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`

Step-by-step 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=([ABC])/([A][B][C])`

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

`K_c=(10^(-2))/((10^(-2))^3)`

`K_c=10^4M^2`

`\Delta G^o=-2.303* RT* \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* 2* 300* \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`