The acid dissociation constant of the 1,10-phenanthrolium ion (phen) was meeasur

ID: 1048540 • Letter: T

Question

The acid dissociation constant of the 1,10-phenanthrolium ion (phen) was meeasured at different temperatures using spectroscopic methods.

The values of pKa measured at 298K and 308K were 4.812 and 4.731, respectively.

Part A: Calculate rGo at 298K. Express your answer in kJ/mol to three significant figures.

Part B: Calculate rGo at 308K. Express your answer in kJ/mol to three significant figures.

Part C: Calculate rHo. Express your answer in kJ/mol to three significant figures.

Part D: Calculate rSo. Express your answer in J/(K.mol) to three significant figures.

Part E: Calculate the pKa of the acid at 359K. Express your answer to three significant figures

We know that pKa = -log Ka

Given pKa (298 K) = 4.812; Ka (298 K) = 10-pKa = 10-4.812 = 1.5417*10-5;

pKa (308 K) = 4.731; Ka (308 K) = 10-4.731 = 1.8578*10-5

A) We know that

G0 = -RTlnKeq

Therefore, rG0298 = -(8.314 J/K.mol)*(298 K)*ln (1.5417*10-5) = 27451.596 J/mol = 27.451 kJ/mol (ans)

B) rG0308 = - (8.314 J/mol.K)*(308 K)*ln (1.8578*10-5) = 27895.199 J/mol = 27.895 kJ/mol (ans).

C) We know, from Vant Hoff equation,

ln (K2/K1) = -H0/R*(1/T2 – 1/T1)

Therefore, for the present problem,

ln (1.8578*10-5/1.5417*10-5) = -rH0/(8.314 J/mol.K)*(1/308 – 1/298) K-1

===> 0.1865 = -rH0/(8.314 J/mol.K)*(-1.0895*10-4) K-1

===> rH0 = 0.1865*8.314/1.0895*10-4 J/mol = 14231.859 J/mol = 14.2318 kJ/mol 14.232 kJ/mol (ans).

D) Take the values at 298 K and use the equation

G0 = H0 – T*S0

Here, we must have,

rG0298 = rH0 – (298 K)*rS0

====> 27.451 kJ/mol = 14.232 kJ/mol – (298 K)*rS0

====> rS0 = (27.451 – 14.232) kJ/mol/298 K = 0.044359 kJ/mol.K = 44.359 J/mol.K (ans).

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