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Calculate the actual, physiological ΔG for the reaction: creatine phosphate + ADP → creatine + ATP at 37°C, as it occurs in the cytosol of neurons with creatine phosphate at 4.7 mM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6mM. Give your answer in kJ/mol to one decimal point.

Respuesta :

ajeigbeibraheem

Answer:

-13.2 kJ/mol

Explanation:

Creatine Phosphate  +  H₂O ------>   Creatine + P₁ (ΔG° = -43 kJ/mol

ADP                             +  Pi    ------->   ATP +  H₂O   ΔG° =  +30.5 kJ/mol    

Creatine Phosphate  + ADP  ------> Creatine  + ATP     ΔG° = -12.5 kJ/mol

Parameters given include:

Creatine Phosphate = 4.7 mM

creatine = 1.0 mM

ADP = 0.73 mM

ATP = 2.6mM

Temperature (T) = 37°C = (37+273)K = 310 K

ΔG° = -12.5 kJ/mol = -12500 J/mol

Rate constant (R) = 8.314

The actual physiological ΔG for the reaction can be determined using the formula:

ΔG = ΔG° + RT㏑[tex][\frac{Products}{Reactants}][/tex]

Replacing our values from above, we have:

ΔG =  - 12500  + (8.314)(310)㏑[tex]\frac{(1.0*10^{-3})(2.6*10^{-3})}{(4.7*10^{-3})(0.73*10^{-3})}[/tex]

ΔG =  - 12500 + 2577.34 × ㏑ [ 0.7578]

ΔG =  - 12500 + 2577.34 ( -0.2773)

ΔG = - 13214.696 J/mol

ΔG = - 13.215 kJ/mol

ΔG = - 13.2 kJ/mol    

∴  the actual, physiological ΔG for the reaction = - 13.2 kJ/mol ( to 1 decimal place)

okpalawalter8

The actual physiological ΔG for the reaction is

  • 0.7kJ/mol

The formula for ΔG is mathematically given as

[tex]\Delta G^0 = -RT ln Keq\\\\ \Delta G^0 = -(8.314)(37+273) In \frac{(ATP)(Creatine)}{(ADP)(phosphate eatine)}\\\\ \Delta G^0 = -(8.314)(37+273) In \frac{(2.6)(1)}{(0.73)(4.7)}\\\\ \Delta G^0 = 714.80 J/mol \\\\ \Delta G^0 = 0.7KJ/mol [/tex]

What is ADP and ATP?

ATP (Adenosine tri-phosphate)

ADP (Adenosine di-phosphate) + phosphate.

For more information on ADP, visit

https://brainly.com/question/11442354

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