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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)+ H20 (l) ->ADP (aq) + HPO4 (negative two overall charge) (aq).




for which ?G�rxn = �30.5 kJ/mol at 37.0 �C and pH 7.0. Calculate the value of ?Grxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.80 mM, and [HPO42�] = 5.0 mM.




A. What is the delta G rxn in kJ/mol?


B. Is the hydrolysis of ATP spontaneous under these conditions?

Respuesta :

Answer:

A) The ΔG of the reaction is -48.9 kJ/mol.

B) yes, reaction will be spontaneous under these condition as ΔG is less than zero.

Explanation:

[tex]ATP (aq)+ H_2O (l) \rightarrow ADP (aq) + HPO_4^{2-}(aq)[/tex]

Given concentration of:

[ATP] = 5.0 mM = 0.005 M, [ADP] = 0.80 mM = 0.0008 M,

[tex][HPO_4^{2-}] = 5.0 mM=0.005 M[/tex]

The reaction quotient is given by:

[tex]Q=\frac{[ADP][HPO_4^{2-}]}{[ATP]}=\frac{0.0008 M\times 0.005 M}{0.005 M}=0.0008 [/tex]

To calculate the ΔG(at 25°C) for given value of Gibbs free energy, we use the relation:

[tex]\Delta G=\Delta G^o+RT\ln Q[/tex]

where,

[tex]\Delta G[/tex] = Gibbs free energy at given conditions

[tex]\Delta G^o[/tex] = Gibbs free energy at equilibrium=-30.5 kJ/mol

R = Gas constant = [tex]8.314J/K mol[/tex]

T = temperature = [tex]37^oC=[273.15+37]K=310.15 K[/tex]

[tex]Q[/tex] = reaction quotient at 37°C   = 0.0008

Putting values in above equation, we get:

[tex]\Delta G=-30500 J/mol+(8.314J/Kmol)\times 310.15 K\times \ln [0.0008][/tex]

[tex]\Delta G=-48900 J/mol= - 48.9 kJ/mol[/tex]

The ΔG of the reaction is -48.9 kJ/mol. Yes, reaction will be spontaneous under these condition as ΔG is less than zero.

pstnonsonjoku

The change in free energy is -48.9 KJ/mol and the hydrolysis is spontaneous under these conditions since ΔG is negative.

Given the reaction;

ATP (aq)+ H20 (l) ----->ADP (aq) + HPO4^2- (aq)

We have the following information;

[ATP] = 5.0 mM or 0.005 M

[ADP] = 0.80 mM or 0.0008 M

[HPO42^2-] = 5.0 mM or 0.005 M

We must now obtain the reaction quotient (Q) from the information provided;

Q =  [HPO42^2-] ×  [ADP]/ [ATP]

Note water is excluded in this equation because it is a pure substance.

Q = [0.005]  × [0.0008]/[0.005]

Q = 0.0008 M

Since we have in the question that ΔG°reaction = -30.5 kJ/mol or 30500 J/mol

But ΔG =  ΔG°reaction  + RTlnQ

Note that;

R = 8.314 J/K

T = 37.0°C + 273 = 310 K

ΔG°reaction = -30.5 kJ/mol or 30500 J/mol

Substituting values;

ΔG = -30.5  × 10^3 J/mol + (8.314 J/K  × 310 K × ln(0.0008 M))

ΔG =  -48.9 KJ/mol

Hence, the hydrolysis is spontaneous under these conditions since ΔG is negative.

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