Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1. Write the rate law for the reaction between HCl and Na 2 S 2 O 3 . 2. Using t

ID: 878696 • Letter: 1

Question

1. Write the rate law for the reaction between HCl and Na2S2O3.

2. Using the rate law that you determined for the reaction, predict the rate of the reaction if the concentration of HCl was changed to 0.8M and the concentration of Na2S2O3is 0.12M.

# 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)

HCl

H2O

Na2S2O3

HCl

Na2S2O3

HCl

Na2S2O3

Trial 1

Trial 2

Average

8

0

12

1M

0.3M

.4

.18

26.6

27.9

27.3

.04

8

6

6

1M

0.3M

.4

.09

64.4

60.9

62.7

.02

8

8

4

1M

0.3M

.4

.06

87.3

82.6

84.9

.01

# 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)

HCl

H2O

Na2S2O3

HCl

Na2S2O3

HCl

Na2S2O3

Trial 1

Trial 2

Average

12

0

8

1M

0.3M

.6

.12

33.7

32.3

33

.03

6

6

8

1M

0.3M

.3

.12

42.6

39

40.8

.02

4

8

8

1M

0.3M

.2

.12

62.9

55.4

59.2

.017

# 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)

HCl

H2O

Na2S2O3

HCl

Na2S2O3

HCl

Na2S2O3

Trial 1

Trial 2

Average

8

0

12

1M

0.3M

.4

.18

26.6

27.9

27.3

.04

8

6

6

1M

0.3M

.4

.09

64.4

60.9

62.7

.02

8

8

4

1M

0.3M

.4

.06

87.3

82.6

84.9

.01

Explanation / Answer

Order with respect to Na2S2O3

r1 = k (0.4)x(0.18)y       --------(1)

r2 = k(0.4)x(0.09)y          --------(2)

Equation 1/Equation 2 = r1/r2 = (0.4)x(0.18)y/(0.4)x(0.09)y = (0.18)x/(0.09)x =

2 = 2y

y= 1

Order with respect to HCl

r3 = k (0.6)x(0.12)y ----------(3)

r4 = k(0.3)x(0.12)y -----------(4)

Equation 3/Equation 4 = r3/r4 = (0.6)x(0.12)y/(0.3)x(0.12)y = (0.6)x/(0.3)x =

1.5 = 2x

x = 0.585

Rate law of the reaction, rate = k[HCl]0.585[Na2S2O3]1

Rate constant k = rate/[HCl]0.585[Na2S2O3]1

= 0.03/(0.60.585 x 0.12) = 0.337

Rate of the reaction if the concentration of HCl was changed to 0.8M and the concentration of Na2S2O3 is 0.12M

Rate = k[HCl]0.585[Na2S2O3]1

= 0.337 x (0.8)0.585(0.12)1

=0.0355/s