# How will the concentrations of the following molecules change given this information? The reaction: 2SO2(g)+o2)g)=2SO3(g) Keq=6.5 1.0 mol of S02,O2 and SO3 are placed in a 1.0 L container 2SO2 + O2 ------>2SO3   Keq = 6.5

1          1                 1

If x moles of SO2 reacted at equilibrium,

Amount of SO2 left   = 1- x

Then amount of O2 reacted = x/2

Therefore amount of O2 left = 1-x/2

Amount of SO3 produced = x

Amount of SO3 at equilibrium...

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2SO2 + O2 ------>2SO3   Keq = 6.5

1          1                 1

If x moles of SO2 reacted at equilibrium,

Amount of SO2 left   = 1- x

Then amount of O2 reacted = x/2

Therefore amount of O2 left = 1-x/2

Amount of SO3 produced = x

Amount of SO3 at equilibrium = 1+x

Since this is a 1 L container, we can use the number of moles as the concentration.

Keq = [SO3]^2/([SO2]^2 x [O2])

6.5 = (1+x)^2/((1-x)^2(1-x/2))

Therefore,

6.5 = 2(1+x)^2/((1-x)^2(2-x))

6.5((1-x)^2(2-x)) = 2(1+x)^2

6.5[(1-2x+x^2)(2-x)] = 2(1+2x+x^2)

6.5[2-5x+4x^2-x^3] = 2(1+2x+x^2)

13-32.5x+26x^2-6.5x^3 = 2+4x+2x^2

This gives,

6.5x^3-24x^2+36.5x-11 = 0

Solving this would give answers for x,

x =0.3914 or x =1.6504

But we know x<1.

Therefore x =0.3914

Therefore amount of SO2 reacted is 0.3914.

The equilibrium concentrations are,

SO2 = 1-0.3914  =0.6086 mol/L

O2 = 1 - 0.1957 = 0.8043 mol/L

SO3 = 1+0.3914 = 1.3914 mol/L

Approved by eNotes Editorial Team