CHEMICAL THERMODYNAMICS 1 chem 16 4 slides per page

H change in enthalpy of system u change in internal

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Unformatted text preview: − [ ∆H o NH 3( g ) ] 298 f ∆H o = {[(472.7) + 3(218)] − [− 46.11]} kJ/mole 298 ∆H o = 1173 kJ/mole NH3 298 average BE N -H = 1173 kJ/mole NH 3 = 391 kJ mol N -H bonds 3 mole N - H mole NH 3 Internal Energy of Gas-phase reactions • Gas-phase reactions may involve changes in volume, and as the volume of the container change (due to expansion or condensation of molecules), Work is involved in the system Relationship of ∆H and ∆U • The total amount of heat energy that a system can provide to its surroundings at constant temperature and pressure is given by ∆H= ∆U + P∆V – which is the relationship between ∆H and ∆U. • ∆H = change in enthalpy of system • ∆U = change in internal energy of system • P∆V = work involved in the system Internal Energy of Gas-phase reactions 1. Then P∆V = 0 ∆ngas = 0 Examples CO (g) + H 2O (g) → H 2 (g) + CO 2(g) 14 244 4 3 14 244 4 3 2 mol gas 2. ∆ngas = (total moles of gaseous products) – (total moles of gaseous reactants) When V2 = V1 3. V2 > V1 P ∆V > 0 ∆ngas > 0 V2 < V1 P ∆V < 0 ∆ngas < 0 2 mol gas Zn (s) + 2 HCl (aq) → ZnCl 2(aq) + H 2(g) 144 44 2 3 144 44 2 3 0 mol gas 1 mol gas N 2(g) + 3 H 2(g) → 2 NH 3(g) 14 244 4 3 123 44 4 mol gas 2 mol gas 12 1/7/2014 Relationship of ∆H and ∆U ∆H= ∆U + P ∆V For reactions with gases: ∆H= ∆U + (∆ngas)RT But ∆U = qv (Are these two definitions compatible?) – Remember ∆U = q + w. – We have also defined w = -P∆V . – Thus ∆U = q + w = q -P∆V – Consequently, ∆H = q- P∆V + P∆V = q Relationship of ∆H and ∆U enthalpy, ∆H: heat measured at constant pressure. internal energy, ∆U: heat measured at constant volume. How much do the ∆H and ∆U for a reaction differ? The difference depends on the amount of work performed by the system or the surroundings. At constant pressure ∆H = qP Relationship of ∆H and ∆U • For reactions in which the volume change is very small or equal to zero (for reactions in which there are no gaseous reactants or products). – ∆V ≈ 0 and P∆V ≈ 0 – Since ∆H = ∆U + P∆V ∴ ∆H ≈ ∆U • Determine the difference between ∆H and ∆U at 100oC for CO(g) + O2(g) CO2(g) • Chlorine trifluoride is a toxic, intensely reactive gas. It used in World War II to make incendiary bombs. It reacts with ammonia and forms nitrogen, chlorine, and hydrogen fluoride gases. When two moles of chlorine trifluoride reacts, 1196 kJ of heat is evolved. – Write the thermochemical equations for the reaction – What is the ∆Hof for ClF3? 13...
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This document was uploaded on 03/26/2014.

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