Question

Chapter 3, problem 21. 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

G = Gproducts - Greactants

G = fG (NH3) - 1/ 2 f G (N2) -3 /2 f G (H2)

= - 16.48 - 1/2 x 0 -3/2 x0

  = - 16.48 kJ / mol

free energy energy = - 16.48 kJ/mol

part B)

If we assume that the enthalpy and the entropy are both independent of temperature,

rG(323) = rG(298) - (323 - 298) rS

rS = S (NH3) - 1/ 2H2 - 3/2 S(N2)

= 192:45 - 1/ 2 x 191:61 - 3 /2 x 130:684

= 99:38 J mol-1 K-1

rG(323) = 16.45 -25 x (- 0:09938)

= -13.97 kJ

free energy = -13.97 kJ /mol

part C)

From the answer to part (a), the maximum amount of work that can be done by the battery is 16 450 J/ mol NH3. For 100 g of H2,

w = 16 450 J/mol NH3 x 2/ 3 mol NH3/ mol H2 x 1/ 2.016 mol H2/ g H2 x 100 g H2

= 544 000 J S

Since the batter is operating at 1 mwatt (or 103 J/sec), the battery can run for 5.44 x 10^8 s or 17 years!

5.44 X 10^8 seconds