Based �on� free� energies� of �hydrolysis�given� below,� for� each �of� the� following�reactions:�
i)�Determine� the� overall� ∆G� of �the�reaction�
ii)�State� whether� it �is �favorable�(spontaneous)�as� written��
∆Ghydrolysis
Phosphocreatine� � � '43.1�kJ/mol�
Glycerolphosphate� � � '9.7�kJ/mol�
Phosphoenolpyruvate� � '61.9�kJ/mol�
Glucose'6'phosphate� � '12.5�kJ/mol�
a. �ATP� + �creatine� <->�Phosphocreatine� + �ADP�
b.� ATP �+� glycerol� <-> glycerolphosphate� +�ADP�
c.�ADP �+ �Phosphoenopyruvate� <-> pyruvate�+ �ATP�
d.�pyruvate �+� glucose'6'phosphate� <-> glucose� +� phosphoenopyruvate
The arrows are supposed to be separate arrows, going in opposite directions. I'm really stuck and I don't understand this question. Please help soon.
To determine the overall ΔG of each reaction and whether it is favorable or spontaneous, you need to use the equation:
ΔG overall = ΔGhydrolysis of product - ΔGhydrolysis of reactant
Firstly, let's calculate the overall ΔG for each reaction:
a. ATP + creatine ↔ Phosphocreatine + ADP
ΔGhydrolysis of ATP = 0 kJ/mol (for simplicity, we assume it is 0)
ΔGhydrolysis of Phosphocreatine = -43.1 kJ/mol
ΔG overall = ΔGhydrolysis of Phosphocreatine - ΔGhydrolysis of ATP
= -43.1 kJ/mol - 0 kJ/mol
= -43.1 kJ/mol
Since the overall ΔG is negative (-43.1 kJ/mol), the reaction is favorable (spontaneous) as written.
b. ATP + glycerol ↔ glycerolphosphate + ADP
ΔGhydrolysis of ATP = 0 kJ/mol
ΔGhydrolysis of glycerolphosphate = -9.7 kJ/mol
ΔG overall = ΔGhydrolysis of glycerolphosphate - ΔGhydrolysis of ATP
= -9.7 kJ/mol - 0 kJ/mol
= -9.7 kJ/mol
Since the overall ΔG is negative (-9.7 kJ/mol), the reaction is also favorable (spontaneous) as written.
c. ADP + Phosphoenolpyruvate ↔ pyruvate + ATP
ΔGhydrolysis of ADP = 0 kJ/mol
ΔGhydrolysis of Phosphoenolpyruvate = -61.9 kJ/mol
ΔG overall = ΔGhydrolysis of pyruvate - ΔGhydrolysis of ATP
= -61.9 kJ/mol - 0 kJ/mol
= -61.9 kJ/mol
Since the overall ΔG is negative (-61.9 kJ/mol), the reaction is favorable (spontaneous) as written.
d. pyruvate + glucose-6-phosphate ↔ glucose + phosphoenolpyruvate
ΔGhydrolysis of pyruvate = 0 kJ/mol
ΔGhydrolysis of glucose-6-phosphate = -12.5 kJ/mol
ΔG overall = ΔGhydrolysis of glucose - ΔGhydrolysis of ATP
= 0 kJ/mol - (-12.5 kJ/mol)
= 12.5 kJ/mol
Since the overall ΔG is positive (12.5 kJ/mol), the reaction is not favorable (not spontaneous) as written.
To summarize:
a. The reaction is favorable (spontaneous).
b. The reaction is favorable (spontaneous).
c. The reaction is favorable (spontaneous).
d. The reaction is not favorable (not spontaneous).
By calculating the overall ΔG using the ΔGhydrolysis values for each component, you can determine the spontaneity of the reactions.