1 answer

When the reaction r(aq) + OC1- (aq) → 10- (aq) + CI+ (aq) was studied the...

Question:

When the reaction r(aq) + OC1- (aq) → 10- (aq) + CI+ (aq) was studied the following data were obtained: [Ilo (mol/L) [OCI),
When the reaction r(aq) + OC1- (aq) → 10- (aq) + CI+ (aq) was studied the following data were obtained: [Ilo (mol/L) [OCI'), (mol/L) Initial Rate (mol/L's) 0.20 0.16 0.075 0.10 0.16 0.038 0.050 0.080 0.0094 0.40 0.080 0.075 What is the rate law? Rate = Calculate the value of the rate constant. k= L mol's Calculate the initial rate for an experiment where both I" and OCI" are initially present at 0.13 mol L. Rate = mol/L's

Answers

1)

see experiment 2 and 1:

[I-] doubles

[OCl-] is constant

rate doubles

so, order of I- is 1

see experiment 3 and 2:

[I-] doubles. This will double the rate as order of I- is 1.

[OCl-] doubles

rate becomes 4 times.

So, rate must have doubled because of OCl- alone

so, order of OCl- is 1

overall order = 1 + 1 = 2

Rate law is:

rate = k*[I-]*[OCl-]

Answer: rate = k[I-][OCl-]

b)

rate = k*[I-]*[OCl-]

Put values from 1st row of table in rate law

rate = k*[I-]*[OCl-]

7.5*10^-2 = k*0.2*0.16

k = 2.3438 M-1.s-1

Answer: 2.34

C)

rate = k*[I-]*[OCl-]

rate =2.3438*0.13*0.13

rate = 0.0396 M/s

Answer: 0.040 M/s

.

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