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Phosgene is a toxic gas prepared by the reaction of carbon
Chapter 16, Problem 28P(choose chapter or problem)
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].