Question

2. The reaction described in problem #1 is the focus of a calorimetry experiment which generates the following data: - A dry coffee-cup calorimeter is weighed on a balance; its mass is 6.092 g - 25.0 mL of 0.317 M sodium sulfate solution is added to the calorimeter; the mass of the calorimeter with this solution volume is measured to be 31.072 g. - 25.0 mL of 0.228 M aluminum nitrate solution is mixed with the sodium sulfate solution already in the calorimeter, after a maximum temperature change is achieved, the mass of the calorimeter and its contents is measured to be 56.023 g. Based on this data and the standard enthalpy of the reaction calculated in problem #1, determine the following (for the values of density and specific heat, assume the aqueous solutions are pure water): a. Calculate the amount of heat generated or absorbed by the reaction based on the moles of sulfate added to the calorimeter. (5 pts) b. Calculate the amount of heat generated or absorbed by the reaction based on the moles of aluminum added to the calorimeter. (5 pts) c. Given that the calorimeter is composed of styrofoam (specific heat is 1.13 Jg' °C and the initial temperature of the reaction mixture is 23.2°C, calculate the final temperature of the reaction mixture after the maximum temperature change is achieved. (7 pts)

Answer 1

4 LUBARU Geren. - mass of Calorimeter taken=6.0929 - concentra? of sodium sulphate takeq = 26 and 0.3177 rol of sockrem sulphate taken = 25 mil. That mean mols of Massou takey a. a concentra? x vel 0-3179 * 26 ond = 0.317 mal x 25 ml 1000 ml -0.007925 med 1 - Giver mass of calorimetat. Mas soy sal - 31.0729 So man of Ma, soy taken-(31-072-6.092) = (24-989]

LALLAHU - They we have added Al CH3), of . concentra? = 0.288m volume .2 25 ml That means mols of Al (403), added =concenton? x vel = 0.228 m * 25 oal das 70.228 mel 28 mil ਹਰ ਘ -0.0057 mels - Mlow the footad man of caloremchert fasad sad after reactom = 56.0289 Then mass of self after neaction = 56-0289 - 6.092g = 149.9369)

UU LUR - New given the initial temp of both . the sool" and the calorimeter betere the reaction takes place a 23-2.6 - Now lets say final temp after the reaction = tric Then the temp change of the solution= = Frond temp - Egétial temp = (tz - 23.2) - Now if we consider the solution to be an aquoir sala, then given mass of sel = 49.936g specific heat of sal? - 4.186 dl go They heat released by the sales reaction i calculated by the famul &u? етках 4т there. ma mass of sol? pa (2 specatic heat of self N ATer change es temp .

+ - ? -Again the heat released by soll es absesbad by the calore meter untill beeth of them come to equal tempo This heat absorbed by calene meter is. Calealated by the famula- | Real-max CX ATT where ca specific heat of calorimeter AT= change is tempo UA- whe

Now the chemical meaefion is | 3 Na₂SO4 + 241 (NO3), Al2(SO4)2 t 6 Nallo, That means by the neaction of 3 moks of Mason with a makes of Al wood We get 1 mole of Allson), salid

And the enthalpy of the meae foon (Ahirzn) ! .350434 kg Sampia That means when I mole of A2 (604) es tamed, total 250.434 K5 of heat will be absorbed. (thene Atran the. So reaction is Endothcomiz. Hence heat will be absorbed by reaction) now we have mali of na sau 20-807925 so neque sed mols of al. (103); for complete neacton = 26 x 0.007925 rocel ? -0.0053 meeles ] And the final mahs of Ala (504), tomed = "X0-007925 orel =10.00264 mals] And final mals of Manlog taimed blix 0-007925 tonel =10001585 mall

WEWNWWLLLLLLLLLL So we can worete the final eqras- 0.007925 0 .0053 - 0.00264 , 0.01585 - Mazsoy Alko) (son) lantos 1 0.007925 (58427)" 0.0053 (tets) 0.002by 'Alz (584)3| so tefal heat absorbed for the formation of 0.00264 mals of H2 (504); - 37.434 ks x 0.00264 ople = 10.925 ks]

h we get moles of (son 2) added = 0.007925 makes so heat absorbed per mole of sulphate a motal heat absorbed mols of sulphate h = 0.925 ks 0.007928 may = 1116.74 kolmal 6 senitaly we get mols of the fs) added = 0.0053 mals so heat absabed per mole of Alt3 is z total heat abscobed molot H t3 = 0.925 kg 0-0053 med =[17452 no/molel

REL ☺ How the tatal heat absorbed by the solitoen after the necefson = tretal heat released total heat released from solution from cal and meter = Qisel & Real - [onxex.AT), + (9* <*47 Jeal - (19.93.69 x 4.435 5 *AT) w ww. dºc . / | . +(6.092901:13 £_*41) BRATION gos = 209.03 AT F 6.88 ATS, = 215:9.41 /

- Sare the heat is released by pater and calisä meter, the final valice will be a negative value so we can worite - 215-9.AT I = 0-925 kg 7-215.9 AT I-9255 - AT = 925 °C -21509 9 (tj - 23.2°c ) = +4.28 €

- Iz =(-4-28 +23-2)'c =[131914) e ethe Fenal teap of men 2 18.916