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    the value of δg° for the phosphorylation of glucose in glycolysis is 13.8 kj/mol.

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    The value of Δ G^∘ for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K.

    Click here👆to get an answer to your question ✍️ The value of Δ G^∘ for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K.

    The value of ΔG

    Question ∘

    for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of K

    c ​ at 298 K.

    A5.8×10

    −5

    B5.8×10

    5

    C3.8×10

    3

    D3.8×10

    −3 Medium Open in App Solution Verified by Toppr

    Correct option is D)

    ΔG

    ∘ =−RTlnK c ​

    Substituting the values, we get

    13800=−8.314×298×lnK

    c ​ ∴K c ​ =3.8×10 −3 Video Explanation

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    The value of Delta G ^(@)for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of K

    Given : Delta G ^(@) = 13. 8 kJ mol ^(-1) = 13. 8 xx 10 ^(3) J mol ^(-1). Delta G ^(@) =- RT ln K (c) implies ln K (c) =- 13. 8 xx 10 ^(3) J mol^(-1) //(8.314 J mol ^(-1) K ^(-1) xx 298K) ln K (c) =- 5.569 implies K(c) = e ^(- 5.569 ) implies K (c) = 3.81 xx 10^(-3)

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    The value of Delta G ^(@)for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of K _(c) at 298 K.

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    Text Solution Solution Given : Δ G ∘ =13.8kJmo l −1 =13.8× 10 3 Jmo l −1 .

    ΔG∘=13.8kJmol-1=13.8×103Jmol-1.

    Δ G ∘ =−RT lnK c ⇒ lnK c =−13.8× 10 3 Jmo l −1 /(8.314Jmo l −1 K −1 ×298K)

    ΔG∘=-RTlnKc⇒lnKc=-13.8×103Jmol-1/(8.314Jmol-1K-1×298K)

    ln K c =−5.569⇒ K c = e −5.569 ⇒ K c =3.81× 10 −3

    Kc=-5.569⇒Kc=e-5.569⇒Kc=3.81×10-3

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    The value of delta G for the phosphorylation of glucose in glycolysis is 13.8 kj/mol. Find the value

    ΔG=-2.303RT log Kcwhere R =8.314jk^-1mol^-1;T=298k;GivenΔG=13.8KJ/Mol but has value of R is in joules,we take it as 13.8×10³j/moltherefore 13.8×10³=-2.303×8.314…

    29.12.2016 Chemistry Secondary School answered

    The value of delta G for the phosphorylation of glucose in glycolysis is 13.8 kj/mol. Find the value of Kc at 298K?

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    Answer

    3.8/5 360 ΔG=-2.303RT log Kc

    where R =8.314jk^-1mol^-1;T=298k;

    Given

    ΔG=13.8KJ/Mol but has value of R is in joules,we take it as 13.8×10³j/mol

    therefore 13.8×10³=-2.303×8.314×298logKc

    log Kc =-2.4185

    antilog=0.2661×10^-2

    izvoru47 and 553 more users found this answer helpful

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    Answer

    4.5/5 100

    Answer: The value of at 298 K is Explanation:Relation between standard Gibbs free energy and equilibrium constant follows:

    where,

    = standard Gibbs free energy = 13.8 kJ/mol = 13800 J/mol  (Conversion factor: 1 kJ = 1000 J)

    R = Gas constant =

    T = temperature = 298 K

    = equilibrium constant in terms of concentration = ?

    Putting values in above equation, we get:

    Hence, the value of at 298 K is

    florianmanteyw and 189 more users found this answer helpful

    4.5 (89 votes)

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