Question
In: Chemistry
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr,...
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl2(g) is injected into the mixture, and the total pressure jumps to 263.0 Torr (at the moment of mixing). The system then re-equilibrates. The appropriate chemical equation is PCl3(g)+Cl2(g) <---> PCl5(g)
Calculate the new partial pressures after equilibrium is reestablished. Answers in Torr
Solutions
queen_honey_blossom answered 2 years agoRelated Solutions
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr,...
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has
partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr,
respectively. A quantity of Cl2(g) is injected into the mixture,
and the total pressure jumps to 263.0 Torr (at the moment of
mixing). The system then re-equilibrates. The appropriate chemical
equation is
PCl3(g) + Cl2(g) <---> PCl5(g)
Calculate the new partial pressures after equilibrium is
reestablished.
P PCl3 = ? torr
P PCl2 = ? torr
P PCl5 = ? torr
An equalibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr,...
An equalibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has
partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr,
respectively. A quantity of Cl2(g) is injected into the mixture,
and the total pressure jumps to 263.0 Torr. The appropriate
chemical equation is PCl3(g) + Cl2(g) forward and reverse arrows
PCl5(g)
Calculate the new partial pressures after equilibrium is
reestablished.
Phosphorus pentachloride decomposes at higher temperatures. PCl5(g) ⇄ PCl3(g) + Cl2(g) An equilibrium mixture at some...
Phosphorus pentachloride decomposes at higher temperatures.
PCl5(g) ⇄ PCl3(g) + Cl2(g)
An equilibrium mixture at some temperature consists of
3.03 g PCl5, 208.23 g/mol
4.86 g PCl3, 137.33 g/mol
3.59 g Cl2, 70.91 g/mol
in a 1.00-L flask.
If you add 1.31 g of Cl2, how will the equilibrium be
affected and what will the concentration of PCl5 be when
equilibrium is reestablished?
shift left
shift right
no shift will occur
[PCl5] = mol/L
Phosphorus pentachloride decomposes at higher temperatures. PCl5(g) ⇄ PCl3(g) + Cl2(g) An equilibrium mixture at some...
Phosphorus pentachloride decomposes at higher
temperatures.
PCl5(g) ⇄ PCl3(g) + Cl2(g)
An equilibrium mixture at some temperature consists of
5.93 g PCl5, 208.23 g/mol
4.86 g PCl3, 137.33 g/mol
3.59 g Cl2, 70.91 g/mol
in a 1.00-L flask.
If you add 1.31 g of Cl2, how will the equilibrium be affected
and what will the concentration of PCl5 be when equilibrium is
reestablished?
shift left
shift right
no shift will occur
[PCl5] = mol/L
For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 4.0 at 228°C. If pure PCl5 is...
For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 4.0 at 228°C.
If pure PCl5 is placed in a 1.00-L container and allowed to come to
equilibrium, and the equilibrium concentration of PCl5(g) is 0.26
M, what is the equilibrium concentration of PCl3?
The equilibrium constant K=0.36 for the reaction PCl5 (g) PCl3 (g) +Cl2 (g). (a) Given...
The equilibrium constant K=0.36 for the reaction PCl5 (g)
PCl3 (g) +Cl2 (g). (a) Given that 2.0 g of PCl5 was initially
placed in the reaction chamber of volume 250 cm3, determine the
molar concentration in the mixture at equilibrium. (b) What is the
percentage of PCl5 decomposed.
PCl5 (g) <--> PCl3 (g) + Cl2 (g) K_p=0.0497 at 500 degrees Celsius. Determine the equilibrium...
PCl5 (g) <--> PCl3 (g) + Cl2 (g) K_p=0.0497 at 500 degrees
Celsius. Determine the equilibrium concentration(M) of all species
for a sealed gas cylinder charged with 1.53 atm of PCl5? Please
show all work.
Phosphorus pentachloride decomposes according to this equation. PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibrium...
Phosphorus pentachloride decomposes according to this equation.
PCl5(g) equilibrium reaction arrow PCl3(g) + Cl2(g) An equilibrium
mixture in a 5.00 L flask at 245°C contains 4.01 g of PCl5, 8.71 g
of PCl3, and 2.87 g of Cl2. How many grams of each will be found if
the mixture is transferred into a 2.00 L flask at the same
temperature? (Enter unrounded values.)
I'm pretty sure I did this problem right but my final answers
were only 3 decimal places...
The reaction: PCl5(g) <-> PCl3(g) + Cl2(g) has Kc=0.0900. A 0.1000 mol sammple of PCl5 is...
The reaction: PCl5(g) <-> PCl3(g) +
Cl2(g) has Kc=0.0900. A 0.1000 mol sammple of
PCl5 is placed in an empty 1.00 flask and the above
reaction is allowed to come to equilibrium at a certain temp. How
manny moles of PCl5, PCl3 and Cl2,
respectively, are present at equilibrium?
9. The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) PCl5(g) is 49 at 230°C....
9. The equilibrium constant Kc for the reaction PCl3(g) +
Cl2(g) PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is
added to 0.70 mol of Cl2 in a 1.00-L reaction vessel at 230°C, what
is the concentration of PCl3 when equilibrium has been
established?
A. 0.049 M
B. 0.11 M
C. 0.30 M
D. 0.59 M
E. 0.83 M
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