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At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, following concentrations are present in the system: [SO2] = 0.10 M; [SO3] = 10. M; [O2] = 0.10 M. Is the mixture at equilibrium? If not at equilibrium, in which direction (as the equation is written), left to right or right to left, will the reaction proceed to reach equilibrium?

Respuesta :

CintiaSalazar

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

[tex]Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

[tex]Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

  • If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
  • If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
  • If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

[tex]Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }[/tex]

[tex]Q=\frac{10^{2} }{0.10^{2} *0.10}[/tex]

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

Q < Kc

The system is not in equilibrium and will evolve left to right to reach equilibrium.

dsdrajlin

When the concentrations are [SO₂] = 0.10 M; [SO₃] = 10. M; [O₂] = 0.10 M; the system is not at equilibrium and will proceed to the right to reach it.

Let's consider the following reaction.

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)   Kc = 4.3 × 10⁶

At a certain instant, following concentrations are present in the system:

  • [SO₂] = 0.10 M
  • [SO₃] = 10. M
  • [O₂] = 0.10 M

To determine if the system is at equilibrium, we need to calculate the reaction quotient (Q), which is the ratio of the product of concentrations of products to the product of the concentrations of reactants, all raised to their stoichiometric coefficients.

  • If Q < Kc, the system is not at equilibrium and will proceed to the right to reach it.
  • If Q = Kc, the system is at equilibrium.
  • If Q > Kc, the system is not at equilibrium and will proceed to the left to reach it.

The reaction quotient is:

[tex]Q = \frac{[SO_3]^{2} }{[SO_2]^{2} [O_2]} = \frac{(10.)^{2} }{(0.10)^{2} (0.10)} = 1.0 \times 10^{5}[/tex]

Since Q < Kc, the system is not at equilibrium and will proceed to the right to reach it.

When the concentrations are [SO₂] = 0.10 M; [SO₃] = 10. M; [O₂] = 0.10 M; the system is not at equilibrium and will proceed to the right to reach it.

Learn more: https://brainly.com/question/5144419

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