The equilibrium constant for the reaction of chlorine gas with phosphorus trichloride to form phosphorus pentachloride is 33 at 250°C. If an experiment is initiated with concentrations of 0.050 M PCI3 and, 0.015 M Cl2 what are the equilibrium concentrations of all three gases? Cl2 (8) + PCl, (g) = PCl (8)

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Check Your Understanding
This reaction can also be considered as the decomposition of PCI5 (g), for which the equilibrium
constant is 0.030 (at the same temperature). If you begin an experiment with 0.040 M PCI5 (g),
what are the expected equilibrium concentrations of all species?
Answer
[CI2] = [PCI3] = 0.036 M, [PC15] = 0.004 M
Transcribed Image Text:Check Your Understanding This reaction can also be considered as the decomposition of PCI5 (g), for which the equilibrium constant is 0.030 (at the same temperature). If you begin an experiment with 0.040 M PCI5 (g), what are the expected equilibrium concentrations of all species? Answer [CI2] = [PCI3] = 0.036 M, [PC15] = 0.004 M
Example Problem 12.7
The equilibrium constant for the reaction of chlorine gas with phosphorus trichloride to form
phosphorus pentachloride is 33 at 250°C. If an experiment is initiated with concentrations of
0.050 M PC13 and, 0.015 M Cl2 what are the equilibrium concentrations of all three gases?
Cl2 (g) + PCI, (g) = PCl (g)
It may seem disconcerting to get two numerical solutions to a problem like this. How do we decide
which root is correct? The resulting equilibrium concentrations must be physically reasonable.
Among other things, that means that none of them can be negative. But if we chose the first value,
æ = 0.087, we would get negative concentrations for both Cl2 and PCl3. So the correct root must be
the second one, æ = 0.0088, which gives us the following results:
[PCI5] = x = 0.0088 M
[ClL] = 0.015 – x = 0.015 – 0.0088 = 0.006 M
%3D
PCI3] = 0.050 – x = 0.050 – 0.0088 = 0.041 M
Analyze Your Answer
One check we can always do for this type of equilibrium problem is to put the final values back into
the equilibrium expression and make sure that they produce a reasonable value for K. In this case,
that gives K = (0.0o88)/(0.006)(0.041) = 36. This number is not quite 33, but because all of our data
have only two significant figures, we should not expect an exact match. Being this close to the
original value of K is reassuring and means that we have probably done the arithmetic correctly.
Transcribed Image Text:Example Problem 12.7 The equilibrium constant for the reaction of chlorine gas with phosphorus trichloride to form phosphorus pentachloride is 33 at 250°C. If an experiment is initiated with concentrations of 0.050 M PC13 and, 0.015 M Cl2 what are the equilibrium concentrations of all three gases? Cl2 (g) + PCI, (g) = PCl (g) It may seem disconcerting to get two numerical solutions to a problem like this. How do we decide which root is correct? The resulting equilibrium concentrations must be physically reasonable. Among other things, that means that none of them can be negative. But if we chose the first value, æ = 0.087, we would get negative concentrations for both Cl2 and PCl3. So the correct root must be the second one, æ = 0.0088, which gives us the following results: [PCI5] = x = 0.0088 M [ClL] = 0.015 – x = 0.015 – 0.0088 = 0.006 M %3D PCI3] = 0.050 – x = 0.050 – 0.0088 = 0.041 M Analyze Your Answer One check we can always do for this type of equilibrium problem is to put the final values back into the equilibrium expression and make sure that they produce a reasonable value for K. In this case, that gives K = (0.0o88)/(0.006)(0.041) = 36. This number is not quite 33, but because all of our data have only two significant figures, we should not expect an exact match. Being this close to the original value of K is reassuring and means that we have probably done the arithmetic correctly.
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