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A 1.00-L flask was filled with 2.00 moles of gaseous SO2 and 2.00 moles of gaseous NO2 and heated. After equilibrium was reached, it was found that 1.30 moles of gaseous NO was present. Assume that the reaction SO2 1 g2 1 NO2 1 g2mSO3 1 g2 1 NO1 g2 occurs under these conditions. Calculate the value of the equilibrium constant, K, for this reaction.

Respuesta :

jufsanabriasa

Answer:

K = 3.45

Explanation:

In the reaction:

SO₂(g) + NO₂(g) ⇄ SO₃(g) + NO(g)

And K is:

K = [SO₃] [NO] / [SO₂] [NO₂]

Where [] are concentrations in equilibrium, as volume is 1L, [] could be taken as moles.

Equilibrium concentrations are:

[SO₂] = 2.00 moles - X

[NO₂] = 2.00 moles - X

[SO₃] = X

[NO] = X

AS moles of NO are 1.3 moles, X = 1.3 moles

Replacing:

[SO₂] = 0.7 moles = 0.7M

[NO₂] = 0.7M

[SO₃] = 1.3M

[NO] = 1.3M

K = 1.3M² / 0.7M²

K = 3.45

pstnonsonjoku

The equilibrium constant is 3.5.

First of all we must obtain the equation of the reaction as follows;

SO₂(g) + NO₂(g) ⇄ SO₃(g) + NO(g)

Next we obtain the initial concentration of each specie;

SO₂(g) =  2.00 moles/1.00-L  = 2 M

NO₂(g)  =  2.00 moles/1.00-L  = 2 M

At equilibrium;

NO =  1.30 moles//1.00-L  = 1.3 M

The we set up the ICE table as follows;

          SO₂(g) + NO₂(g) ⇄ SO₃(g) + NO(g)

I          2             2                0          0

C        -x             -x               +x          +x

E       2 - x         2 - x            0 + x     0 + x

When x = 1.3

Concentration of each specie at equilibrium;

SO₂(g) = 2 - 1.3 = 0.7 M

NO₂(g) = 2 - 1.3 = 0.7 M

SO₃(g)  = 1.3 M

NO(g) M

K = (1.3)^2/(0.7)^2

K = 3.5

Learn more about equilibrium constant: https://brainly.com/question/17960050

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