The chloronation of propene by HCl to 2-chloropropane in the gas phase HCl(g) +

Question

The chloronation of propene by HCl to 2-chloropropane in the gas phase

HCl(g) + CH3CH=CH2(g) CH3CHClCH3(g)

is assumed to have the following rate expression (in terms of partial pressures)

rate = kPmHClPnCH3CH=CH2

where m and n are integers. The reaction is thought to obey the following mechanism:

HCl + HCl <--> (HCl)2    K1

HCl + CH3CH=CH2 <--> complex K2

(HCl)2 + complex --> CH3CHClCH3 + 2HCl k3 (slow)

Verify that the mechanism is consistent with the reaction. Derive a rate expression from thismechanism and determine the values of m and n.

Explanation / Answer

we know that

rate is determined by the slowest reaction

so

rate = k3 x [ p(HC1)2] [ pComplex]

now

consider the equilibrium

HCl + CH3CH=CH2 ---> Complex

K2 = [pcomplex] / [ pHCl] [ pCH3CH=CH2]

[pcomplex ] = K2 x [ pHCl] [ pCH3CH=CH2]

now

consider the equilibrium

HCl + HCl ---> (HCl)2

K1 = [p(HCl)2] / [ pHCl] [pHCl]

[p(HCl)2] = K1 x [pHCl] [pHCl]

so

the rate is given by

rate = k3 x K1 x [pHCl] [pHCl] x K2 x [pHCl] [ pCH3CH=CH2]

rate = ( k3 x K1 x K2) [pHCl]^3 [ pCH3CH=CH2]

now

consider k3 x K1 x K2 = k

so

rate = k [pHCl]^3 [ pCH3CH=CH2]

so

the values of m and n are

m = 3

n = 1

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