ERT 108/3 PHYSICAL CHEMISTRY FIRST LAW OF THERMODYNAMICS Prepared by: Pn. Hairul Nazirah Abdul Halim

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Transcript of ERT 108/3 PHYSICAL CHEMISTRY FIRST LAW OF THERMODYNAMICS Prepared by: Pn. Hairul Nazirah Abdul Halim

  • ERT 108/3PHYSICAL CHEMISTRY

    FIRST LAW OF THERMODYNAMICSPrepared by:Pn. Hairul Nazirah Abdul Halim

  • ThermochemistryThermochemistry is the study of heat produced or required by chemical reactions.

    If we know the U and H for a reaction, we can predict the heat of reaction.

    Enthalpy of a system decreases when heat is releases. H < 0. (Exothermic process).

    Enthalpy of a system increases when heat is absorbed. H > 0. (Endothermic process).

  • Standard Enthalpy ChangesStandard enthalpy change, is the change in enthalpy for a process in which the initial and final substances are in their standard states.Standard states of a substance at a specified temperature is its pure form at 1 bar.For example;- the standard state of liquid ethanol at 298K is pure liquid ethanol at 298K and 1 bar.

  • Example of standard enthalpy change, ;

    the standard enthalpy of vaporization, is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar.

  • Enthalpy of Physical Change

    The standard enthalpy transition, is the standard enthalpy change that accompanies a change of physical state.

    Example, - standard enthalpy of vaporization, - standard enthalpy of fusion,

  • Change of enthalpy is independent of the path between the two states.

  • For example, the direct conversion from solid to vapor;

    Or as two steps, first fusion (melting) and then vaporization;

    Conclusion;

  • The standard enthalpy changes of a forward process and its reverse must differ only in sign.

    For example, at 298 K; - enthalpy of vaporization of water is +44 kJ/mol - enthalpy of condensation is -44 kJ/mol

  • b) Enthalpy of Chemical Change

    Thermochemical equation is a combination of a chemical equation and the corresponding change in standard enthalpy:

    = the standard enthalpy change

  • Alternatively, we can write the chemical equation with standard reaction enthalpy,

    For combustion of methane, we write;

  • The standard reaction enthalpy is;

    Where, v = stoichiometric coefficients= the standard molar enthalpy of species J at the temp. of interest.

  • For the reaction;

    The standard reaction enthalpy is;

    Where is the standard molar enthalpy of species J at the temp. of interest.

  • Standard enthalpy of combustion, is the standard reaction enthalpy for the complete oxidation of an organic compound to CO2 gas and liquid H2O.

    Organic compound contains C, H, O and N2 (if N is present).

    Example, the combustion of glucose;

  • Hesss Law

    Standard enthalpies of individual reactions can be combined to obtain the enthalpy of another reaction.

    This application of the First Law is called Hesss Law:The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided.

  • ExampleUsing Hesss Law

    The standard reaction enthalpy for the hydrogenation of propene is -124 kJ mol-1,

    The standard reaction enthalpy for the combustion ofPropane is -2220 kJ mol-1;

    Calculate the standard enthalpy of combustion of propene.

  • Solution

    Add or subtract the reactions given, together with any others needed, so as to reproduce the reaction required.

    Then add or subtract the reaction enthalpies in the same way. Additional data are in Table 2.5.

    The combustion reaction we require is;

  • This reaction can be recreated from the following sum:

  • Standard Enthalpy of FormationThe standard enthalpy of formation, of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states.

    The reference state of an element is its most stable state at the specified temp. and 1 bar.

    For example, at 298K the reference state of nitrogen is a gas of N2 molecules, mercury is liquid mercury, carbon is graphite.

  • The reaction enthalpy in terms of enthalpies of formation

  • Example The standard reaction enthalpy of:

    is calculated as follows:

  • The temperature dependence of reaction enthalpiesStandard reaction enthalpies at different temperatures may be estimated from heat capacities and the reaction enthalpy.

    Kirchhoffs Law;

  • is the difference of the molar heat capacities of products and reactants under standard conditions weighted by the stoichiometric numbers;

  • ExampleUsing Kirchhoffs Law

    The standard enthalpy of formation of gaseous H20 at 298 K is -241.82 kJ mol-1. Estimate its value at 100C given the following values of the molar heat capacities at constant pressure: H20(g): 33.58 J K-1mol-1 ; H2(g): 28.84 J K -1mol-1; 02 (g): 29.37 J K-1 mol-1. Assume that the heat capacities are independent of temperature.

  • Solution

    When is independent of temperature in the range T1 to T2,. Therefore,

  • To proceed, write the chemical equation, identify the stoichiometric coefficients, and calculate from the data.The reaction is;

    So;

  • ExampleCalculate for the following reaction at 1450K and 1 bar; H2 (g) + Cl2(g) HCl(g)Given that (HCl,g) = -92.3 kJ/mol at 298.15K and that;

    over this temperature range.

  • Solution

    Use the following equation;

    T1 = 298.15 KT2 = 1450 K