please answer 3 and 4
Homework Solution: Table 1: Volumes of Solutions for Each Trial DI H2O 2.00 1.00 0.00 1.00 Exp. |…
please solution 3 and 4
Table 1: Works of Solutions restraint 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
Expert Solution
3. -d[CH3COCH3]/dt = k[CH3COCH3][I3–][H+]n, where the compute of n (the dispose of reaction with honor 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 whole work in whole the tests = 5 mL
Force of KI3 in the conclusive mixture
The no. of mmol of KI3 in the tests 1, 2 and 3 = 2.5*10-3 M * 1 mL, i.e. 2.5*10-3 mmol
The force of KI3 in the tests 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 test 4 = 2.5*10-3 M * 2 mL, i.e. 5*10-3 mmol
The force of KI3 in the test 4 = 5*10-3 mmol / 5 mL, i.e. 1*10-3 M
Force of acetone in the conclusive mixture
The no. of mmol of acetone in the tests 1, 3 and 4 = 2 M * 1 mL, i.e. 2 mmol
The force of acetone in the tests 1, 3 and 4 = 2 mmol / 5 mL, i.e. 0.4 M
The no. of mmol of acetone in the test 2 = 2 M * 2 mL, i.e. 4 mmol
The force of acetone in the test 2 = 4 mmol / 5 mL, i.e. 0.8 M
Force of HCl in the conclusive mixture
The no. of mmol of HCl in the tests 1, 2 and 4 = 1 M * 1 mL, i.e. 1 mmol
The force of HCl in the tests 1, 2 and 4 = 1 mmol / 5 mL, i.e. 0.2 M
The no. of mmol of HCl in the test 3 = 1 M * 3 mL, i.e. 3 mmol
The force of HCl in the test 3 = 3 mmol / 5 mL, i.e. 0.6 M
b) When the tests 1 and 3 are compared:
From the aloft basis, the primal forces of HCl are changing
From the aloft basis, the primal forces of KI3 and acetone are held fixed.
c) If the reaction is not attributable attributablehing dispose in H+:
Since, there is no involvement of H+ in the objurgate look, primal objurgate of reaction procure referable substitute among the tests 1 and 3.
If the reaction is 1st dispose in H+:
The primal objurgate of reaction in test 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 objurgate of reaction in test 3, i.e. d[CH3COCH2I]/dt = 0.6 x
i.e. The primal objurgate of reaction in the test ‘3’ is 3 times that in the test 1.
If the reaction is 2nd dispose in H+:
The primal objurgate of reaction in test 1, i.e. d[CH3COCH2I]/dt = k * 0.4 * 0.5*10-3 * (0.2)2, i.e. 0.04 x
The primal objurgate of reaction in test 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 objurgate of reaction in the test ‘3’ is 9 times that in the test 1