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.

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.

queen_honey_blossom answered 1 year ago

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