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butalik [34]
4 years ago
12

As an approximation, we can assume that proteins exist either in the native (or physiologically functioning) state or the denatu

red state. The standard molar enthalpy and entropy of the denaturation of a certain protein are 545 kJ·mol−1 and 1.55 kJ·K−1·mol−1, respectively. Comment on the signs and magnitudes of these quantities.
Chemistry
1 answer:
Andrews [41]4 years ago
7 0

Answer:

Reaction is spontaneous at high temperature and nonspontaneous at low temperature

Explanation:

Given:

Enthalpy change \Delta H= 545 \frac{KJ}{mol}

Entropy change \Delta S = 1.55 \frac{KJ }{K. mol}

From the formula of change in free energy,

  \Delta G = \Delta H - T\Delta S

But for spontaneous process the values of quantities are given below

For spontaneous process value of \Delta G is negative

For nonspontaneous process value of \Delta G is positive

Here values of \Delta H and \Delta S are positive, so reaction is spontaneous at

high temperature and nonspontaneous at low temperature

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