contestada

2SO(g) <-> 2SO2(g) + O2(g)

After the eq. represented above is established, some pure O2(g) is injected into the reaction vessel at a constant temperature. After equilibrium is reestablished, which of the following has a lower value compared to its value at the original equilibrium?

A. Keq for the reaction
B. The amount of SO3(g) in the reaction vessel
C. The amount of O2(g) in the reaction vessel.
D. The amount of SO2(g) in the reaction vessel.

Respuesta :

IthaloAbreu

The correct reaction is:

2SO₃(g) ⇄ 2SO₂(g) + O₂(g)

Answer:

D.

Explanation:

The equilibrium is reached when the velocity of the formation of products is equal to the velocity of the formation of the reactants, and it's can be represented by the constant of equilibrium Keq, which depends on the temperature.

For a generic reaction:

aA + bB ⇄ cC + dD Keq = [tex]\frac{[C]^c*[D]^d}{[A]^a*[B]^b}[/tex]

Then, for the reaction given,

Keq = ([O₂]*[SO₂]²)/[SO₃]²

When more O₂ is injected into the reaction, the equilibrium is disturbed, and to solve this, the O₂ must be consumed. So, the amount of SO₂ will be reduced, the amount of SO₃ will be increased, and the value of Keq must remain constant because there's no change in the temperature.

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