Question
In: Chemistry
The equilibrium constant, K, for the following reaction is 1.87×10-2 at 511 K. PCl5(g) <--------->>>PCl3(g) +...
The equilibrium constant, K, for the following reaction is 1.87×10-2 at 511 K.
PCl5(g) <--------->>>PCl3(g) + Cl2(g)
An equilibrium mixture of the three gases in a
13.1 L container at 511 K
contains 0.209 M PCl5,
6.24×10-2 M
PCl3 and
6.24×10-2 M
Cl2. What will be the concentrations of
the three gases once equilibrium has been reestablished, if the
equilibrium mixture is compressed at constant temperature to a
volume of 7.01 L?
| [PCl5] | = | M |
| [PCl3] | = | M |
| [Cl2] | = | M |
Solutions
Expert Solution
Volume is decreased from 13.1 L to 7.01 L at constant
temperature. The concentrations will increase by a factor of
These concentrations will change to restablish the equilibrium. To determine concentrations at new equilibrium, an ICE table is prepared.
 |
 |
 |
|
| Initial concentration (M) | 0.391 | 0.117 | 0.117 |
| Change in concentration (M) | -x | x | x |
| Equilibrium concentration (M) | 0.391-x | 0.117+x | 0.117+x |
The equilibrium constant
This is quadratic equation with solution
or
The value -0.224 is discarded as it will lead to negative value for chlorine concentration.
These are the equilibrium concentrations.
queen_honey_blossom answered 2 years agoRelated Solutions
The equilibrium constant, K, for the following reaction is 1.42×10-2 at 504 K. PCl5(g) --> PCl3(g)...
The equilibrium constant, K, for the following reaction is
1.42×10-2 at 504 K.
PCl5(g) --> PCl3(g) + Cl2(g)
An equilibrium mixture of the three gases in a 10.9 L container
at 504 K contains 0.279 M PCl5, 6.29×10-2 M PCl3 and 6.29×10-2 M
Cl2. What will be the concentrations of the three gases once
equilibrium has been reestablished, if the equilibrium mixture is
compressed at constant temperature to a volume of 4.95 L?
[PCl5] = M
[PCl3] = M
[Cl2]...
The equilibrium constant, K, for the following reaction is 3.16×10-2 at 525 K. PCl5(g) <----->PCl3(g) +...
The equilibrium constant, K, for the following reaction is
3.16×10-2 at 525
K.
PCl5(g)
<----->PCl3(g) +
Cl2(g)
An equilibrium mixture of the three gases in a
4.56 L container at 525 K
contains 0.243 M PCl5,
8.77×10-2 M
PCl3 and
8.77×10-2 M
Cl2. What will be the concentrations of
the three gases once equilibrium has been reestablished, if the
volume of the container is increased to 10.3
L?
[PCl5]
=
M
[PCl3]
=
M
[Cl2]
=
M
The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) +...
The equilibrium constant, K, for the following reaction is
1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture
of the three gases in a 1.00 L flask at 500 K contains 0.218 M
PCl5, 5.11×10-2 M PCl3 and 5.11×10-2 M Cl2. What will be the
concentrations of the three gases once equilibrium has been
reestablished, if 3.92×10-2 mol of PCl3(g) is added to the flask?
[PCl5] = M
[PCl3] = M
[Cl2] = M
The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) +...
The equilibrium constant, K, for the following reaction is
1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture
of the three gases in a 1.00 L flask at 500 K contains 0.200 M
PCl5, 4.90×10-2 M PCl3 and 4.90×10-2 M Cl2. What will be the
concentrations of the three gases once equilibrium has been
reestablished, if 3.12×10-2 mol of Cl2(g) is added to the
flask?
[PCl5] =
[PCl3] =
[Cl2] =
The equilibrium constant, Kc, for the following reaction is 1.05×10-3 at 446 K. PCl5(g) PCl3(g) +...
The equilibrium constant, Kc, for the following reaction is
1.05×10-3 at 446 K. PCl5(g) PCl3(g) + Cl2(g) When a sufficiently
large sample of PCl5(g) is introduced into an evacuated vessel at
446 K, the equilibrium concentration of Cl2(g) is found to be 0.498
M. Calculate the concentration of PCl5 in the equilibrium mixture.
____M
_______________________________________________________________________________________________________________
A student ran the following reaction in the laboratory at 689 K:
N2(g) + 3H2(g) 2NH3(g) When she introduced 4.06×10-2 moles of N2(g)
and 5.20×10-2...
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.
1) The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <-----...
1) The equilibrium constant, K, for the following reaction is
1.20×10-2 at 500 K. PCl5(g) <----- ------> PCl3(g) + Cl2(g)
An equilibrium mixture of the three gases in a 1.00 L flask at 500
K contains 0.201 M PCl5, 4.91×10-2 M PCl3 and 4.91×10-2 M Cl2. What
will be the concentrations of the three gases once equilibrium has
been reestablished, if 3.27×10-2 mol of Cl2(g) is added to the
flask? [PCl5] = M [PCl3] = M [Cl2] = M
2)...
1) The equilibrium constant, K, for the following reaction is 2.02×10-2 at 513 K. PCl5(g) <------...
1) The equilibrium constant, K, for the following reaction is
2.02×10-2 at 513 K. PCl5(g) <------ ------> PCl3(g) + Cl2(g)
An equilibrium mixture of the three gases in a 18.3 L container at
513 K contains 0.216 M PCl5, 6.60×10-2 M PCl3 and 6.60×10-2 M Cl2.
What will be the concentrations of the three gases once equilibrium
has been reestablished, if the equilibrium mixture is compressed at
constant temperature to a volume of 8.25 L?
[PCl5] = ______M
[PCl3] =...
1) The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <---...
1) The equilibrium constant, Kc, for the following reaction is
1.20×10-2 at 500 K. PCl5(g) <--- --->PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of reactant and products
when 0.279 moles of PCl5(g) are introduced into a 1.00 L vessel at
500 K. [PCl5] = M [PCl3] = M [Cl2] = M 2) The equilibrium constant,
Kc, for the following reaction is 55.6 at 698 K. H2 (g) + I2 (g)
<---- ----> 2 HI (g) Calculate the equilibrium concentrations
of...
1) The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) <---...
1) The equilibrium constant, Kc, for the following reaction is
1.20×10-2 at 500 K. PCl5(g) <--- --->PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of reactant and products
when 0.279 moles of PCl5(g) are introduced into a 1.00 L vessel at
500 K. [PCl5] = M [PCl3] = M [Cl2] = M 2) The equilibrium constant,
Kc, for the following reaction is 55.6 at 698 K. H2 (g) + I2 (g)
<---- ----> 2 HI (g) Calculate the equilibrium concentrations
of...
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