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Enthalpy changes. The enthalpy change for a process is the heat energy exchanged with the surroundings at constant pressure. Enthalpy change is given the symbol ∆ H . The units are kJ mol –1.
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Enthalpy changes The enthalpy change for a process is the heat energy exchanged with the surroundings at constant pressure. Enthalpy change is given the symbol ∆H.The units are kJmol–1. Enthalpy changes are frequently measured under standard conditions, i.e. 298K and 100kPa. If an enthalpy change is measured under standard conditions, the symbol өis used in superscript, ∆Hө.
Bond enthalpies When a chemical reaction takes place, bonds are broken in the reactants and bonds are formed in the products.Breaking bondsis an endothermic process.Making bonds is an exothermic process. The enthalpy change for a reaction can be calculated by working out the enthalpy changes for bonds made and bonds broken during the reaction using mean bond enthalpies. The mean bond enthalpy is the average (mean) bond dissociation enthalpy for a particular bond in a range of different compounds. Precisely, it is the average enthalpy change for breaking 1 mole of a particular bond in a range of different compounds in the gas phase.
Born–Haber cycle for MgCl2 Mg2+(g) + 2Cl(g) + 2e– 2 × DHeaө (Cl) = 2 × (–364) DHi(2nd)ө (Mg) = +1450 Mg+(g) + 2Cl(g) + e– Mg2+(g) + 2Cl–(g) DHi(1st)ө (Mg) = +736 Mg(g) + 2Cl(g) 2 × DHaө (Cl) = 2 × (+121) Mg(g) + Cl2(g) DHlө (MgCl2) = –2493 DHaө (Mg) = +150 Mg(s) + Cl2(g) DHfө (MgCl2) MgCl2(s)
Lattice formation enthalpies N3– O2– F– increasing lattice formation enthalpy K+ Na+ Li+ When an ionic lattice is formed, the oppositely charged ions are attracted to each other. The stronger the attraction, the higher the lattice formation enthalpy. Two factors increase the attraction and therefore the lattice formation enthalpy: • high charge • small size.
Enthalpies of solution and hydration The standard enthalpy of solution (DHsolө) is the enthalpy change when one mole of an ionic compound is dissolved in water to produce aqueous ions. NaCl(s) Na+(aq) + Cl–(aq) The standard enthalpy of hydration (DHhydө) is the enthalpy change when one mole of gaseous ions is converted to one mole of aqueous ions. Na+(g) Na+(aq)
Factors affecting enthalpy of hydration Ion DHhyd (kJmol–1) Ion DHhyd (kJmol–1) Ion DHhyd (kJmol–1) Ion DHhyd (kJmol–1) The size of the enthalpy of hydration depends on: • The size of the ion. The smaller the ion, the larger the enthalpy of hydration. Li+ –519 F– –506 increasing size Na+ –406 Cl– –364 K+ –322 Br– –335 • The charge on the ion. The larger the charge on the ion, the larger the enthalpy of hydration. Fe2+ –1950 Fe3+ –4430
Entropy Entropy is a measure of disorder, and is given the symbol S. The units of S are: JK–1mol–1. • ordered • disordered • regular arrangement of particles • random arrangement of particles • low entropy • high entropy
Calculating entropy changes Standard entropy changes for any chemical reaction or physical change can be calculated using the following simple expression: DS = SSөproducts – SSөreactants Remember the following points: • the units of entropy, S, are JK–1mol–1 • entropies of elements are not zero like DHf values, so they should be included in calculations.
Gibbs free energy Whether a reaction is spontaneous depends on: • the entropy change of the system • the enthalpy change of the system • the temperature. The change in a quantity called the Gibbs free energy provides a measure of whether a reaction is spontaneous. The Gibbs free energy change is given the symbol DG and can be calculated for a reaction using the expression: A reaction will be spontaneous if DG< 0. DG = DH – TDS
Feasibility of reactions DH DS As temp. increases… Feasible? Even if DG is positive at room temperature, there may be a higher temperature at which a reaction becomes feasible. DG = DH – TDS If S is positive, there may be a point at which TS is big enough to outweigh the enthalpy factor. yes, above a certain temp. positive positive makes TDS > DH negative positive makes DG more negative always positive negative no effect: DG always positive never negative negative usually unlikely to make TDS > DH
Finding the temperature Consider the reduction of aluminium oxide with carbon: Al2O3(s) + 3C(s) 2Al(s) + 3CO(g) DH = +1336 kJmol–1 DS = +581JK–1mol–1 As both DH and DS are positive, DG will become negative if TDS > DH. The temperature at which this reaction becomes feasible can be calculated. This will be when DG = 0. If DG = 0, then T = DH / DS T = 1336 / (581/1000) T = 2299K
Solubility DG / feasibility DHsol DSsol In the same way that reactions are only feasible if DG < 0, a substance will be soluble in water at a specific temperature if DGsol < 0. DGsol = DHsol – TDSsol feasible if T is large enough to make DG negative positive positive negative positive always feasible positive negative never feasible usually feasible (T is unlikely to be large enough to make DG positive) negative negative
Thermodynamics vs. kinetics C(s) + O2(g) CO2(g) Just because a reaction is spontaneous does not mean that it appears to happen. It may be that the reaction is so slow or has such a high activation energy that it is not generally observed. Consider the reaction involving the combustion of carbon: DH = –394kJmol–1 DS = +3JK–1mol–1 This reaction has a negative value for DG, so is feasible, but a piece of carbon does not spontaneously burn if left on the table. Energy would need to be put in for the reaction to begin. A reaction that is thermodynamically feasible is not necessarily kinetically feasible.
