Question

An electrochemical battery is used to provide 1 milliwatt (mW) of power for a (small) light. The chemical reaction in the battery is: 1/2 N2 (g) + 3/2 H2 (g) → NH3 (g)

a. What is the free-energy change for the reaction at 25o C, 1 bar?

b. Calculate the free-energy change for the reaction at 50o C, 1 bar. State any assumptions made for this calculation.

c. The limiting reactant in the battery is 100 g of H2. Calculate the maximum length of time (seconds) the light can operate at 25o C

Answer 1

N₂ (g) + 3H₂ (g) → 2NH₃ (g)

ΔH = [2ΔHf(NH3 (g))] - [ΔHf(N2 (g)) + 3ΔHf(H2 (g))]
= [2(-46.11)] - [1(0) + 3(0)] = -92.22 kJ

ΔH = -92.22 kJ/mol of N₂ that reacts

ΔS = [2ΔSf(NH3 (g))] - [ΔSf(N2 (g)) + 3ΔSf(H2 (g))]
[2(192.34)] - [(191.5) + 3(130.59)] = -198.59 J/K

ΔS = -198.59 J/K-mol of N₂ that reacts

Free energy change, ΔG = ΔH - TΔS

a)

ΔG = (1/2 mol) * [(-92.22 * 1000 J/mol) - (25 + 273.15 K) * (-198.59 J/K-mol)]

= -16505.2 J or -16.5 kJ

b)

Assuming that ΔH^o and ΔS^o are independent of temperature

ΔG = (1/2 mol) * [(-92.22 * 1000 J/mol) - (50 + 273.15 K) * (-198.59 J/K-mol)]

= -14022.8 J or -14 kJ

c)

Gibbs energy is the same as the maximum non-PdV work (i.e. electrical) we can get from a chemical reaction when run at constant T and P.

Moles of N₂ that reacts = 1/3 * Moles of H₂ that reacts

= 1/3 * 100/2 = 50/3 moles

At T=25 ^oC

ΔG = (50/3 mol) * [(-92.22 * 1000 J/mol) - (25 + 273.15 K) * (-198.59 J/K-mol)]

= -467427.36 J

Power = Rate of electrical work done

1 * 10^-3 W = (467427.36 J) / time (seconds)

Time = 467427.36 / (10^-3)

= 4.67 x 10⁸ seconds