please answer 3 and 4
please counter-argument 3 and 4
Table 1: Bulks of Solutions control Each Trial DI H2O 2.00 1.00 0.00 1.00 Exp. | # Trials | 2.5×103 M K13 | 2.0 M Acetone | 1.0 M HCI | (mL) (mL)(mL) 1.00 1.00 1.00 2.00 1.00 3.00 1.00 1.00 2.00 1.00 1.00 4
3. -d[CH3COCH3]/dt = k[CH3COCH3][I3–][H+]n, where the appreciate of n (the direct of reaction with regard to H+)is referable unreserved.
-d[I3–]/dt = k[CH3COCH3][I3–][H+]n
-d[H+]/dt = k[CH3COCH3][I3–][H+]n
d[CH3COCH2I]/dt = k[CH3COCH3][I3–][H+]n
4. a) The sum bulk in complete the illustrations = 5 mL
Eagerness of KI3 in the decisive mixture
The no. of mmol of KI3 in the illustrations 1, 2 and 3 = 2.5*10-3 M * 1 mL, i.e. 2.5*10-3 mmol
The eagerness of KI3 in the illustrations 1, 2 and 3 = 2.5*10-3 mmol / 5 mL, i.e. 0.5*10-3 M
The no. of mmol of KI3 in the illustration 4 = 2.5*10-3 M * 2 mL, i.e. 5*10-3 mmol
The eagerness of KI3 in the illustration 4 = 5*10-3 mmol / 5 mL, i.e. 1*10-3 M
Eagerness of acetone in the decisive mixture
The no. of mmol of acetone in the illustrations 1, 3 and 4 = 2 M * 1 mL, i.e. 2 mmol
The eagerness of acetone in the illustrations 1, 3 and 4 = 2 mmol / 5 mL, i.e. 0.4 M
The no. of mmol of acetone in the illustration 2 = 2 M * 2 mL, i.e. 4 mmol
The eagerness of acetone in the illustration 2 = 4 mmol / 5 mL, i.e. 0.8 M
Eagerness of HCl in the decisive mixture
The no. of mmol of HCl in the illustrations 1, 2 and 4 = 1 M * 1 mL, i.e. 1 mmol
The eagerness of HCl in the illustrations 1, 2 and 4 = 1 mmol / 5 mL, i.e. 0.2 M
The no. of mmol of HCl in the illustration 3 = 1 M * 3 mL, i.e. 3 mmol
The eagerness of HCl in the illustration 3 = 3 mmol / 5 mL, i.e. 0.6 M
b) When the illustrations 1 and 3 are compared:
From the aloft postulates, the primal eagernesss of HCl are changing
From the aloft postulates, the primal eagernesss of KI3 and acetone are held regular.
c) If the reaction is not attributable attributablehing direct in H+:
Since, there is no involvement of H+ in the reprove indication, primal reprove of reaction accomplish referable vary among the illustrations 1 and 3.
If the reaction is 1st direct in H+:
The primal reprove of reaction in illustration 1, i.e. d[CH3COCH2I]/dt = k * 0.4 * 0.5*10-3 * 0.2, i.e. 0.2 x, where x = k * 0.4 * 0.5*10-3
The primal reprove of reaction in illustration 3, i.e. d[CH3COCH2I]/dt = 0.6 x
i.e. The primal reprove of reaction in the illustration ‘3’ is 3 times that in the illustration 1.
If the reaction is 2nd direct in H+:
The primal reprove of reaction in illustration 1, i.e. d[CH3COCH2I]/dt = k * 0.4 * 0.5*10-3 * (0.2)2, i.e. 0.04 x
The primal reprove of reaction in illustration 3, i.e. d[CH3COCH2I]/dt = k * 0.4 * 0.5*10-3 * (0.6)2, i.e. 0.36 x
i.e. The primal reprove of reaction in the illustration ‘3’ is 9 times that in the illustration 1