Cytochrome, a complicated molecule that we will represent as CyFe2+, reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). At pH 7.0 the following reduction potentials pertain to this oxidation of CyFe2+:
O2(g) + 4 H+(aq) + 4 e- --> 2 H2O(l);
E= +0.82 V
CyFe3+(aq) + e- --> CyFe2+(aq);
E= +0.22 V
a) What is delta G for the oxidation of CyFe2+ by air?
b) If the synthesis of 1.00 mol of ATP from (ADP) requires a delta G of 37.7 kJ, now many moles of ATP are synthesized per mole of O2.
I know for part a you have to add the two Es but after that I'm not sure
To calculate ΔG for the oxidation of CyFe2+ by air, we can use the equation:
ΔG = -nFΔE
where:
- ΔG is the standard free energy change
- n is the number of moles of electrons transferred (in this case, 4)
- F is Faraday's constant (96,485 C/mol)
- ΔE is the difference in reduction potentials of the two half-reactions
a) ΔG for the oxidation of CyFe2+ by air:
ΔG = -4 × 96,485 × (0.82 - 0.22) = -4 × 96,485 × 0.6 = -231,924 J/mol
Converting to kJ/mol: -231,924 J/mol ÷ 1000 = -231.924 kJ/mol
b) To find how many moles of ATP are synthesized per mole of O2, we can use the equation:
ΔG = -nFΔE = -ΔG'
nF = ΔG' / ΔE
where:
- ΔG' is the standard free energy change for the synthesis of 1 mole of ATP (37.7 kJ/mol)
- n is the number of moles of electrons transferred (in this case, 4)
- F is Faraday's constant (96,485 C/mol)
- ΔE is the reduction potential for the oxidation of CyFe2+ by air (0.6 V)
Substituting the values:
nF = (37.7 × 10^3) / 0.6 = 62,833.33 C/mol
Converting to moles of ATP per mole of O2:
1 mole of ATP / 62,833.33 C = 1.59 × 10^-5 mol ATP/mol O2
Therefore, approximately 1.59 × 10^-5 moles of ATP are synthesized per mole of O2.
To answer part a) of the question, you are correct that you need to add the reduction potentials (E values) of the two reactions in order to calculate the overall potential. However, before that, we need to identify the oxidation and reduction half-reactions.
From the given reduction reactions:
O2(g) + 4 H+(aq) + 4 e- --> 2 H2O(l);
E = +0.82 V (reduction potential for the oxidation of O2)
CyFe3+(aq) + e- --> CyFe2+(aq);
E = +0.22 V (reduction potential for the reduction of CyFe3+)
We can see that the oxidation half-reaction involves the reduction potential for O2, while the reduction half-reaction involves the reduction potential for CyFe3+. Therefore, we need to multiply the reduction potential of the oxidation reaction by -1 to reverse the sign, since it is now acting as an oxidation.
Now, let's calculate the overall potential through the following steps:
1. Reverse the sign of the reduction potential for the oxidation half-reaction:
O2(g) + 4 H+(aq) + 4 e- --> 2 H2O(l);
E = -0.82 V
2. Add the two reduction potentials:
E_total = E_reduction + E_oxidation
E_total = +0.22 V + (-0.82 V)
E_total = -0.60 V
Now, we can use the formula ΔG = -nFE, where ΔG is the change in Gibbs free energy, n is the number of electrons transferred, F is Faraday's constant (96485 C/mol), and E is the overall potential.
3. Determine the number of electrons transferred:
From the balanced reactions above, we see that 4 electrons are transferred in the oxidation of CyFe2+ and in the reduction of O2.
4. Calculate ΔG:
ΔG = -nFE
ΔG = -(4 mol)(96485 C/mol)(-0.60 V)
ΔG = 231,564 J or 231.6 kJ (rounded to one decimal place)
Therefore, the ΔG for the oxidation of CyFe2+ by air is 231.6 kJ.
Moving on to part b) of the question, we need to determine how many moles of ATP are synthesized per mole of O2 using the ΔG value given.
We can use the formula ΔG = -nFE to find the moles of electrons transferred in order to synthesize ATP.
1. Rearrange the formula to solve for n:
n = -ΔG / (FE)
n = -(-37.7 kJ) / ((4 mol)(96485 C/mol)(-0.60 V))
n = 0.0195 mol (rounded to four decimal places)
Since we see that 4 moles of electrons are transferred for 1 mole of O2, we can say that 0.0195 moles of electrons are transferred per mole of O2.
2. Convert the moles of electrons to moles of ATP:
From the balanced reaction for ATP synthesis, we observe that 2 moles of ATP are synthesized for each mole of electrons.
Therefore, the number of moles of ATP synthesized per mole of O2 is:
0.0195 mol electrons x (2 mol ATP / 1 mol electrons) = 0.0390 mol ATP
Hence, approximately 0.0390 moles of ATP are synthesized per mole of O2.