Question

At a particular temperature, 12.0 moles of so3 is placed into a 3.0-l rigid container, and the so3 dissociates by the reaction 2so3(g) 34 2so2(g) 1 o2(g) at equilibrium, 3.0 moles of so2 is present. calculate k for this reaction.

Answer 1

            2 SO₃ --> 2 SO₂ + O₂
I             12             0          0
C           -2x           +2x      +x
---------------------------------------------
E         12-2x          2x         x

Since the moles of SO₂ at equilibrium is 3 mol, 2x = 3. Then, x = 1.5 mol. So, the amounts at equilibrium is:
SO₃: 12 - 2(1.5) = 9
SO₂: 2(1.5) = 3
O₂: 1.5

The formula for K basing on the stoichiometric reaction is:
K = [SO₂]²[O₂]/[SO₃]² 
where the unit used is conc in mol/L.

K = [3 mol/3 L]²[1.5 mol/3 L]/[9 mol/3 L]²
K = 0.0556

Answer 2

The value of k for the reaction is 0.055.

In the reaction, the $\rm SO_3$ dissociate to form $\rm SO_2$ and $\rm O_2$.

2 moles of sulfur trioxide will form 2 moles of sulfur dioxide and 1 mole of oxygen.

According to the ICE setup :

$\rm 2\; SO_3\;\rightarrow\;2\;SO_2\;+\;O_2$

4                  0           0     I

-2x               +2x        x     C

                     1                  E

Since there were 3 moles of  $\rm SO_2$ at the equilibrium.

2x = 3 moles.

x = 1.5 moles.

The initial amount of $\rm SO_3$ = 12 moles

At equilibrium, there has been a 2x reduction.

The amount of  $\rm SO_3$ = 12 - 2x

= 12 - 3

= 9 moles.

The amount of $\rm SO_2$ at equilibrium is 2x

= 3 moles.

The amount of $\rm O_2$ at equilibrium is x.

= 1.5 moles.

k can be calculated as: $\rm \dfrac{[SO_2]^2\;[O_2]}{[SO_3]^2}$

k = $\rm \dfrac{[3\;moles/\;3\;L]^2\;[1.5\;moles\;/\;3\;L]}{[9\;moles\;/3\;L]^2}$

k = 0.055.

The value of k for the reaction is 0.055.

For more information, refer to the link:

https://brainly.com/question/19671384