Consider the equilibrium between molecule A and molecule B, pictured at the righ
ID: 1020144 • Letter: C
Question
Consider the equilibrium between molecule A and molecule B, pictured at the right. We represent the energies of molecule A and molecule B as energy ladders. As shown in the diagram, the ground state energies are related by: epsilon_b, o + delta epsilon = epsilon_a, 0 For energy ladder A the spacing between energy levels is epsilon_A and for energy ladder B the spacing between energy levels is epsilon_B Using energy ladder expressions for the partition functions, write out the expression for the equilibrium constant K = qb/qa in terms of epsilon_A, epsilon_B, delta epsilon, and T. The constant kb will also be in the expression. Using your result in part a, calculate the equilibrium constant at T=10K. Calculate also the standard molar Gibbs energy change delta G degree Does the equilibrium lie toward A or B in this case? Calculate the standard molar enthalpy delta H degree, and the standard molar entropy delta S degree. Is the equilibrium is determined mainly by the enthalpy change or the entropy change. Assume delta epsilon= 1.00 times 10^20J. Also assume epsilon_B = 1.00 epsilon 10^-20 J and epsilon_a = 1.00 times l0^21y. Raise the temperature to T=1000K and repeat the calculation. Is A or B favored now, and is this due mainly to the entropy term or the enthalpy term?Explanation / Answer
if we examine the energy ladder, from statistical thermodynamics , we are able to express partical population with e as
N2 / N1 = e ^ (– /kT)
we know that equalibrium constant K = [A]/[B] = qA/ qB =e ^(– G0 /RT)
jusby applying logarithm for two eqations
ln(EA/EB) = – /kT
ln( qA/ qB) = – G0 /RT
by simply rearranging this equations you will get qA/ qB in required terms