Question

why is the first one wrong and what are the answers to the rest of the problems?

ADDED PROBLEMS: M Using Bond Enthalples: It-i-u- HA CH4 + H2 ---> CHAH = -124 kl/mol (34) +(973) . 137 -> (317) + 6") 2431-gods = -394 b) HC-CH2(g) + H2(g) -> H,C-CH(R) AH -138 kl/mol What is the enthalpy change during the process in which 100.0 g of water at 50.0°C is cooled to ice at - 30.0°C? How much energy is needed to convert 180 grams of ice at 0°C to liquid water at the same temperature? How much energy is needed to convert 180 grams of ice at -12°C to liquid water at 25°C?

Answer 1

1. (a) CH4 + H₂ ~ G Ho It H H-C-(-H CH + H - H - I H 14 AH = [BEreactant - Beproclucts = BEC-e +4BE-H + BEH-H – BE-, 6BE-H = 614 kJ/mol +4 (413 kJ/m) + 436 kJ/mon - 348 kJ/mol -6(413 kJ/mor) = 614 kJ/mol + 1652 kJ/mol +436 kJ/mo) – 348 kJ/mol -2478 kJ/mon -|-124 kJ/mon 2. Enthalpy change for cooling 100.og of water to ice at -30°C : Heat released to cooled from 50°C to 0°C: Q, = 100g 4.184 3/goc *(50-0)'( = 20920 J

Heat released during phase change from equid to ice Q = 4H fus * 100.0g 3 334 J/g x 1000g = 33400 J Heat released in cooling from 0°C to -30°C Q3 3 1009x 2.05 5/9 x 30°C (heat capacity of I = 6150 J 1. ice = 2:053/9° C) Total heat released = Q,+Q₂ +Q3 - (20920 + 33400 + 6150) J - 60470 J = 60470 kJ = 60.47 kJ 100 Enthalpy change = - Toral heat released -- 6047 kJ 1

1. 3. Heat need to convert 180g ice to liquid at 0°C = 1909 * oH furion = 180 g * 3343/ = 601201) Heat need to increase temperature from 12 to 0°C: Q, = 180gx Cice * 12°C = 1809 * 2.057/goc * 12°C = 4428 J Heat need to convert liquid into ice at 0°C Q₂ = 180g x AA fun = 1809 * 334J/g = 60120) Heat needed to increase temperature from 0°C 40 25°C - Q₂ = 180gx Cwater * (25-0)°C = 1809 x 4.184 7/goc X25°C = 18828 J Now, total heat required= Q , +Q2 +Q, - (4428 + 60120+18828) = 833 76 5