Thermochemistry Chapter 5

Heat - the transfer of thermal energy between two bodies that are at different temperatures Energy Changes in Chemical Reactions Temperature - a measure of the thermal energy 90 °C 40 °C greater thermal energy Temperature = Thermal Energy greater temperature (intensive) (extensive)

Exothermic process - gives off heat – transfers thermal energy from the system to the surroundings. 2H 2 (g) + O 2 (g) 2H 2 O (l) + energy H 2 O (g) H 2 O (l) + energy

Endothermic process - heat has to be supplied to the system from the surroundings. energy + CaCO 3 (s) CaO (s) + CO 2 (g) energy + H 2 O (s) H 2 O (l)

 E system +  E surroundings = 0 or  E system = −  E surroundings C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. Internal energy of system

Exothermic Energy relationships Fig 5.4 (a)

Endothermic Energy relationships Fig 5.4 (b)

Fig 5.5 Energy Diagram for the Interconversion of H 2 (g), O 2 (g), and H 2 O (l)

Another form of the first law for  E system  E = q + w  E is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system Table 5.1 pg 171

Sign Conventions for Heat and Work Fig 5.6

Thermodynamics State functions - properties that are determined by the state of the system, regardless of how that condition was achieved. energy, pressure, volume, temperature  E = E final - E initial Fig 5.8

Work Done By the System on the Surroundings  V = V f – V i > 0 -P  V < 0 w sys < 0  w = w final - w initial initialfinal Work is not a state function! Fig 5.12 Since the opposing pressure, P can vary: ViVi VfVf Expansion of a Gas:

 H =  E + P  V  H = (q+w) − w  H = q p So, at constant pressure, the change in enthalpy is the heat gained or lost. H = E + PV Enthalpy (H) - the heat flow into or out of a system in a process that occurs at constant pressure.

 H = heat given off or absorbed during a reaction at constant pressure H products < H reactants H products > H reactants

Enthalpies of Reaction H 2 O (s) H 2 O (l)  H = 6.01 kJ System absorbs heat Endothermic  H > kJ are absorbed for every 1 mole of ice that melts at 0 ° C and 1 atm.  H = H (products) – H (reactants) = q p

CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l)  H = kJ System gives off heat Exothermic  H < kJ are released for every 1 mole of methane that is combusted at 25 ° C and 1 atm. Enthalpies of Reaction

Enthalpy is an extensive physical property.  H for a reaction in the forward direction is equal in magnitude, but opposite in sign, to  H for the reverse reaction. 2H 2 O (s) 2H 2 O (l)  H = 2 x 6.01 = 12.0 kJ H 2 O (s) H 2 O (l)  H = 6.01 kJ H 2 O (l) H 2 O (s)  H = kJ The Truth about Enthalpy

 H for a reaction depends on the state of the products and the state of the reactants. The Truth about Enthalpy H 2 O (s) H 2 O (l)  H = 6.01 kJ H 2 O (l) H 2 O (g)  H = 44.0 kJ

How much heat is evolved when 155 g of iron undergoes complete oxidation in air? 4Fe (s) + 3O 2 (g) 2Fe 2 O 3 (s)  H = kJ 155 g Fe 1 mol Fe g Fe x = -776 kJ kJ 4 mol Fe x

The specific heat (s) - the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius: Heat (q) absorbed or released: q = m·C s ·  T where  T = T final - T initial Table 5.2

How much heat is given off when an 869 g iron bar cools from 94.0 ° C to 5.0 ° C? C s of Fe = 0.45 J/(g ° C)  T = T final – T initial = 5.0 ° C – 94.0 ° C = ‒ 89.0 ° C q = m·C s ·  T = (869 g) · (0.45 J/(g ° C)) · (– 89.0 ° C) = ‒ 34,800 J ≈ ‒ 3.5 x 10 4 J = ‒ 35 kJ

Calorimetry: Measurement of Heat Changes H cannot be determined absolutely, however, ΔH can be measured experimentally Device used is called a calorimeter Typically, ΔT ∝ ΔH rxn

Constant-Pressure Calorimetry No heat enters or leaves! q solution = m·C s ·  T = ‒ q rxn Because reaction at constant P:  H = q rxn Figure 5.17

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.) Standard enthalpy of reaction ( ) - the enthalpy of a reaction carried out at 1 atm. n  H o (products) f =  m  H o (reactants) f  - Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

Fig 5.20 An enthalpy diagram comparing a one-step and a two-step process for a reaction.

Must I measure the enthalpy change for every reaction of interest? The standard enthalpy of formation of any element in its most stable form is zero:  H o (O 2 ) ≡ 0 f  H o (O 3 ) = 142 kJ/mol f  H o (C, graphite) ≡ 0 f  H o (C, diamond) = 1.90 kJ/mol f Establish an arbitrary scale with the standard enthalpy of formation ( ) as a reference point for all enthalpy expressions. NO!! Standard enthalpy of formation ( ) - the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

Table 5.3 Standard Enthalpies of Formation, at 298 K

Calcium carbide (CaC 2 ) reacts with water to form acetylene (C 2 H 2 ) and Ca(OH) 2. How much heat is released per mole of calcium carbide reacted? The standard enthalpy of formation of CaC 2 is kJ/mol. CaC 2 (l) + 2H 2 O (l) C 2 H 2 (g) + Ca(OH) 2 (s) HoHo rxn n  H o (products) f =  m  H o (reactants) f  ‒ f = HoHo rxn [  H o (C 2 H 2 ) +  H o (Ca(OH) 2 ] –[  H o f (CaC 2 ) +  H o f (H 2 O)] f HoHo rxn = [( kJ) + (–986.2 kJ) ] – [( kJ) + 2( ] = kJ kJ 1 mol CaC 2 = kJ/mol CaC 2