Question

In: Chemistry

Chemical kinetics For the reaction 2A + B → C + D + 2E, data for...

Chemical kinetics

For the reaction 2A + B → C + D + 2E, data for a run with [A]0 = 800 mmol/L and [B]0 = 2.00 mmol/L are

t/ks

8

14

20

30

50

90

[B]/[B]0

0.836

.745

.680

.582

.452

.318

and data for a run with [A]0 = 600 mmol/L and [B]0 = 2.00 mmol/L are

t/ks

8

20

50

90

[B]/[B]0

0.901

0.787

0.598

0.453


Find the order with respect to each reactant (A and B) and the rate constant.

Solutions

Expert Solution

Reaction rate

d[A]/2dt = - d[B]/dt = k[A]x[B]y      

t/ks

0

8

14

20

30

50

90

Exp. 1

[B]/[B]0

1

0.836

.745

.680

.582

.452

.318

[B] used

0

0.328

0.51

0.64

0.836

1.096

1.364

Exp.2

[B]/[B]0

1

0.901

0.787

0.598

0.453

[B] used

0

0.198

0.426

0.804

1.094

Ratio exp1/exp2

1.656566

1.502347

1.363184

1.246801

OBSERVE THAT AT t=14ks [B] used is 0.51 mol/L .It means for A 2x0.51 mol A /L consumed that is MORE than available.

I will continue for a model solution.

Observe that :

[B]0 = 2.00 mmol/L is the same in both experiments.

Verify and observe that:

x is not 0 (the reaction rate depends on [A])

Observe that the ratio of [B] consumed in first and second experiments during the first time interval (8 ks) is 1.65 that aprox. Let’s assume that the concentrantions of [B] consumed during this interval are close to the initial reaction rates.

1.66 is close to (8 mmol/L / 6 mmol/L)2 = 1.332 = 1.77

It means that x=2

t/ks

0

8

14

20

30

50

90

Exp. 1

[B]/[B]0

1

0.836

.745

.680

.582

.452

.318

[B] used

0

0.328

0.51

0.64

0.836

1.096

1.364

d[B]/dt[A]2

0.5125

0.796875

1

1.30625

1.7125

2.13125

1/[ d[B]/dt[A]2 ]

1.95122

1.254902

1

0.76555

0.583942

0.469208

Observe in this table that 1/[ d[B]/dt[A]2 ] is in a liner relation to t. Then y=0

Reaction rate is now

d[A]/2dt = - d[B]/dt = k[A]2               

For the first time interval:

[A] consumed is 2x0.328 = 0.656 mol/L          

1/[A] = kt + 1/[Ao] as for a second reaction rate

1/0.656 = 8000 k + 1/0.800

1.524 = 8000 k + 1.25

0.274 M-1 = 8000 k

k = 0.000034 M-1 s-1

This is a provisional solution. I will revise it in the next 24 h.

Observe that :

[B]0 = 2.00 mmol/L is the same in both experiments.

Verify and observe that:

x is not 0 (the reaction rate depends on [A])

Observe that the ratio of [B] consumed in first and second experiments during the first time interval (8 ks) is 1.65 that aprox. Let’s assume that the concentrantions of [B] consumed during this interval are close to the initial reaction rates.

1.66 is close to (8 mmol/L / 6 mmol/L)2 = 1.332 = 1.77

It means that x=2

t/ks

0

8

14

20

30

50

90

Exp. 1

[B]/[B]0

1

0.836

.745

.680

.582

.452

.318

[B] used

0

0.328

0.51

0.64

0.836

1.096

1.364

d[B]/dt[A]2

0.5125

0.796875

1

1.30625

1.7125

2.13125

1/[ d[B]/dt[A]2 ]

1.95122

1.254902

1

0.76555

0.583942

0.469208

Observe in this table that 1/[ d[B]/dt[A]2 ] is in a liner relation to t. Then y=0

Reaction rate is now

d[A]/2dt = - d[B]/dt = k[A]2               

For the first time interval:

[A] consumed is 2x0.328 = 0.656 mol/L          

1/[A] = kt + 1/[Ao] as for a second reaction rate

1/0.656 = 8000 k + 1/0.800

1.524 = 8000 k + 1.25

0.274 M-1 = 8000 k

k = 0.000034 M-1 s-1

This is a provisional solution. I will revise it in the next 24 h.


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