Answer to Question #74457 in General Chemistry for john

Question

Canada produces about 1.5 million tonnes of chlorine per year by the electrolysis of brine (NaCl(aq)). If the required electrolytic cells are operated at a potential difference of 4.5 V, calculate how many total kilowatt hours (kWh) of electrical power are required. 1 kWh = 3.6 × 106 J.

Answer 1

Solution:
2Cl2e= Cl2; n = 2
F – 96500 Q
1 mole of Cl2 (70.9 g) require 2 x 96500 Q of electricity
1.5 x 1012 g require x
X = 2 x 96500 Q x 1.5 x 10^12 g / 70.9 g = 4.08 x 10^15 Q
Per 1 hour: Q = It; I = Q / t = 4.08 x 1015 Q / 3600 s = 1.13 x 10^12 A
The power: P = IU = 1.13 x 1012 A x 4.5 V = 5.1 x 10^12 W = 5.1 x 10^9 kW.

For 1 year process: 365 days = 8760 hours
5.1 x 10^9
kW x 8760 h = 4.47 x 10^16 kW
Answer: 4.47 x 10^16 kW.