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Katyanochek1 [597]
3 years ago
10

Assume that some protein molecule, in its folded native state, has one favored conformation. But when it is denatured, it become

s a "random coil", with many possible conformations.
How will the contribution of ΔS for native → denatured affect the favorability of the process? What apparent requirement does this impose on ΔH if proteins are to be stable structures?
Chemistry
1 answer:
san4es73 [151]3 years ago
3 0

Answer:

Gibbs Free Energy is given by the next equation: ΔG = ΔH - TΔS, where

ΔH - <u>change</u> in enthalpy

T - temperature in Kelvin

ΔS - <u>change</u> in entropy

ΔG can be:

  • ΔG>0 positive(not spontaneous reaction)
  • ΔG<0 negative(spontaneous reaction)
  • ΔG=0(in equilibrium).

→ When the proteins are denatured, the entropy of the system increases. Therefore, the ΔS also increases and becomes more positive.  

→ From the equation we see that positive value of ΔS contributes to negative value of ΔG.

The proteins are stable when there is no spontaneous reaction, thus, the ΔG>0 is required. Which means that ΔH has to be large and positive to counteract the influence of ΔS.

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3 0
3 years ago
The Haber Process is the main industrial procedure to produce ammonia. The reaction combines nitrogen from air with hydrogen mai
Firdavs [7]

Answer:

A) N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g).

B) Kc=0.0933.

C) 0.9 mol.

D) Increasing both temperature and pressure.

Explanation:

Hello,

In this case, given the information, we proceed as follows:

A)

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

B) For the calculation of Kc, we rate the equilibrium expression:

Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}

Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent x, we have:

[NH_3]=0.6M=2*x

x=\frac{0.6M}{2}=0.3M

Next, the concentrations of nitrogen and hydrogen at equilibrium are:

[N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M

[H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M

Therefore, the equilibrium constant is:

Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933

C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.

D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.

Best regards.

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