Question

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.

Answer 1

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

Explanation:

Equilibrium concentration of $A$ = $10^{-2}M$

Equilibrium concentration of $B$ = $10^{-2}M$

Equilibrium concentration of $C$ = $10^{-2}M$

Equilibrium concentration of $ABC$ = $10^{-2}M$

The given balanced equilibrium reaction is,

                            $A+B+C\rightleftharpoons ABC$

The expression for equilibrium constant for this reaction will be,

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

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

$K_c=\frac{10^{-2}}{(10^{-2})^3}$

$K_c=10^4M^2$

$\Delta G^o=-2.303\times RT\times \log K_c$

where,

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

T = temperature = 300 K

$K_c$ = equilibrium constant = $10^4$

$\Delta G^o=-2.303\times 2\times 300\times \log (10^4)$

$\Delta G^o=-5527.2cal/mol=-5.5kcal/mol$

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