czas[min] |
temp[°C] |
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1.Obliczenie pojemności cieplnej kalorymetru |
0,5 |
18,5 |
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1 |
18,52 |
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1,5 |
18,52 |
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2 |
18,54 |
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2,5 |
18,54 |
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3 |
18,55 |
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3,5 |
18,57 |
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4 |
18,58 |
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4,5 |
18,6 |
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5 |
18,97 |
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5,5 |
19,57 |
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6 |
19,6 |
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6,5 |
19,62 |
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7 |
19,63 |
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7,5 |
19,65 |
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8 |
19,66 |
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8,5 |
19,66 |
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9 |
19,68 |
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9,5 |
19,68 |
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10 |
19,69 |
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10,5 |
19,72 |
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11 |
19,71 |
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5,15 |
19,6027 |
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11,5 |
19,72 |
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5,15 |
18,59845 |
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12 |
19,74 |
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12,5 |
19,75 |
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13 |
19,74 |
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13,5 |
19,75 |
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14 |
19,75 |
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14,5 |
19,76 |
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15 |
19,77 |
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Wykres zależności temperatury kalorymetru od czasu. |
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( ∆T +∑v )=19,61-18,59= 1,02 K |
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mkwasu=3,18g M=98 g/mol |
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nwoda/nkwas |
qr [kJ/mol] |
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nkwasu = 3,18 g / 98 g/mol =0,032 mola |
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20 |
71,4 |
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Vwody = 600ml = 600cm 3 |
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50 |
72,78 |
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mwody=ρ* V = 1g/cm3 * 600cm3 = 600 g |
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100 |
73,49 |
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Mwody=18g/mol |
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200 |
74,37 |
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nwody = m/M = 600g/18g/mol = 33,33mola |
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600 |
76,8 |
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1000 |
78,19 |
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4000 |
83 |
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8000 |
85,52 |
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20000 |
87,95 |
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Efekt cieplny procesu rozpuszczania wyznaczam graficznie |
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nwody/ nkwasu =33,33/0,032 =1041,5625 |
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qr= 78,3 kJ/mol = 78300 J/mol |
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Q kal=nkw * qr = 0,032mola * 78300J/mol = 2505,6 J= 2,5 kJ |
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Obliczam pojemność cieplną kalorymetru: |
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K=Qkal /( ∆T +∑v ) |
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K=2,5 kJ / 1,02K =2,45 kJ/K |
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czas[min] |
temp[°C] |
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2. Obliczenie ciepła zobojętniania |
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0,5 |
19 |
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1 |
19,03 |
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1,5 |
19,03 |
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2 |
19,04 |
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2,5 |
19,04 |
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3 |
19,06 |
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3,5 |
19,06 |
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4 |
19,07 |
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4,5 |
19,08 |
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5 |
19,09 |
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5,5 |
21,31 |
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6 |
21,5 |
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6,5 |
21,52 |
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7 |
21,53 |
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7,5 |
21,54 |
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8 |
21,55 |
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8,5 |
21,56 |
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9 |
21,55 |
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9,5 |
21,55 |
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10 |
21,55 |
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5,75 |
19,098 |
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10,5 |
21,56 |
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5,75 |
21,53 |
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11 |
21,55 |
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11,5 |
21,55 |
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12 |
21,55 |
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12,5 |
21,55 |
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13 |
21,56 |
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13,5 |
21,55 |
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14 |
21,55 |
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14,5 |
21,56 |
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15 |
21,56 |
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Wykres zmian temperatur kalorymetru w czasie |
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( ∆T +∑v ) = 21,53-19,1 =2,43 K |
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Obliczam ciepło procesu Q korzystając z wyznaczonej pojemności cieplej kalorymetru K: |
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Q=[K*( ∆T +∑v )*Mkwasu]/mkwasu= [2450J/K * 2,43 K *98g/mol]/ 3,18g=183,5kJ/mol |
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Q=qrozp + qzob |
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Obliczenie ciepła rozpuszczania: |
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mkwasu=3,41g |
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nkwasu=3,24g /98 g/mol =0,033 mola |
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nwody=600g/18g/mol = 33,33 mola |
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nwody/nkwasu = 33,33mola / 0,033mola =1010 |
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qr=78,2 KJ = 78200 J |
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qzob =Q - qr |
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qzob= 183,5kJ-78,2 kJ=105,3kJ ( ciepło zobojętniania na dwa mole wody) |
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qzob=105,3 / 2 =52,65kJ/mol ( ciepło zobojętniania na jeden mol) |
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qzob=- 52,65kJ/mol (uwzględniając typ reakcji- reakcja egzotermiczna) |
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Obliczam procent błędu wartości wyliczonej w stosunku do literaturowej ciepła zobojętniania: |
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literaturowa wartość ciepła zobojętniania qzob=-56,61kJ/mol |
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X= (56,61-52,65)/ 56,61 =0,07 = 7 % |
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Podsumowanie i wnioski: |
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Ciepło tej reakcji nie zależy od rodzaju kwasu i zasady biorących udział w reakcji i wynosi |
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ono literaturowo –56,61 kJ na mol utworzonej wody (qzob = -56,61 kJ/mol). |
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niewielkie różnice wielkości ciepła zobojętniania obliczonego i literaturowego (7%) wynikają z |
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odpowiednio dużej dokładności pomiarów a także poprawności wykonania zadania. |
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