Question

1.

Consider the following generic reaction:

2A + B + C -> 2D +3E

From the following initial rate determinations, write the rate law for the reaction, determine the overall order for the reaction and calculate the rate constant, k.

Initial rate [A] [B] [C]

1.27x10^-4 Ms^-1 0.0125M 0.0125M 0.0125M

2.56x10^-4 Ms^-1 0.0250M   0.0125M 0.0125M

1.27x10^-4 Ms^-1 0.0125M   0.0250M 0.0125M

5.06x10^-4 Ms^-1 0.0125M 0.0125M 0.0250M

2. Consider the following reaction and its rate law:

2N2O5 -> 4NO2 + O2; rate = k[N2O5]

At 25 degrees Celsius, the half-life for this reaction is 96.3 minutes; calculate the initial N2O5 concentration, [N2O5]o, if [N2O5] = 1.80x10^-2 M after 60.0 minutes when the reaction proceeds at 25 degrees Celsius.

Answer 1

1 )

rate= k [A]^x [B]^y [C]^z

from data

1.27x10^-4 =k [0.0125]^x [0.0125]^y [0.0125]^z ------------->(1)

2.56x10^-4 =k [0.0250]^x [0.0125]^y [0.0125]^z ------------------> (2)

1.27x10^-4 =k [0.0125]^x [0.0250]^y [0.0125]^z -------------> (3)

5.06x10^-4 =k [0.0125]^x [0.0125]^y [0.0250]^z ----------------> (4)

by solving 1 and 2

x = 1

by solving 2 and 3

y = 1/2

by solving 3 and 4

z = 2

rate law = k [A]¹ [B]^1/2 [C]²

overall order = 1+ 1/2 + 2 = 3.5

from any of above equations we can solve rate constant k value

1.27x10^-4 =k [0.0125]^x [0.0125]^y [0.0125]^z ------------->(1)

1.27x10^-4 =k [0.0125]¹ [0.0125]^1/2 [0.0125]²

k = 581.6 mol^-7/2 lit^7/2 sec^1

rate constant = k = 581.6 mol^-7/2 lit^7/2 sec^1

2)

rate constant = k = 0.693 / half life

                           = 0.693 / 96.3

                            = 7.196 x 10^-3 min^-1

for first order

initila concentration = A

after time t concentration = At

k = (2.303 / t ) * log (A / At)

7.196 x 10^-3 = (2.303 / 60 ) log (A /1.8 x10^-2)

0.1875 = log A - log (1.8 x10^-2)

0.1875 - 1.745 = log A

-1.557 = logA

A = 10^-1.557

A = 0.0277

A = 2.77 x 10^-2 M

initial concentration of N2O2 = 2.77 x 10^-2 M