1. PCl5(g)->/<-PCl3(g)+Cl2(g) Kd=.0420 The concentrations of the products at equ
ID: 845225 • Letter: 1
Question
1. PCl5(g)->/<-PCl3(g)+Cl2(g) Kd=.0420 The concentrations of the products at equilbrium are PCl3=.120M and Cl2=.290M. -What is the concentration of the reactatn PCL5 atequilbrium?
2.At a certain temperature the equilibrium constant Kc is equal to 53.3 H2(g)+I2(g) ->/<- 2HI(g Kc=53.3 At this temperature .300M of H2 and .30M of I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?
3.Note the following 2 reactions. A+2B ->/<- 2C where Kc=2.15 2C ->/<- D where Kc=.126 -Calculate the equilibrium constant(Kc) for the reaction D ->/<-A+2B Please note that in the previous question the ->/<- means reactions occurs both ways.
Explanation / Answer
(a)
Reaction equation:
PCl?(g) + Cl?(g) ? PCl?(g)
Equilibrium equation0
Kc = [PCl?] / ( [PCl?]?[Cl?] )
with
Kc = 49
The initial concentration for phosphorus trichloride and is:
c? = 0.12 mol M
The initial concentration for chlorine is:
c = 0.29 M
ICE Table:
............ [PCl?]........... [Cl?].......... [PCl?]
I............. 0.12.............. 0.29............. 0
C........... -x................. -x.............. +x
E......... 0.12 - x.......... 0.29 - x........... x
Hence,
49 = x / (0.12-x)(0.29-x)
<=>
49?(0.0348-0.41x+x2 ) = x
<=>
49?x
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