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Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol − 1 −5.980 kJ·mol−1 . Calculate the equilibrium constant for the reaction at 25 °C. K ′ e q = Keq′= What is ΔG for the reaction at body temperature (37.0 °C) if the concentration of A is 1.8 M 1.8 M and the concentration of B is 0.55 M ?

Respuesta :

Answer : The value of [tex]K_{eq}[/tex] is, 11.2

The value of [tex]\Delta G_{rxn}[/tex] is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

[tex]\Delta G^o=-RT\times \ln K_{eq}[/tex]

where,

[tex]\Delta G^o[/tex] = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = [tex]25^oC=273+25=298K[/tex]

[tex]K_{eq}[/tex]  = equilibrium constant  = ?

Now put all the given values in the above formula, we get:

[tex]\Delta G^o=-RT\times \ln K_{eq}[/tex]

[tex]-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}[/tex]

[tex]K_{eq}=11.2[/tex]

Thus, the value of [tex]K_{eq}[/tex] is, 11.2

Now we have to calculate the [tex]\Delta G_{rxn}[/tex].

The formula used for [tex]\Delta G_{rxn}[/tex] is:

The given reaction is:

[tex]A(aq)\rightleftharpoons B(aq)[/tex]

[tex]\Delta G_{rxn}=\Delta G^o+RT\ln Q[/tex]

[tex]\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}[/tex]    ............(1)

where,

[tex]\Delta G_{rxn}[/tex] = Gibbs free energy for the reaction  = ?

[tex]\Delta G_^o[/tex] =  standard Gibbs free energy  = -30.5 kJ/mol

R = gas constant = [tex]8.314\times 10^{-3}kJ/mole.K[/tex]

T = temperature = [tex]37.0^oC=273+37.0=310K[/tex]

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

[tex]\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})[/tex]

[tex]\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol[/tex]

Therefore, the value of [tex]\Delta G_{rxn}[/tex] is -9.04 kJ/mol

The equilibrium constant for the reaction at 25 °C, [tex]K_{eq}[/tex] is 11.2.

ΔG for the reaction at body temperature (37.0 °C) is -9.04 kJ/mol.

Relation between equilibrium constant and standard Gibbs free energy:

[tex]\triangle G^o= -RT*lnK_{eq}[/tex]

where,

[tex]\triangle G ^o[/tex] = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298K

[tex]K_{eq}[/tex]   = equilibrium constant = ?

→ Calculation for  [tex]K_{eq}[/tex] :

Now, substituting the values in the above formula:

[tex]\triangle G^o= -RT*lnK_{eq}\\\\-5980J/mol=-(8.314J/K.mol)*(298K)* lnK_{eq}\\\\K_{eq}=11.2[/tex]

Thus, the value of [tex]K_{eq}[/tex] is 11.2.

→ Calculation for [tex]\triangle G_{rxn}[/tex]:

Chemical reaction:

A(aq) ⇌B(aq)

The formula used is:

[tex]\triangle G_{rxn}=\traingle G^o+RT lnQ\\\\\triangle G_{rxn}=\traingle G^o+RT ln\frac{[B]}{[A]}[/tex]

where,

  • [tex]\triangle G_{rxn}[/tex] = Gibbs free energy for the reaction  = ?
  • [tex]\triangle G^o[/tex] =  standard Gibbs free energy  = -30.5 kJ/mol
  • R = gas constant =  8.314 J/K.mol
  • T = temperature = 310 K
  • Q = reaction quotient
  • [A] = concentration of A = 1.8 M
  • [B] = concentration of B = 0.55 M

On substituting the values in the above formula:

[tex]\triangle G_{rxn}=\traingle G^o+RT ln\frac{[B]}{[A]}\\\\\triangle G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)*(310K)*ln\frac{0.55}{1.8}] \\\\\triangle G_{rxn}=-9035.75J/mole=-9.04kJ/mol[/tex]

Therefore, the value of [tex]\triangle G_{rxn}[/tex] is -9.04 kJ/mol.

Find more information about Gibbs free energy here:

brainly.com/question/10012881

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