5) Ca¹2(aq) + Ba(s) = Ca(s) + Ba*²(aq) T (K) Ke 198 6.5 248 12.3 298 18.7 Use the experimental data above to calculate AH xn and As°rxn for the reaction. Calculate theoretical values for AHºrn and As xn using AH(Ba¹2(aq)) = -537.6kJ/mol, AH°(Ca¹2(aq)) = -542.8kJ/mol, S°(Ba(s)) = 62.5J/molK, S°(Ba¹2(aq)) = 9.6J/molK, S°(Ca(s)) = 41.6J/molk and S°(Cat2(aq)) = -53.1J/molK,. Do the experimental data agree with the theoretical values?
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
Explanation
Experimental determination of ∆Ho and ∆So
Since we know from the relation between ∆Go and Kc
∆Go = -RTlnKc........................................(i)
Also
∆Go = ∆Ho - T∆So...............................(ii)
from eq (i) and (ii)
∆Ho - T∆So = -RTlnKc
rearranging
lnKc = -(∆Ho / RT) + (∆So / R)
Therefor plot between lnKc and (1/T) gives the values of ∆Ho and ∆So
The ∆Ho is calculated from slope which is given as -∆Ho / R. And ∆So can be calculated from intercept of plot and given as ∆So / R.
Theoretical determination of ∆Ho and ∆So
For given reaction, ∆Ho is given as
∆Hrxno = (∆HBa2+o + ∆HCao) - (∆HCa2+o + ∆HBao) ,
Similarly
For reaction ∆So is given as
∆Srxno = (SBa2+o + SCao) - (SCa2+o + SBao)
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