Chemical Thermodynamics 2013/2014

of 18 /18
Chemical Chemical Thermodynamics Thermodynamics 2013/2014 2013/2014 5 th Lecture:Thermochemistry Valentim M B Nunes, UD de Engenharia

Embed Size (px)

description

Chemical Thermodynamics 2013/2014. 5 th Lecture:Thermochemistry Valentim M B Nunes, UD de Engenharia. Thermochemistry. The study oh heat required or produced by chemical transformations is called thermochemistry . Its an application of the first Law. Remember: q p = Δ H and q V = Δ U. - PowerPoint PPT Presentation

Transcript of Chemical Thermodynamics 2013/2014

  • Chemical Thermodynamics2013/2014

    5th Lecture:ThermochemistryValentim M B Nunes, UD de Engenharia

  • *ThermochemistryThe study oh heat required or produced by chemical transformations is called thermochemistry. Its an application of the first Law. Remember: qp = H and qV = U Thermochemistry is a branch of thermodynamics. We can measure (indirectly, for instance work or temperature) the energy a reaction produces as heat, and identify q, depending on the conditions, with a change in internal energy or enthalpy.

  • *Standard enthalpy changesChanges in enthalpy when a system undergoes a physical or chemical change are normally reported to a set of standard conditions. We will consider the standard enthalpy change, H, the change of enthalpy for a process in which the initial and final states are in their standard states. Standard states: The standard state of a substance is its pure more stable form at the pressure of p = 1 bar, and specified temperature means standard state!Normally values are tabled at given temperatures, usually 298.15 K

  • Examples: phase transitions*Standard enthalpy of vaporization, HvapH2O(l) H2O(g): Hvap(373.15 K) = 44.7 kJ.mol-1 Standard enthalpy of fusion, HfusH2O(s) H2O(l): Hfus(273.15 K) = 6.01 kJ.mol-1 Standard enthalpy of sublimation, HsublC(s, graphite) C(g): Hsubl(298 K) = 716.7 kJ.mol-1

  • Standard enthalpy change in chemical reactions *The standard reaction enthalpy, Hr ,at a given temperature T, is the enthalpy of reaction when all reagents and products of reaction are in their standard states. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l): Hr = - 890 kJ.mol-1 The - signal means the process is EXOTHERMIC!

  • *Standard enthalpy of formation The standard enthalpy of formation, Hf , of a substance is the standard enthalpy for the reaction of its formation from its constituent elements in their reference states. 6 C(s, graphite) + 2 H2(g) C6H6(l): Hf (benzene, l) = 49 kJ.mol-1 As a consequence, by definition, the standard enthalpy of formation of any element in their reference state is zero at all temperaturesHf (C, graphite) = 0Hf (O2, g) = 0 ...

  • *Standard enthalpy of formation

  • *Hess's LawWe can now combine standard enthalpies of individual reactions to obtain the enthalpy of another reaction. This is an application of the first Law, resulting from the fact that enthalpy is a state function, is called the Hesss Law: The enthalpy of an overall reaction is the sum of reaction enthalpies of individual reactions into which a reaction may be divided.As a consequence, we may calculate the enthalpy of a chemical reaction, if we now the standard enthalpy of formation of all the substances (see for instance table of slide 7)i are the stoichiometric coefficients

  • *Exothermic and Endothermic processesAt constant pressure, if Hr < 0, qp < 0, and heat flows from the reaction to the surroundings and the process is exothermic.If Hr > 0, qp > 0 , and heat flows into the reaction from the surroundings and the process is endothermic.

  • Combustion reactions*Combustion reactions are of particular importance in thermodynamics and applications (thermoelectric power stations, motors,). The standard enthalpy of combustion, Hc is the standard reaction enthalpy for the complete oxidation of organic compounds to CO2(g) and H2O(l) and N2 if nitrogen is also present. Take for instance the combustion of natural gas (mainly methane):CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)Zero!All combustions are extremely exothermic!If we obtain water vapor (instead of liquid): Vaporization of two moles of water!

  • *CalorimetryConstant volumeqr = UH = U + (pV) Assume only contributions from gasesAdiabatic!qV = (qwater + qpump + qr) = 0 qr = -(qwater + qpump)qwater = mwater 4.184 J.g-1.K-1 Tqpump = Cpump t

  • *CalorimetryConstant pressureqp = H

  • *The variation of enthalpy with temperatureThe majority of industrial or biological reaction with interest occurs at temperatures different from 298 K. Knowing the Hr at one temperature, how to calculate it at other temperatures?Consider the following thermodynamic cycle: Tables!?

  • *The variation of enthalpy with temperatureRecallorIntegrating over two temperatures, T1 and T2:This equation is especially simple when Cp is independent of T:

  • *The variation of enthalpy with temperatureSince the previous equation applies to each substance in the reaction, the standard reaction enthalpy at temperature T2 is:This equation is known as Kirchhoffs law. Cp is calculated by;

  • *Temperature dependence of heat capacitiesIf heat capacity is temperature independent previous equations are very simple. If not, the temperature dependence of heat capacity is taken into account by writing:For organic compounds:For inorganic gases:Characteristic coefficients for each substance!

  • *Adiabatic flame temperatureFor a combustion process that takes place adiabatically with no shaft work, temperature of the products is referred to as the adiabatic flame temperature. This is the maximum temperature that can be achieved for given reactants. The maximum temperature for a given fuel and oxidizer combination occurs with a stoichiometric mixture.

  • *Adiabatic flame temperature

    Adiabatic Flame Temperature (K)FuelOxygen as oxidizerAir as oxidizerHydrogen, H230792384Methane, CH430542227Propane, C3H830952268Octane, C8H1831082277