Question

Sulfur trioxide decomposes into oxygen & sulfur dioxide (all gases). To this 500 mL container, 396.5 grams of reactant is heated to 527^oC. Equilibrium is reestablished and 0.300 moles of SO₂ is in the container. Find concentrations of all substances at the reestablished equilibrium to complete any of your calculations.

a) Write the equilibrium expression for this reaction.     Expression: K_c =

b) Calculate K_c

c) Calculate the K_p value.

Answer 1

Answer : expression Kc = [SO2(g)] [O2(g)]/ [SO3(g)]

kc = 0.038 , Kp =2.12

explanation:

The decompositions of SO3 into SO2 and O2 is given below:-

SO3(g) <=======> SO2(g) + O2(g)

We have been given,

Volume of container = 500 ml = 500/1000 L = 0.5 L

Reactant ( SO3) amount = 396.5 g = 396.5 g/80 g/mol = 4.95 Mole.

At equilibrium, mole of SO2 = 0.300 mole.

Now , we constructing the ICE TABLE.

	SO3(g)	SO2(g)	O2(g)
Initial amount	4.95	0	0
Change	?	?	?
Equilibrium amount	?	0.300	?

Now, we calculate the change in mole of SO2 , and O2 by this formula.

Equilibrium amount= initial amount + change

Change = + 0.300 mole for SO2,O2, AND -0.300 mole for SO3.

then, the equilibrium amount of SO3(g) = 4.95-0.300 =4.65 mole

We constructing another ICE TABLE.

	SO3(g)	SO2(g)	O2(g)
Initial amount	4.95	0	0
Change	-0.300	+0.300	+0.300
Equilibrium amount	4.65	+0.300	+0.300

Writing expression for Kc,

Kc= [SO2(g)][O2(g)] / [SO3(g) ]

calculation of Kc:-

We shall first calculate the equilibrium concentration ,

Thus , at equilibrium, concentration of [SO3 ] = 4.65/0.5 =9.3 M

[SO2] = [O2] = 0.300/0.5 = 0.6 M

Kc = 0.6*0.6/9.3 = 0.36 /9.3 = 0.038

​​​​​​Kc= 0.038

Calculation of Kp,

we know the relation,

Kp= Kc(RT)^∆n

∆n = 2-1 =1, T = 273+396.5 = 669.5 K

Kp= 0.038* 0.0821*669.5

Kp=2.12

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