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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

1 Answer

5 votes

Answer:
K_c = 0.0046

Step-by-step explanation:

Moles of
SO_3 = 0.740 mole

Volume of solution = 2.50 L

Initial concentration of
SO_3 =
(moles)/(volume)=(0.740)/(2.50)=0.296M

The given balanced equilibrium reaction is,


2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)

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,


K_c=([SO_2]^2* [O_2])/([SO_3]^2)

Given : moles of
O_2 at equilibrium = 0.180M

Concentration of
O_2 at equilibrium=
(moles)/(volume)=(0.180)/(2.50)=0.072M

x = 0.072 M

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


K=((2x)^2* (x)^2)/((0.296-2x)^2)


K=((2* 0.072)^2* (0.072)^2)/((0.296-2* 0.072)^2)


K=0.0046

Thus the value of
K_c is 0.0046

User Sathish Murugesan
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