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At a certain temperature, 0.620 mol of SO3 is placed in a 3.00 L containter. `2SO3 (g)...

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bjackie | eNoter

Posted September 12, 2013 at 9:45 AM via web

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At a certain temperature, 0.620 mol of SO3 is placed in a 3.00 L containter.

`2SO3 (g) harr 2SO2 (g) + O2 (g)`

At equilibrium, 0.130 mol of O2 is present. Calculate Kc.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 12, 2013 at 9:57 AM (Answer #1)

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`2SO_3 harr 2SO_2+O_2`

Initial `[SO_3] = 0.62/3M = 0.2067M`

Let us say xM of `SO_3` has dissociated.

Mole ratio

`SO_3:SO_2 = 2:2 = 1:1`

`SO_3:O_2 = 2:1`

So xM of SO_2 and x/2M of O_2 will be produced.

Final concentrations

`[SO_3] = 0.2067-x`

`[SO_2] = x`

`[O_2] = x/2`

`K_C = ([SO_2]^2[O_2])/[SO_3]^2`

According to the given data x = 0.13

` K_C = ([0.13]^2[0.13/2])/[0.0767]^2`

`K_c = 0.1867`

So `K_c` for the reaction is 0.1867.

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