Calculate δ g for atp hydrolysis in liver at 18 °c. use the liver concentrations from the first question.

This question is incomplete and the full question can be seen below:  

b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.

The equation for the ATP hydrolysis is:  

           H₂0  

ATP ------------> ADP + P₁                      ΔG = -30.5  

a) calculate ΔG for ATP hydrolysis to rank the following conditions from most favorable to least favorable. Assume a temperature of 37.0° C, R = 8.315  

muscle: [ATP]= 8.1mM; [ADP]= 0.9mM; [P₁]= 8.1mM  

liver: [ATP]= 3.4mM; [ADP]= 1.3mM; [P₁]= 4.8mM  

brain: [ATP}= 2.6mM; [ADP}= 0.7mM; [P₁]= 2.7mM  

b) Calculate ΔG for ATP hydrolysis in liver at 18° C. Use the liver concentration From part a.

−45.8 KJ/mol

Explanation:  

Equilibrium constant (k) is defined as a measure of the ratio of the equilibrium concentration of the products of a reaction, to the Equilibrium concentration of the reactants with each concentration raised to the power corresponding to the coefficient in the balanced equation of the reaction.

In the reaction in the question given above;  

For muscle;

ADP= 0.9 × 10⁻³

P₁= 8.1 × 10⁻³

ATP= 8.1 × 10⁻³

∴ K =

  K = 0.9 × 10⁻³

For liver;

ADP= 1.3 × 10⁻³

P₁= 4.8 × 10⁻³

ATP= 3.4 × 10⁻³

∴ K =

  K = 1.8 × 10⁻³

For brain;

ADP= 0.7 × 10⁻³

P₁= 2.7 × 10⁻³

ATP= 2.6 × 10⁻³

∴ K =

  K = 7.3 × 10⁻⁴

b) Since we are concerned about calculating ΔG for ATP hydrolysis in liver at 18° C and we've already obtained the liver concentration from part a; we can therefore calculate ΔG as:

ΔG = ΔG° + RTInK

ΔG° = -30.5

R= 8.315  

T= 18° C = 18 + 273.15k = 291.15k

K= 1.8 × 10⁻³

InK = In(1.8 × 10⁻³ )

      ≅ -6.32

∴ ΔG = -30.5 +  8.315 × 291.15k × (-6.32)

   ΔG = -30.5 +  (−15.30016542)

   ΔG = −45.80016542

   ΔG ≅ −45.8 KJ/mol

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.