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ss7ja [257]
3 years ago
15

The following initial rate data are for the reaction of mercury(II) chloride with oxalate ion: 2 HgCl2 + C2O42- 2 Cl- + Hg2Cl2 +

2 CO2
Experiment [HgCl2] [C2O42-]o, M Initial Rate
1 0.124 0.115 1.61E-5
2 0.248 0.015 3.23E-5
3 0.124 0.229 6.40E-5
4 0.248 0.229 1.28E-4
1) Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n.
2) From these data, the rate constant is ____ M-2s-1?
Chemistry
1 answer:
Sever21 [200]3 years ago
3 0

Answer:

Explanation:

2 HgCl₂ + C₂O₄²⁻   =  2 Cl⁻ + Hg₂Cl₂ + 2CO₂

1 )

Rate of reaction = k[HgCl_2]^m[C_2O_4^{-2}]^n

             [HgCl₂]        [C₂O₄²⁻ ]           Rate  

1 .              .124             .115               1.61 x 10⁻⁵

2 .             .248             .115             3.23 x 10⁻⁵

3 .              .124             .229              6.4 x 10⁻⁵

4 .              .248             .229            1.28 x 10⁻⁴

comparing 1 and 3 , when concentration of HgCl₂ remains constant and concentration of C₂O₄²⁻  becomes twice , rate becomes 4 times so rate is proportional to square of concentration of C₂O₄²⁻  .

Hence n = 2

comparing 1 and 2 , when concentration of HgCl₂ becomes twice  and concentration of C₂O₄²⁻  remains constant  , rate becomes 2 times so rate is proportional to simply  concentration of C₂O₄²⁻  .

Hence m = 1

Putting the data of  1 in the rate equation found

 1.61 x 10⁻⁵ = k x .124 x  .115²

k = 11.3 x 10⁻⁴ M⁻² s⁻¹

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