# Chemistry

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A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by

ATP(aq) +H2O (l) --->ADP(aq) +HPO4^2-

for which ΔG°rxn = –30.5 kJ/mol at 37.0 °C and pH 7.0. Calculate the value of ΔGrxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.30 mM, and [HPO4^2–] = 5.0 mM.

I got an answer of -15.527 kJ/mol and it marked me wrong. Also I got an answer of -27.3968 kJ/mol and it is wrong. Could you please tell me your answer and how you worked it out thanks.

• Chemistry -

Did you use dG = dGo + RTlnQ?

• Chemistry -

Yes and I'm not getting it...what did u get?

• Chemistry -

I used that..its the change in Gibbs energy formula.

• Chemistry -

If we use
dG = -30.5*1000 + 8.324*310*ln(ADP)*(H2PO4^-)/(ATP)= ?
I don't know how the pH 7 fits in unless that changes the concn (H2PO4^-)

• Chemistry -

I got an answer of -33603.04727....does that sound right?

• Chemistry -

and is the above answer in kJ/mol...cause it needs to be those units.

• Chemistry -

I checked the answer and it marked me incorrect. I attempted this problem 3 times....can u please tell me the correct answer!!

• Chemistry -

I keep getting about 51.4 kJ/mol
30,500 + 8.314*310*ln(0.0003*.005/.005)
30500 + (-20,907) = 51.4 kJ. Check that carefully.

• Chemistry -

Its actually -51.4! Thanks(:

• Chemistry -

Of course. I slipped the negative sign TWICE in the same problem.

• Chemistry -

It's fine! I really appreciate your help! :)

• Chemistry -

I am working on a problem like this one. I don't understand how you got to 51.4 from 30500 + (-20,907) = 51.4 kJ. Can you explain?

• Chemistry -

make sure to convert millimolar to molar

• Chemistry -

"I am working on a problem like this one. I don't understand how you got to 51.4 from 30500 + (-20,907) = 51.4 kJ. Can you explain?"

The 30500 should be negative: -30500+(-20907)
This gives you -51407 J/mol.
Convert to kJ: -51.407 kJ/mol or -51.4 kJ/mol.

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