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Proteins A, B and C bind to each other to form a complex, ABC. Under equilibrium the concentrations of A, B, C and ABC are 10-2 M. The equilibrium constant and the standard free energy of this association reaction at T=300 K are, respectively,

a) 10-6 M2 and -8.3 kcal/mol.
b) 10-4 M2 and -5.5 kcal/mol.
c) 10-3 M2 and -4.1 kcal/mol.
d) 10-2 M2 and -2.8 kcal/mol.
e) 10-1 M2 and -1.4 kcal/mol.

Respuesta :

Answer: b) [tex]10^{4}M2[/tex] and [tex]-5.5 kcal/mol[/tex]

Explanation:

Equilibrium concentration of [tex]A[/tex] = [tex]10^{-2}M[/tex]

Equilibrium concentration of [tex]B[/tex] = [tex]10^{-2}M[/tex]

Equilibrium concentration of [tex]C[/tex] = [tex]10^{-2}M[/tex]

Equilibrium concentration of [tex]ABC[/tex] = [tex]10^{-2}M[/tex]

The given balanced equilibrium reaction is,

                            [tex]A+B+C\rightleftharpoons ABC[/tex]

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[ABC]}{[A][B][C]}[/tex]

Now put all the given values in this expression, we get :

[tex]K_c=\frac{10^{-2}}{(10^{-2})^3}[/tex]

[tex]K_c=10^4M^2[/tex]

[tex]\Delta G^o=-2.303\times RT\times \log K_c[/tex]

where,

R = universal gas constant = 2 cal/K/mole

T = temperature = 300 K

[tex]K_c[/tex] = equilibrium constant = [tex]10^4[/tex]

[tex]\Delta G^o=-2.303\times 2\times 300\times \log (10^4)[/tex]

[tex]\Delta G^o=-5527.2cal/mol=-5.5kcal/mol[/tex]

Thus the equilibrium constant and the standard free energy of this association reaction at T=300 K are [tex]10^4M^2[/tex]  and [tex]-5.5kcal/mol[/tex]

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