An expression for the equilibrium constant, Keq, based in statistical thermodynamics is shown below. Answer the following questions. Ke, - i) How is the sign of vj different for reactants and products? ii) What is the effect on Keq if the reactants are higher in energy than the products? Explain Would this case be exothermic or endothermic? iii) How is the Keq effected if the reactants have a greater number of thermally accessible states than the products? How is this related to the expression for Keq? case. Predict the sign of ArS in this
Solution
vi is actually called the stoichiometric number which is the difference of the stoichiometric factor for the reactents minus the products..
If the reactents are higher in energy then the delE is large.. The exponential term dominates but the minus sign before guarantees that the K is very very less than 1. This in turn illustrates that the concentration of the product is very low. In the equilibrium. As the energy states ( electronic) are higher for the products than the reactents the reaction is endothermic.
K is directly propotional to the ratio of the partition function of the product and the reactent then the higher number of the thermally accesible states of reactent means that the ratio will decrease, as the denominator is high.. Thus the K will decrease and reactent side will be predominating in the equilibrium. The entropy on the reactant side will be greater than that of the product side because of the higher number of accessible states (as S =k lnw) and the delS will be negative for the overall reaction.
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An expression for the equilibrium constant- Keq- based in statistical.docx
1. An expression for the equilibrium constant, Keq, based in statistical thermodynamics is shown
below. Answer the following questions. Ke, - i) How is the sign of vj different for reactants and
products? ii) What is the effect on Keq if the reactants are higher in energy than the products?
Explain Would this case be exothermic or endothermic? iii) How is the Keq effected if the
reactants have a greater number of thermally accessible states than the products? How is this
related to the expression for Keq? case. Predict the sign of ArS in this
Solution
vi is actually called the stoichiometric number which is the difference of the stoichiometric factor
for the reactents minus the products..
If the reactents are higher in energy then the delE is large.. The exponential term dominates but
the minus sign before guarantees that the K is very very less than 1. This in turn illustrates that
the concentration of the product is very low. In the equilibrium. As the energy states ( electronic)
are higher for the products than the reactents the reaction is endothermic.
K is directly propotional to the ratio of the partition function of the product and the reactent then
the higher number of the thermally accesible states of reactent means that the ratio will decrease,
as the denominator is high.. Thus the K will decrease and reactent side will be predominating in
the equilibrium. The entropy on the reactant side will be greater than that of the product side
because of the higher number of accessible states (as S =k lnw) and the delS will be negative for
the overall reaction.
