must be interpreted as numbers of moles. 1 mol of C 5 H 12 reacts with 8 mol of O 2 to produce 5 mol of CO 2, 6 mol of H 2 O, and releasing 3523 kJ is referred to as one mole of reactions.

© 2006 Brooks/Cole - Thomson Hess’s Law Hess’s Law of Heat Summation,  H rxn =  H 1 +  H 2 +  H , states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps. –Hess’s Law is true because  H is a state function. If we know the following  H o ’s

© 2006 Brooks/Cole - Thomson Hess’s Law For example, we can calculate the  H o for reaction [1] by properly adding (or subtracting) the  H o ’s for reactions [2] and [3]. Notice that reaction [1] has FeO and O 2 as reactants and Fe 2 O 3 as a product. –Arrange reactions [2] and [3] so that they also have FeO and O 2 as reactants and Fe 2 O 3 as a product. Each reaction can be doubled, tripled, or multiplied by half, etc. The  H o values are also doubled, tripled, etc. If a reaction is reversed the sign of the  H o is changed.

© 2006 Brooks/Cole - Thomson Hess’s Law Given the following equations and  H o  values calculate  H o for the reaction below.

© 2006 Brooks/Cole - Thomson Hess’s Law Use a little algebra and Hess’s Law to get the appropriate  H o  values

© 2006 Brooks/Cole - Thomson Thermochemical Equations This is an equivalent method of writing thermochemical equations.  H < 0 designates an exothermic reaction.  H > 0 designates an endothermic reaction

© 2006 Brooks/Cole - Thomson Standard Molar Enthalpies of Formation,  H f o The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements. –The symbol for standard molar enthalpy of formation is  H f o. The standard molar enthalpy of formation for MgCl 2 is:

© 2006 Brooks/Cole - Thomson Standard Molar Enthalpies of Formation,  H f o Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 15-1 and Appendix K in the text. Standard molar enthalpies of elements in their most stable forms at K and atm are zero. Example 15-4: The standard molar enthalpy of formation for phosphoric acid is kJ/mol. Write the equation for the reaction for which  H o rxn = kJ. P in standard state is P 4 Phosphoric acid in standard state is H 3 PO 4(s)

© 2006 Brooks/Cole - Thomson Hess’s Law Hess’s Law in a more useful form. –For any chemical reaction at standard conditions, the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products (each multiplied by its coefficient in the balanced chemical equation) minus the corresponding sum for the reactants.

© 2006 Brooks/Cole - Thomson Hess’s Law

∆H f o, standard molar enthalpy of formation ½˚ H 2(g) + ½ O 2(g) → H 2 O (g) ∆H f ˚ (H 2 O, g )= kJ/mol C (s) + ½ O 2(g) → CO (g) ∆H f ˚ of CO = kJ/mol By definition, ∆H f o = 0 for elements in their standard states. Use ∆H˚’s to calculate enthalpy change for H 2 O(g) + C(graphite) → H 2 (g) + CO(g)

Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆H o rxn for CH 3 OH(g) + 3/2 O 2 (g) → CO 2 (g) + 2 H 2 O(g) ∆H o rxn =  ∆H f o (prod) -  ∆H f o (react)

© 2006 Brooks/Cole - Thomson Thermochemical Equations Write the thermochemical equation for the reaction for CuSO 4(aq) + 2NaOH (aq) Cu(OH) 2(s) + Na 2 SO 4(aq) 50.0mL of M CuSO4 at o C T final o C 50.0mL of M NaOH at o C Density final solution = 1.02 g/mL C H 2 O = J/g o C

© 2006 Brooks/Cole - Thomson Standard Molar Enthalpies of Formation,  H f o Calculate the enthalpy change for the reaction of one mole of H 2(g) with one mole of F 2(g) to form two moles of HF (g) at 25 o C and one atmosphere.

© 2006 Brooks/Cole - Thomson Standard Molar Enthalpies of Formation,  H f o Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al 2 O 3 at 25 o C and one atmosphere.

© 2006 Brooks/Cole - Thomson Hess’s Law Calculate the  H o 298 for the following reaction from data in Appendix K.

© 2006 Brooks/Cole - Thomson Hess’s Law Application of Hess’s Law and more algebra allows us to calculate the  H f o  for a substance participating in a reaction for which we know  H rxn o, if we also know  H f o  for all other substances in the reaction. Given the following information, calculate  H f o for H 2 S (g).