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Thermochemistry. Chapter 5. Temperature = Thermal Energy. 90 ° C. 40 ° C. Energy Changes in Chemical Reactions. Heat - the transfer of thermal energy between two bodies that are at different temperatures. Temperature - a measure of the thermal energy. (intensive). (extensive).
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Thermochemistry Chapter 5
Temperature = Thermal Energy 90 °C 40 °C Energy Changes in Chemical Reactions Heat - the transfer of thermal energy between two bodies that are at different temperatures Temperature - a measure of the thermal energy (intensive) (extensive) greater temperature greater thermal energy
2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy Exothermic process - gives off heat – transfers thermal energy from the system to the surroundings.
energy + CaCO3(s) CaO (s) + CO2(g) energy + H2O (s) H2O (l) Endothermic process - heat has to be supplied to the system from the surroundings.
C3H8 + 5O2 3CO2 + 4H2O 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 DEsystem + DEsurroundings = 0 or DEsystem = −DEsurroundings Exothermic chemical reaction!
Energy relationships Fig 5.4 (a) Exothermic
Energy relationships Fig 5.4 (b) Endothermic
Energy Diagram for the Interconversion of H2 (g), O2 (g), and H2O (l) Fig 5.5
Another form of the first law for DEsystem DE = q + w DEis the change in internal energy of a system qis the heat exchange between the system and the surroundings w is the work done on (or by) the system Table 5.1 pg 171
Thermodynamics State functions - properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Fig 5.8 DE = Efinal - Einitial
DV = Vf – Vi > 0 -PDV < 0 wsys < 0 Dw = wfinal - winitial Work Done By the System on the Surroundings Expansion of a Gas: Since the opposing pressure, P can vary: Work is not a state function! Vf Vi initial final Fig 5.12
Enthalpy (H) - the heat flow into or out of a system in a process that occurs at constant pressure. H = E + PV H = E + PV H = (q+w) −w H = qp • So, at constant pressure, the change in enthalpy is the heat gained or lost.
Enthalpy (H) - the heat flow into or out of a system in a process that occurs at constant pressure. DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants
H2O (s) H2O (l) DH = 6.01 kJ Enthalpies of Reaction DH = H (products) – H (reactants) = qp System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0°C and 1 atm.
DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Enthalpies of Reaction System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25°C and 1 atm.
2H2O (s) 2H2O (l) H2O (s) H2O (l) H2O (l) H2O (s) DH = -6.01 kJ DH = 2 x 6.01= 12.0 kJ DH = 6.01 kJ The Truth about Enthalpy • 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.
H2O (l) H2O (g) H2O (s) H2O (l) DH = 6.01 kJ DH = 44.0 kJ The Truth about Enthalpy • H for a reaction depends on the state of the products and the state of the reactants.
How much heat is evolved when 155 g of iron undergoes complete oxidation in air? 4Fe (s) + 3O2(g) 2Fe2O3(s)DH = -1118.4 kJ x -1118.4 kJ 1 mol Fe x 4 mol Fe 55.85 g Fe = -776 kJ 155 g Fe
The specific heat (s) - the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius: Table 5.2 Heat (q) absorbed or released: q = m·Cs·DT where DT = Tfinal - Tinitial
How much heat is given off when an 869 g iron bar cools from 94.0°C to 5.0°C? Cs of Fe = 0.45 J/(g • °C) DT = Tfinal – Tinitial = 5.0 °C – 94.0 °C = ‒89.0 °C q = m·Cs·DT = (869 g) · (0.45 J/(g • °C)) · (– 89.0 °C) = ‒ 34,800 J ≈ ‒ 3.5 x 104 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 ∝ ΔHrxn
Constant-Pressure Calorimetry Figure 5.17 qsolution = m·Cs·DT = ‒ qrxn Because reaction at constant P: DH = qrxn No heat enters or leaves!
S S - = nDHo (products) mDHo (reactants) f f Standard enthalpy of reaction () - the enthalpy of a reaction carried out at 1 atm. 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. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
Fig 5.20 An enthalpy diagram comparing a one-step and a two-step process for a reaction.
Establish an arbitrary scale with the standard enthalpy of formation ( ) as a reference point for all enthalpy expressions. 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. DHo (O2) ≡ 0 DHo (O3) = 142 kJ/mol DHo (C, graphite) ≡ 0 DHo (C, diamond) = 1.90 kJ/mol f f f f Must I measure the enthalpy change for every reaction of interest? NO!! The standard enthalpy of formation of any element in its most stable form is zero:
CaC2(l) + 2H2O (l) C2H2(g) + Ca(OH)2(s) S ‒ S = DHo DHo DHo rxn rxn rxn [(226.77 kJ) + (–986.2 kJ) ] – [(-62.67 kJ) + 2(-285.83 ] = -125.1 kJ = [DHo (C2H2) + DHo (Ca(OH)2] –[DHof(CaC2) +DHof (H2O)] = -125.1 kJ/mol CaC2 -125.1 kJ f 1 mol CaC2 mDHo (reactants) nDHo (products) f f Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) and Ca(OH)2. How much heat is released per mole of calcium carbide reacted? The standard enthalpy of formation of CaC2 is -62.76 kJ/mol. = f