Question

Below is a dataset from an experiment where equilibrium constants of unfolding for a protein were determined at several temperatures. Use the data to calculate the enthalpy of unfolding (∆Hu) for this protein. Explain your work!

T(°C)	Ku(M-1)
40	0.021997
50	0.050024
60	0.108285
70	0.224085
80	0.445008
90	0.850968

Answer 1

Use the relation

ln K_eq = -ΔH⁰/RT + ΔS⁰/R where R = gas constant.

Prepare a table of ln K_u against 1/T as below.

t (⁰C)	T (K) = (273 + t)	1/T (K-1)	Ku (M-1)	ln Ku
40	313	0.003195	0.021997	-3.8168
50	323	0.003096	0.050024	-2.9952
60	333	0.003003	0.108285	-2.2232
70	343	0.002915	0.224085	-1.4957
80	353	0.002833	0.445008	-0.8097
90	363	0.002755	0.850968	-0.1614

Plot the graph of ln K_u vs 1/T.

Plot of ln K_u vs 1/T

The left side is dimensionless; 1/T has the unit K^-1; therefore the slope must have unit K. The slope of the plot is -8306.8 K.

We have

-ΔH_u/R = -8306.8 K

===> ΔH_u = (8306.8 K)*R = (8306.8 K)*(8.314 J/mol.K) = 69062.7352 J/mol = (69062.7352 J/mol)*(1 kJ/1000 J) = 69.0627352 kJ/mol ≈ 69.063 kJ/mol (ans).