At a certain temperature, 0.740 mol SO 3 0.740 mol SO3 is placed in a 2.50 L 2.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) 2SO3(g)↽−−⇀2SO2(g)+O2(g) At equilibrium, 0.180 mol O 2 0.180 mol O2 is present. Calculate K c

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kobenhavn kobenhavn · Answer 1 · 2020-02-04T05:45:46+01:00

Answer: [tex]K_c[/tex] = 0.0046

Explanation:

Moles of [tex]SO_3[/tex] = 0.740 mole

Volume of solution = 2.50 L

Initial concentration of [tex]SO_3[/tex] = [tex]\frac{moles}{volume}=\frac{0.740}{2.50}=0.296M[/tex]

The given balanced equilibrium reaction is,

[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]

Initial conc. 0.296 M 0M 0M

At eqm. conc. (0.296-2x)M (2x) M (x) M

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[SO_2]^2\times [O_2]}{[SO_3]^2}[/tex]

Given : moles of [tex]O_2[/tex] at equilibrium = 0.180M

Concentration of [tex]O_2[/tex] at equilibrium= [tex]\frac{moles}{volume}=\frac{0.180}{2.50}=0.072M[/tex]

x = 0.072 M

Now put all the given values in this expression, we get :

[tex]K=\frac{(2x)^2\times (x)^2}{(0.296-2x)^2}[/tex]

[tex]K=\frac{(2\times 0.072)^2\times (0.072)^2}{(0.296-2\times 0.072)^2}[/tex]

[tex]K=0.0046[/tex]

Thus the value of [tex]K_c[/tex] is 0.0046