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 −

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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 ?

Chemistry

1 answer:

Tems11 [2.7K] · Answer 1 · 2026-02-22T23:11:40+03:00

Answer: The value of $K_{eq}$ is 11.2

The value of $\Delta G_{rxn}$ is -9.04 kJ/mol

Explanation:

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

$\Delta G^o=-RT\times \ln K_{eq}$

where,

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

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant =?

Substituting all available values into the formula, we find:

$\Delta G^o=-RT\times \ln K_{eq}$

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

$K_{eq}=11.2$

Thus, $K_{eq}$ is 11.2

We now need to determine $\Delta G_{rxn}$ .

The equation used for $\Delta G_{rxn}$ is:

The reaction in question is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

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

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction =?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

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

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Plugging all values into formula (1), we derive:

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

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

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