Phosgene is a toxic gas prepared by the reaction of carbon

Chapter 16, Problem 28P

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QUESTION:

Problem 28P

Phosgene is a toxic gas prepared by the reaction of carbon monoxide with chlorine:

CO(g) + Cl2(g)→COCl2(g)

These data were obtained in a kinetics study of its formation:

Experiment

Initial [CO] (mol/L)

Initial [Cl2]  (mol/L)

Initial Rate (mol/L·s)

1

1.00

0.100

1.29 × 10?29

2

0.100

0.100

1.33 × 10?30

3

0.100

1.00

1.30 × 10?29

4

0.100

0.0100

1.32 × 10?31

(a)        Write the rate law for the formation of phosgene.

(b)        Calculate the average value of the rate constant.

Questions & Answers

QUESTION:

Problem 28P

Phosgene is a toxic gas prepared by the reaction of carbon monoxide with chlorine:

CO(g) + Cl2(g)→COCl2(g)

These data were obtained in a kinetics study of its formation:

Experiment

Initial [CO] (mol/L)

Initial [Cl2]  (mol/L)

Initial Rate (mol/L·s)

1

1.00

0.100

1.29 × 10?29

2

0.100

0.100

1.33 × 10?30

3

0.100

1.00

1.30 × 10?29

4

0.100

0.0100

1.32 × 10?31

(a)        Write the rate law for the formation of phosgene.

(b)        Calculate the average value of the rate constant.

ANSWER:

Solution 28P

Step 1

(a) We have to write the rate law for the formation of phosgene.

Phosgene is a toxic gas prepared by the reaction of carbon monoxide with chlorine:

CO(g) + Cl2(g)→COCl2(g)

These data were obtained in a kinetics study of its formation:

Experiment

Initial [CO] (mol/L)

Initial [Cl2]  (mol/L)

Initial Rate (mol/L·s)

1

1.00

0.100

1.29 × 1029

2

0.100

0.100

1.33 × 1030

3

0.100

1.00

1.30 × 1029

4

0.100

0.0100

1.32 × 1031

The rate law for the above chemical equation can be written as,

Rate = k [CO]m [Cl2]n

Order of reactant [CO]:-

Let us consider the concentration of [CO] is m.

1st we have to identify the reaction experiment in which [CO] changes but [Cl2] does not i.e in experiment 1 and 2.

Rate expt 1/ Rate expt 2 = ([CO expt 1] / [CO expt 2])m

1.29 10-29 mol L-1 min-1 / 1.3310-30 mol L-1 min-1 = (1.00 mol L-1/ 0.100 mol L-1)m

9.699 = (10.00)m

Taking log of both sides,

log (9.699) =log (10.00)m

log (9.699) = m log (10.00)

m = 0.9861 1

Thus the reaction is 1st order with respect to [CO].

Order of reactant [Cl2]:-

Let us consider the concentration of [Cl2] is n.

1st we have to identify the reaction experiment in which [Cl2] changes but [CO] does not i.e in experiment 2 and 3.

Rate expt 3/ Rate expt 2 = ([Cl2 expt 3] / [Cl2 expt 2])n

1.3010-29 mol L-1 min-1 / 1.3310-30 mol L-1 min-1 = (1.00 mol L-1/ 0.100 mol L-1)n

9.774= (10.00)n

Taking log of both sides,

log (9.774) =log (10.00)n

log (9.774) = n log (10.00)

n = 0.9901 = 1

The rate is 1st order with respect to [Cl2] 

Thus the rate law can be written as, Rate = k [CO] [Cl2].

 

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