# 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 ΔGQuestion ∘

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

c at 298 K.

**A**5.8×10

−5

**B**5.8×10

5

**C**3.8×10

3

**D**3.8×10

−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

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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

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