Home
Class 12
CHEMISTRY
The value of DeltaG^(ϴ) for the phosphor...

The value of `DeltaG^(ϴ)` for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K.

Text Solution

Verified by Experts

`Delta G^(ϴ)=13.8` kJ/mol`=13.8xx10^(3)` J/mol
Also `DeltaG^(ϴ)=-RT In K_(c)`
Hence In `K_(c)=-13.8xx10^(3)`J/mol (`8.314 J "mol"^(-1)K^(-1)xx298K)`
In `K_(c)=-5.569`
`K_(c)=e^(-5.569)`
`K_(c)=3.81xx10^(-3)`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL EQUILIBRIUM AND ACIDS-BASES

    VGS PUBLICATION-BRILLIANT|Exercise NUMERICAL PROBLEMS|71 Videos

  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    VGS PUBLICATION-BRILLIANT|Exercise LONG ANSWER QUESTIONS|9 Videos

  • CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

    VGS PUBLICATION-BRILLIANT|Exercise Long Answer Questions|24 Videos

Similar Questions

Explore conceptually related problems

The value of DeltaG^(oplus) for the phosphorylation of glucose in glucloysis is 13.8kJ//mol . Find the value of K_(c) of 298K

Find the Value : 8+7/3

Find the value of tan ""(13pi)/(12).

Find the value of "sec" (13 (pi)/3)

What must be true of value of DeltaG^(@) for a reaction if K=1

The value of Delta_fG^(Theta) for formation of Cr_2 O_3 is – 540 kJ "mol"^(-1) and that of Al_2 O_3 is – 827 kJ "mol"^(-1) . Is the reduction of Cr_2 O_3 possible with Al ?

The value of DeltaG^(@) for a reaction is 7.97 KJ//"mole" . Its equilibrium constant K_(c) at 298K is 4 xx 10^(-x) What is x?