Question

Given the concentrations, calculate the equilibrium constant for this reaction: 2SO2(g) +O2(g)⇌2SO3(g) At equilibrium, the molar concentrations for reactants and products are found to be [SO2]=0.48 M, [O2]=0.40 M, and [SO3]=1.12 M. What is the equilibrium constant (Kc) for this reaction?

Answer 1

The equilibrium constant for this reactions is 13.61

Step-by-step explanation:

Step 1: Data given

Concentrations at the equilibrium are:

[SO2]=0.48 M

[O2]=0.40 M

[SO3]=1.12 M.

Step 2: The balanced equation

2SO2(g) +O2(g) ⇌ 2SO3(g)

Step 3: Calculate Kc

Kc = [SO3]²/[O2]*[SO2]²

Kc = (1.12²) / (0.40 * 0.48²)

Kc = 13.61

The equilibrium constant for this reactions is 13.61

Answer 2

Kc=13.61

Step-by-step explanation:

A chemical equilibrium is the situation in which the ratio between the amounts of reagents and products in a chemical reaction remains constant over time.

The equilibrium constant (Kc) is useful for the study of chemical equilibrium, being a constant that indicates whether the reaction is favored to the formation of products or to the formation of reactants.

Being:

aA + bB ⇔ cC + dD

where a, b, c, d are the stoichiometric coefficients of the reaction and A, B, C, D are the symbols or formulas of the different substances involved

Then:

`Kc=([C]^(c) *[D]^(d) )/([A]^(a) *[B]^(b) )`

where [] is the Molar concentration of each of the substances in equilibrium.

In this case, you know: 2 SO₂(g) +O₂(g)⇌2 SO₃(g)

So:
`Kc=([SO_(3) ]^(2) )/([SO_(2)] ^(2) *[O_(2) ])`

Being: [SO2]=0.48 M, [O2]=0.40 M, and [SO3]=1.12 M

`Kc=([1.12]^(2) )/([0.48] ^(2) *[0.40 ])`

Kc=13.61