Question

The following reaction was performed in a sealed vessel at 727 ∘C :

H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.45M and [I2]=2.50M. The equilibrium concentration of I2 is 0.0500 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Express your answer numerically.

Answer 1

ICE table

	H2	I2	2HI
initial concentration	3.45 M	2.5 M	0
change in concentration	-x	-x	+2x
equilibrium concentration	3.45-x	2.5-x	2x

construct ICE chart given above.
In that fill the given datas.Since intially there was no product ,HI concentration is taken as
zero.

After some time x M have been reacted, to form 2x M of HI at equilibrium

equilibrium concentration of I2 = 2.5-x which is given as 0.05 M
thus calculate x first
0.05 M= 2.5-x
x= 2.5-0.05 = 2.45 M

Thus equilibrium concentrations of other species too can be calculated using this x value

[H2] = 3.45M- 2.45 M= 1M

[2HI] = 2*2.45 = 4.9 M
thus we got numbers for equilibrium concentrations of all species we can get equilibrium
constant using following expression

K= [2HI]^2/[H2][I2] = (4.9)^2/1*0.05 = 480.2

Equilibrium constant K= 480.2
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thank you