Chemistry Math Question 3

Question

Consider a fuel cell that works by reacting 33.8 g of liquid methanol with oxygen to form liquid water and CO₂(g) at constant pressure. Compute the maximum amount of electrical work that can be performed by the fuel cell under standard conditions at 25 C.

SpeciesΔfG0	kJ/mol
CH3OH(l)	-166.6
H2O(l)	-237.1
CO2(g)	-394.4

Answer 1

This problem can be broken down into a thermodynamic calculation and a small stoichiometric one.

We are given the free energy of formation (Δ­_fG^o) at standard conditions for methanol, water and carbon dioxide from their elements

C +½ O₂ → CH₃OH_(l)  Δ­_fG^o= -166.6 kJ/mole

H₂ + ½ O₂ → H₂O_(l)  Δ­_fG^o= -237.1 kJ/mole

C + O₂ → CO_{2 (g)}       Δ­_fG^o= -394.4 kJ/mole

We combine these three equations such a way to yield the desired equation for reaction in the fuel cell, namely the reaction of methanol with oxygen to form carbon dioxide and water (combustion). To do this, we (a) invert the first reaction, and reverse the sign of the free energy. If a reaction is exergonic, its reverse is endergonic (b) multiply the second reaction by 2. This will result in the release of twice as much energy as the original reaction, therefore we multiply the free energy by 2 as well (c) add up all 3 reactions to get the free energy of combustion of methanol

CH₃OH_(l) →   C +½ O₂  Δ­_fG^o= 166.6 kJ/mole

2H₂ + O₂ → 2H₂O_(l)  Δ­_fG^o= -474.2 kJ/mole

C + O₂ → CO₂              Δ­_fG^o= -394.4 kJ/mole

Adding these 3 reactions together yields:

CH₃OH­_(l) + 3/2O₂ → 2H₂O_(l) + CO_{2 (g)}    Δ­G^o = -702 kJ/mole

Notice that the equation is balanced.

This Δ­G^o is the maximum amount of work that can be done by the combustion of 1 mole of methanol. So now, we need to figure out how many moles of methanol we have in 33.8 grams, as given in the problem. This is the stoichiometric part.

We divide 33.8 grams by the formula weight of methanol, which is 32.04 g/mole:

33.8 g/32.04 g/mole = 1.05 moles of methanol

Multiplying this by the work that can be obtained from the combustion of 1 mole of methanol gives us the answer to the problem.

Δ­G = (-702 kJ/mole)(1.05 moles) = -737 kJ of work ( rounded to the correct number of sigfigs)