Sulfur trioxide decomposes into oxygen & sulfur dioxide (all gases). To this 500 mL container, 396.5 grams of reactant is heated to 527oC. 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: Kc =
b) Calculate Kc
c) Calculate the Kp value.
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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