A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to NleftharpoonsD . At 60 °C, the concentrations of both N and D are equal at equilibrium, and the standard

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Quantrex Academy · Accepted Answer

Correct answer: B. 2.0

For the reversible thermal denaturation

\text{N} \rightleftharpoons \text{D}

, the equilibrium constant is given by

K = \dfrac{[\text{D}]}{[\text{N}]}

.

At

60

^\circ

C, the concentrations of

\text{N}

and

\text{D}

are equal, which means

K = 1

.

The standard Gibbs free energy change is

\Delta G^\circ = -RT \ln K

. Since

K = 1

,

\Delta G^\circ = 0

.

Using the thermodynamic relation

\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ

, we get:

\Delta H^\circ = T \Delta S^\circ

Given

\Delta H^\circ = 666

kJ mol

^{-1}

and

T = 60 + 273 = 333

K.

\Delta S^\circ = \dfrac{\Delta H^\circ}{T} = \dfrac{666}{333} = 2.0

kJ K

^{-1}

mol

^{-1}

Answer:

2.0

A protein undergoes reversible thermal denaturation from its...

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