Question

The formation of a dipeptide is the first step toward the synthesis of a

protein molecule. Consider the following reaction:

glycine (aq) + glycine (aq) à glycylglycine (aq) + H2O (l)

Given that the standard formation free energies for glycylglycine (aq), glycine (aq) and

water are respectively – 487.9 kJ/mol, – 315.9 kJ/mol and – 237.13 kJ/mol, calculate

DRGo’ and the equilibrium constant at 298 K, keeping in mind that the reaction is carried

out in an aqueous buffer solution. Assume that the value is essentially the same at 310K.

What conclusion can you draw from your result?

Answer 1

From the question , the given reaction is -

Glycine + Glycine ----> Glycylglycine + Water

The Gibb's free energy for a reaction can be calculated as the difference the the standard gibb's energy of product and reactant.

So,

ΔG ( reaction ) =[ Δ⁰G(glycylglycine ) + Δ⁰G(water) ]- [ 2Δ⁰G ( glycine )]

Given the date -

Δ⁰G(glycylglycine) = - 487.9 KJ/mol

Δ⁰G(water) = - 237.13 KJ/mol

Δ⁰G ( glycine ) = -315.9 KJ/mol

Calculating the value from the above data -

ΔG ( reaction ) = [ Δ⁰G(glycylglycine ) + Δ⁰G(water) ]- [ 2Δ⁰G ( glycine )]

ΔG ( reaction ) = [- 487.9 KJ/mol + - 237.13KJ/mol] - [ 2*(-315.9 KJ/mol) ]

ΔG ( reaction ) = [-725.03 KJ/mol ] -[-631.8 KJ/mol ]

ΔG ( reaction ) = [-725.03 KJ/mol + 631.8 KJ/mol ]

ΔG ( reaction ) = -93.23 KJ/mol

The reaction between gibb's free energy and equilibrium constant is -

ΔG = RT ln K

where,

ΔG = gibbs free energy

K = equilibrium constant

R = 8.314 * 10 ^-3 KJ/Kmol

T = 298 K ( given in the question )

applying the formula,

ΔG = RT ln K

-93.23 KJ/mol = 8.314 * 10 ^-3 KJ/Kmol * 298 K ln K

ln K = -93.23 KJ/mol / 8.314 * 10 ^-3 KJ/Kmol * 298 K

ln K = -37.6295

K= 4.546 * 10 ^-17

Now, from the question, keeping the value of gibb's free energy to be same , the value of equilibrium constant at 310 K is calculated as -

ΔG = RT ln K

-93.23 KJ/mol = 8.314 * 10 ^-3 KJ/Kmol * 310 K ln K

ln K = -93.23 KJ/mol / 8.314 * 10 ^-3 KJ/Kmol * 310 K

ln K = - 36.1729

K= 1.9512 * 10 ^-16

So, when the value of gibbs free energy is the same , the value for the equilibrium constant will increase with increasing temperature .