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Sprawozdanie nr 6

 

                                      Sprawozdanie nr 6

 

Doświadczenie 4.1 Badanie wpływu stężenia na szybkość reakcji chemicznej.

 

1.      Celem doświadczenia jest zbadanie wpływu stężenia na szybkość reakcji chemicznej.

2.      Korzystałam z probówek, cylindrów miarowych, stopera oraz z następujących odczynników: 0,25 M H2SO4, Na2S2O3, H2O2, H2S2O3

 

Najpierw przygotowałam roztwory Na2S2O3 o różnych stężeniach. Do probówek z roztworem Na2S2O3 dodałam po 3 cm3 0,25 M H2SO4. Następnie za pomocą stopera zmierzyłam czas reakcji, a wyniki umieściłam w tabeli.

NR

OBJĘTOŚĆ ROZTWORU         Na2S2O3 [ cm3 ]

         c = 0,5 mol/dm3

OBJĘTOŚĆ  WODY

      [cm3 ]

         STĘŻENIE 

  c H2S2O3 = c Na2S2O3

         [mol/dm3]

CZAS REAKCJI

           [s]

SZYBKOŚĆ  REAKCJI

[mol/dm3·s-1]

1

10

0

0,5

25

0,02

2

8

2

0,4

35

0,0114

3

6

4

0,3

55

0,0054

4

4

6

0,2

70

0,0028

5

2

8

0,1

130

0,0007

6

1

9

0,05

250

0,0002

 

I . Obliczam stężenie  H2S2O3

1.

Dane:                                                                                                                             Szukane:

V1 = 10 cm3 = 0,01 dm3                                                                                                                                                  n = ?

C1 = 0,5 mol/dm3                                                                                                                                                                  c H2S2O3 =   ?                                                           

V2 = 10 cm3 = 0,01 dm3

                                                                C =  n ¸ V

                                                                n = V1 · C1

                                                                n = 0,01 dm3 · 0,5 mol/dm3

                                                                      n = 0,005 mol

                                                            c H2S2O3 = n ¸ V2

                                                            c H2S2O3 =  0,005 mol ¸ 0,01 dm3

                                                            c H2S2O3 = 0,5 mol/dm3

2.

Dane:                                                                                                                             Szukane:

V1 = 8 cm3 = 0,008 dm3                                                                                                                                                 n = ?

C1 = 0,5 mol/dm3                                                                                                                                                                c H2S2O3 =   ?                                                           

V2 =10 cm3 = 0,01 dm3

                                                                C = n  ¸ V

                                                                n = V1 · C1

                                                                n = 0,008 dm3 · 0,5 mol/dm3

                                                                      n = 0,0004 mol

                                                             c H2S2O3 = n ¸ V2

                                                             c H2S2O3 = 0,004 mol ¸ 0,01 dm3

                                                             c H2S2O3 = 0,4 mol/dm3

3.

Dane:                                                                                                                             Szukane:

V1 = 6 cm3 = 0,006 dm3                                                                                                                                                 n = ?

C1 = 0,5 mol/dm3                                                                                                                                                                c H2S2O3 =   ?                                                           

V2 =10 cm3 = 0,01 dm3

                                                               C = n  ¸ V

                                                                n = V1 · C1

                                                                n = 0,006 dm3 · 0,5 mol/dm3

                                                                      n = 0,0003 mol

                                                           c H...

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