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The Thermochemistry of Physical Change 6.9 Vaporization 6.10

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The Thermochemistry of Physical Change 6.9 Vaporization 6.10 The Thermochemistry of Physical • Enthalpy of vaporization (DHvap) - the Change enthalpy change accompanying the conversion of one mole of a liquid into vapor – phase changes: DH = H – H solid « liquid liquid « gas solid « gas vap vapor liquid DH > 0 (endothermic) 6.9 Vaporization vap • Enthalpy of condensation (DH ) – vaporization (liquid ® gas) cond DH = H – H = – DH – condensation (gas ® liquid) cond liquid vapor vap • Vaporization is endothermic - the system DHcond < 0 (exothermic) absorbs energy needed to overcome the • The DH for the reverse of any process is the attractive forces between the molecules in the negative of the DH for the forward process liquid in order to spread them apart in the vapor (because H is a state function) Example: The vaporization of 23 g of ethanol 6.10 Melting and Sublimation at 78.5ºC and 1 atm requires 22 kJ of energy – melting, fusion (solid ® liquid) supplied as heat. Calculate the enthalpy of – freezing (liquid ® solid) vaporization of ethanol. • Enthalpy of fusion (DHfus) - the enthalpy P = constant Þ DH = q = 22 kJ change accompanying the conversion of one need DH per 1 mol of ethanol mole of a solid into liquid DH = H – H = – DH æ 1 mol ethanol ö fus liquid solid freez n = 23 g ethanol ´ ç ÷ = 0.50 mol ethanol • Melting is endothermic - energy is needed to è 46.1 g ethanolø overcome the forces holding the particles of a 22 kJ DH = = 44 kJ/mol solid in position and form the mobile liquid vap 0.50 mol ÞDHfus > 0 (endothermic); DHfreez < 0 (exothermic) – sublimation (solid ® gas) DHvap = Hvapor – Hliquid – deposition (gas ® solid) DHfus = Hliquid – Hsolid • Enthalpy of sublimation (DHsub) - the enthalpy change accompanying the conversion of one DHsub = Hvapor – Hsolid mole of a solid into vapor DHvap + DHfus = Hvapor – DHsub = Hvapor – Hsolid = – DHdepos Hliquid + Hliquid – Hsolid = • Sublimation is endothermic - energy is needed = H – H = DH to overcome the forces holding the particles of a vapor solid sub solid in position and form the gas phase Þ DHvap + DHfus = DHsub ÞDHsub > 0 (endothermic); DHdepos < 0 (exothermic) (true only at constant T) For H2O DHfus = 6.0 kJ/mol and DHsub = 50 kJ/mol Fig. 6.26 1 6.11 Heating and Cooling Curves • Heating (or cooling) curve - a graph of the temperature of a sample as a function of the heat added to (or removed from) it. • Regions in the heating (cooling) curves – sloped regions - correspond to temperature changes in the pure solid, liquid or gas phases (slope depends on the heat capacity of each phase) – flat regions - correspond to phase changes (temperature remains constant) Fig. 6.27 The Enthalpy of Chemical Change • Reaction enthalpy (DHr) - DH reported per one mol of a given reactant or product 6.12 Reaction Enthalpies • Properties of thermochemical equations • Thermochemical equations - chemical equation – multiplying an equation by a number multiplies the with physical states + enthalpy change (DH) DH value by the same number CH (g) + 2O (g) ® CO (g) + 2H O(l) – reversing the direction of a reaction changes the 4 2 2 2 sign of the DH value DH = – 890 kJ 2CH4(g) + 4O2(g) ® 2CO2(g) + 4H2O(l) the DH is for 1 mol CH and 2 mol of O 4 2 DH = – 2´890 kJ • The value of DH depends on the way the CO (g) + 2H O(l) ® CH (g) + 2O (g) chemical equation is written (stoichiometric 2 2 4 2 coefficients) DH = + 890 kJ • Example: When 2.31 g of solid P reacts with æ 1 mol P ö -2 gaseous Cl2 to form liquid PCl3 in a constant 2.31 g P ´ ç ÷ = 7.46 ´10 mol P pressure calorimeter with C = 2.16 kJ/ºC, the è 30.97 g Pø temperature rises by 11.06ºC. Write the - 23.9 kJ thermochemical equation. DH = = -320. kJ/mol P r 7.46 ´10-2 mol P qsurr = C´DT = 2.16 kJ/ºC´ 11.06ºC = 23.9 kJ P = constant - 320 kJ DH = ´ 2 mol P = -641 kJ Þ DH = qsys = – 23.9 kJ (for 2.31 g P) 1 mol P Chem. Eq. 2P(s) + 3Cl2(g) ® 2PCl3 (g) Þ need DH for 2 mol P 2P(s) + 3Cl2(g) ® 2PCl3 (g) DH = – 641 kJ 2 CH4(g) + 2O2(g) ® CO2(g) + 2H2O(g) 6.13 Standard Reaction Enthalpies DH = – 802 kJ (g) (g) (g) (l) • DHr depends on the physical states of reactants CH4 + 2O2 ® CO2 + 2H2O an products, P and T DHº = – 890 kJ • Standard state - the state of a substance in its pure form at 1 atm and a given temperature (usually 298 K). For substances in solutions, the standard state is at concentrations 1 mol/L • Standard reaction enthalpy (DHrº) - DHr for a reaction in which all reactants and products are in their standard states Fig. 6.29 3.
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