1) The reaction I+ OCI.01-+ Cl. was studied in basic aqueous media. The data bel

Question

1) The reaction I+ OCI.01-+ Cl. was studied in basic aqueous media. The data below was collected at 25°C. (None of the solutions initially contained OI or CI.) Initial Composition of I (M) Initial Composition of OCI (M) Initial Composition of OH (M) Initial Rate (mol L1 s') 2 x 10-3 4 x 10 2 x 10 1.5 x 10-3 1.5 x 103 3 x 10 3 x 10 1.00 1.00 2.00 1.00 1.8 x 104 3.6 x 10 1.8 x 104 7.2 x 104 -3 -3 a) The rate law can be expressed as R [ITTOciT[OHJe. Enter the values of a, b, and c in that order, to two significant figures. b) Enter the value of k to two significant figures and no units. Next, enter the letter without the semicolon corresponding to the right units (a: s'; b: M's '; c: Ms c) Is the rate law consistent with the following mechanism? (fast, equilibrium) (slow) (fast, equilibrium)

Explanation / Answer

a)

keep [OCl-] amd [OH-] constant

now

increasing the conc of I- by 2 times increases the rate by 2 times

so

order with respect to I- is 1

a = 1

now

keep [I-] and [OH-] = constant

increasing the conc of OCl- by 2 times increased the rate by 2 times

so

order with respect to OCl- is 1

b = 1

now

the rate law becomes

rate = k [I-] [OCl-] [OH-]^c

now

keep [OCl-] constant for the last two readings

7.2 x 10-4 / 1.8 x 10-4 = [ 4 x 10-3 / 2 x 10-3] [ 1 /2]^c

c = -1

so

the rate law is

rate = k [I-] [OCl-] / [OH-]

b)

now

consider the last reading

7.2 x 10-4 = k [ 4 x 10-3 ] [3 x 10-3] / [1]

k = 60

so

the value of k is 60

and the unit is s-1

c)

consider the slow reaction

rate - k2 [I-] [HOCl]

now

consider the equilibrium

K1 = [HOCl ] [OH-] / [OCl-]

[HOCl ] = K1 [OCl-] / [OH-]

so

rate = k2 x [I-] x K1 x [OCl-] / [OH-]

rate = ( k2 x K1) [I-][OCl-] / [OH-]

consider

k2 x K1 = k

then

rate = k [I-] [OCl-] /[OH-]

so

yes , the rate law is consistent with the given mechanism

d)

we know that

ln (k2/k1) = (Ea /R) ( 1/T1 - 1/T2)

ln 160 = (Ea / 8.314) ( 1/298 - 1/ 398)

Ea = 50.045 x 1000

Ea = 50.045 kJ /mol

so

the activation energy is 50.045 kJ /mol

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