Question

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Write you answers on separate paper. Show your work for full credit. Include the version by your name. Given the following data Ca(s)+2C(graphite) → Cac, (s) AH = -62.8 kJ Ca(s)+10,(g)>CaO(s) AH; = -634.9 kJ Cao(s)+H,0(1) + Ca(OH), (aq) AH; = -653.1 kJ C,H,(g)+{0,(8) ► 200,(g)+H,0(1) AH; =-1300.2 kJ C(graphite) +0,(g) → CO2(g) AH; = -393.5 kJ calculate AH for the reaction Cac (s)+2H,0(1) Ca(OH), (aq)+C,H,(8) Using standard enthalpies of formation along with the enthalpy change for one of the reactions above, find the AH for aqueous calcium hydroxide. It may help noting that the enthalpy changes of several of the five given reactions are standard enthalpies of formation • Consider the reaction of 32.0 g of calcium carbide and 27.0 g water. o How much heat is absorbed or released? o Bonus: If the reaction runs at 1.00 atm and 32.0 °C, find the volume of gas generated, the P-V work for the system, and E. o Bonus: If a 70.5%-efficient heat exchanger connects that reaction to a container of 5.05 kg water at 24.3 °C, find the final temperature of the water.

Answer 1

using Hess's lawi oH for Calat 24hoch Ca Coma aay + C2H₂ (9) OH = H₂ + OH3 + 2OHS - SHY - AH = 6634.9 kJ)+(-653.1 ks) +2-393.5 kJ) - (-1300.2 k) -(-62.8kJ) [SH= -712 kJ (OH Con = Di + DHg 11), aga - 1291 kJ moles of caCa = 32g 64gmol = mol = 0.5 mol moles & cetera 27 g = 30 1.5 mol. - 18 gimol so limiting Reagent is = CaC • Heat released = 356 kJ • OE = OH - ong RT ; AV song. RT/P a. ang= ² 1€ = (356-**8.314,X905 Jos DE= 354•73 kJ DV = x 8-314 * 305 - 0.0125 ms. Tav= 12.516 et] 1.013 X105

Heat exchange a 356 x 1000 x 70 = 5aos xya186 X 1000 X (7–24,39). Too T= 34.17°C