Question

				Concentrations
	# of drops	# of drops	# of drops	Stock Solution	Stock Solution	Reaction (after mixing in well)	Reaction (after mixing in well)	Reaction Time (seconds)			Reaction Rate (sec-1)
Well #	HCl	H2O	Na2S2O3	HCl	Na2S2O3	HCl	Na2S2O3	Trial 1	Trial 2	Average
C1, D1	12	0	8	1M	0.3M	Yellow, odor	Yellow, odor	36	34	35	0.03
C2, D2	6	6	8	1M	0.3M	Yellow, odor	Yellow, odor	42	41	42.5	0.02
C3, D3	4	8	8	1M	0.3M	Yellow, odor	Yellow, odor	42	44	43	0.02

				Concentrations
	# of drops	# of drops	# of drops	Stock Solution	Stock Solution	Reaction (after mixing in well)	Reaction (after mixing in well)	Reaction Time (seconds)			Reaction Rate (sec-1)
Well #	HCl	H2O	Na2S2O3	HCl	Na2S2O3	HCl	Na2S2O3	Trial 1	Trial 2	Average
C4, D4	8	0	12	1M	0.3M	Yellow, odor	Yellow, odor	21	23	22	0.05
C5, D5	8	6	6	1M	0.3M	Yellow, odor	Yellow, odor	47	50	48.5	0.02
C6, D6	8	8	4	1M	0.3M	Yellow, odor	Yellow, odor	80	90	85	0.01

Determine the Reaction Order for HCl using calculations described in the Background section. Show your work. Note that your answer will probably not be an even whole number as it is in the examples, so round to the nearest whole number.

B.    Determine the Reaction Order for Na₂S₂O₃ using calculations described in the Background section. Show your work. Note that your answer will probably not be an even whole number as it is in the examples.

Answer 1

Solution : -

Calculating order with respect to HCl

From the volume of the HCl used and the concentration of the stock solution of the HCl and volume of Na2S2O3 used we need to calculate new concentrations of the HCl in each trial after mixing the solutions.

So lets calculate the final concentration of the HCl

Trial 1

Drops of HCl = 12 drop

Drops of Na2S2O3 = 8 drop

Water drops = 0

Final concentration of HCl = initial concentration of HCl * initial volume / total volume

                                               = 1 M * 12 frop / (8 drop +12 drop)

                                                = 0.6 M

Trial 2

Drops of HCl = 6 drop

Drops of Na2S2O3 = 8 drop

Water drops = 6

Final concentration of HCl = initial concentration of HCl * initial volume / total volume

                                               = 1 M * 6 drop / ( 6 drop +6 drop+8 drop )

                                                = 0.3 M

Trial 3

Drops of HCl = 4 drop

Drops of Na2S2O3 = 8 drop

Water drops = 8

Final concentration of HCl = initial concentration of HCl * initial volume / total volume

                                               = 1 M * 4drop / ( 4 drop + 8 drop+8 drop )

                                                = 0.2 M

Now lets calculate the order of the HCl

Using the following set up

rate 1 / rate 2 = [HCl]1^m   / [HCl]2^m

0.03/0.02 = (0.6/0.3)^m

1.5 = 2^m

Log 1.5 = m * log 2

Log 1.5 / log 2 = m

0.6 = m

Therefore order with respect to HCl is half order

Now lets calculate the order with respect to Na2S2O3

Trial 1

Drops of HCl = 8 drop

Drops of Na2S2O3 = 12 drop

Water drops = 0

Final concentration of Na2S2O3 = initial concentration of Na2S2O3 * initial volume / total volume

                                               = 0.3 M * 12 frop / (8 drop +12 drop)

                                                = 0.18 M

Trial 2

Drops of HCl = 8 drop

Drops of Na2S2O3 = 6 drop

Water drops = 6

Final concentration of Na2S2O3 = initial concentration of Na2S2O3 * initial volume / total volume

                                               = 0.3 M * 6 drop / ( 6 drop +6 drop+8 drop )

                                                = 0.09 M

Trial 3

Drops of HCl = 8 drop

Drops of Na2S2O3 =4 drop

Water drops = 8

Final concentration of Na2S2O3= initial concentration of Na2S2O3* initial volume / total volume

                                               = 0.3 M * 4drop / ( 4 drop + 8 drop+8 drop )

                                                = 0.06 M

Now lets calculate the order with respect to Na2S2O3 by using the following set up

( rate 1 / rate 2) = [Na2S2O3]1ⁿ   / [Na2S2O3]2ⁿ

0.05/0.02 = (0.18/0.09)ⁿ

2.5 = 2ⁿ

Log 2.5 = n * log 2

Log 2.5/log 2 = n

n = 1.3

so we can round it n = 1

therefore order with respect to Na2S2O3 is first order

Therefore overall rate equation for the reaction is

Rate = K [ HCl]^1/2 [Na2S2O3]