1.) Determine the concentration of each component (C2) in the final mixture for each run (refer to Table 4-1 for
the total volume (V2) of the mixture, and the initial concentration (C1) and volume (V1) of each component).
SAMPLE CALCULATION:
C1V1=C2V2
[I2] =
(0.00120 𝑀)(2.00 𝑚𝐿)
25 𝑚𝐿
[I2] = 0.000096 M
Table 1. Final mixture of acetone, I2, and HCl for each run.
Run
1
2
3
4
5
Final Concentration (mol L-1)
C3H6O
I2
HCl
0.0312
0.0624
0.0312
0.0312
0.0468
0.000096
0.000096
0.000096
0.000192
0.000144
0.0792
0.0792
0.1584
0.0792
0.1188
2. Calculate for the reaction rate per run using the concentration of iodine and time elapsed.
SAMPLE CALCULATION:
Reaction rate = [I2] In the final mixture/time elapsed
Reaction rate =
0.000096 𝑀
258 𝑠
Reaction rate = 0.0000003721
Table 2. Reaction rate of Iodine (I2) for each run.
Run
Reaction rate (mol L-1 sec-1)
1
2
3
4
5
0.000000372
0.000000738
0.000000727
0.000000376
0.000000791
3. Determine the reaction order with respect to I2. Also determine the reaction order with respect to C3H6O and
the reaction order with respect to HCl. Round off your answers to the nearest whole number. Refer to the sample
calculation in the Prelab SG I as your guide.
SAMPLE CALCULATION:
C3H6O [Run 1 & Run 2]:
𝑟1 𝑘(0.0312 M)x (0.000096 M)y (0.0792)z
=
𝑟2 𝑘(0.0624 M)x (0.000096 M)y (0.0792)z
0.000000372 M/s 𝑘(0.0312 M)x (0.000096 M)y (0.0792)z
=
0.000000738 M/s 𝑘(0.0624 M)x (0.000096 M)y (0.0792)z
0.000000372 (0.0312)x
=
0.000000738 (0.0624)x
(0.0312)
0.000000372
=(
)
(0.0624)
0.000000738
𝑥
(0.0312)
0.000000372
𝑙𝑛 (
) = 𝑥 𝑙𝑛 (
)
(0.0624)
0.000000738
𝑥=
0.000000372
𝑙𝑛 (0.000000738)
(0.0312)
𝑙𝑛 (
)
(0.0624)
𝑥 = 0.988859442 ≈ 1
I2 [Run 1 & Run 4]:
𝑟1 𝑘(0.0312 M)x (0.000096 M)y (0.0792 M)z
=
𝑟4 𝑘(0.0312 M)x (0.000192 M)y (0.0792 M)z
0.000000372 M/s 𝑘(0.0312 M)x (0.000096 M)y (0.0792 M)z
=
0.000000376 M/s 𝑘(0.0312 M)x (0.000192 M)y (0.0792 M)z
0.000000372 (0.000096)y
=
0.000000376 (0.000192)y
0.000000372
0.000096 M 𝑦
=(
)
0.000000376
0.000192 M
𝑙𝑛 (
0.000000372
0.000096 M
) = 𝑦 𝑙𝑛 (
)
0.000000376
0.000192 M
𝑦 = 0.016873819 ≈ 0
HCl [Run 1 & Run 3]:
𝑟1 𝑘(0.0312 M)x (0.000096 M)y (0.0792 M)z
=
𝑟3 𝑘(0.0312 M)x (0.000096 M)y (0.1584 M)z
0.000000372 M/s 𝑘(0.0312 M)x (0.000096 M)y (0.0792 M)z
=
0.000000727 M/s 𝑘(0.0312 M)x (0.000096 M)y (0.1584 M)z
0.000000372 (0.0792)z
=
(0.1584)z
0.000000727
0.000000372
0.0792 𝑧
=(
)
0.000000727
0.1584
0.000000372
0.0792
𝑙𝑛 (
) = 𝑧 ln (
)
0.000000727
0.1584
𝑧 = 0.966833136 ≈ 1
4. Calculate for the rate constant for each run using the reaction orders obtained (use rounded off values of
reaction orders).
Run
I2 Reaction rate (mol L-1 sec-1)
Rate Constant ( L sec-1)
1
2
3
4
5
0.000000372
0.000000738
0.000000727
0.000000376
0.000000791
0.00015058
0.00014942
0.00014716
0.00015235
0.00014231
5. Rate
Rate = 1.48 ∙ 10−4 𝐿2 𝑚𝑜𝑙 −2 sec −1 [𝐶3 𝐻6 𝑂]1 [𝐼2 ]0 [𝐻𝑙𝐶]1