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At 850 K, the equilibrium constant for the reaction 2 SO 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 SO 3 ( g ) is K c = 15 . If the given concentrations of the three gases are mixed, predict in which direction the net reaction will proceed toward equilibrium.

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ajeigbeibraheem

At 850 K, the equilibrium constant for the reaction 2 SO 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 SO 3 ( g ) is K c = 15 . If the given concentrations of the three gases are mixed, predict in which direction the net reaction will proceed toward equilibrium.

Given Concentration of the three gases at the three phases is given as:

LEFT                                   NO NET REACTION                  RIGHT

[SO₂] = 0.16 M                     [SO₂] = 0.20 M                          [SO₂] = 0.50 M

[O₂] = 0.20 M                      [O₂] =  0.60 M                           [O₂] =   0.60 M

[SO₃] = 0.50 M                    [SO₃] = 0.60 M                          [SO₃] = 0.15 M

Answer:

Therefore, since [tex]K__C_2} = K_c[/tex] ; NO NET REACTION and the reaction doesn't proceed to either left or right

Explanation:

The equation for the reaction is shown as:

2 SO₂ (g)    +      O₂ (g) ⇄    2 SO₃ (g)  

For the gas reaction at LEFT

[tex]K_c = 15[/tex]

[tex]K__C_1}= \frac{[S0_3]^2}{[SO_2]^2[O_2]}[/tex]

[tex]K__C_1}= \frac{[0.50]^2}{[0.16]^2[0.20]}[/tex]

[tex]K_C_1= 48.82[/tex]

Therefore, since [tex]K__C_1} > K_c[/tex] ; the reaction definitely proceeds to the left.

For the gas reaction at NO NET REACTION

[tex]K_c = 15[/tex]

[tex]K__C_2}= \frac{[S0_3]^2}{[SO_2]^2[O_2]}[/tex]

[tex]K__C_2}= \frac{[0.60]^2}{[0.20]^2[0.60]}[/tex]

[tex]K__C_2}=15[/tex]

Therefore, since [tex]K__C_2} = K_c[/tex] ; NO NET REACTION and the reaction doesn't proceed to either left or right.

For the gas reaction at RIGHT

[tex]K_c = 15[/tex]

[tex]K__C_3}= \frac{[S0_3]^2}{[SO_2]^2[O_2]}[/tex]

[tex]K__C_3}= \frac{[0.15]^2}{[0.50]^2[0.60]}[/tex]

[tex]K__C_3}=0.15[/tex]

Therefore, since [tex]K__C_3} < K_c[/tex] ; the reaction definitely proceeds to the RIGHT.

Answer:

[SO2] = 0.20 M

[O2] = 0.60 M

[SO3] = 0.60 M

Qc = 15

Qc = Kc, the reaction mixture is in equilibrium.

[SO2] = 0.14 M

[O2] = 0.20 M

[SO3] = 0.40 M

Qc = 40.8

Qc > Kc, the reaction will shift to the left.

[SO2] = 0.50 M

[O2] = 0.50 M

[SO3] = 0.10 M

Qc = 0.08

Qc < Kc, the reaction will shift to the right

Explanation:

At 850 K, the equilibrium constant for the reaction

2SO2 (g) + O2 (g)⇔ 2SO3 (g)

Kc = 15 . If the given concentrations of the three gases are mixed, predict in which direction the net reaction will proceed toward equilibrium.

[SO2] = 0.20 M

[O2] = 0.60 M

[SO3] = 0.60 M

[SO2] = 0.14 M

[O2] = 0.20 M

[SO3] = 0.40 M

[SO2] = 0.50 M

[O2] = 0.50 M

[SO3] = 0.10 M

Will each go to the left, right, or will there be no net reaction?

Step 1:

From the Law of Mass Action, the equilibrium constant for this reaction is defined as:

Kc = [SO3]² / [SO2]² [O2] = 15.0

If Qc = Kc, the reaction mixture is in equilibrium.

If Qc > Kc, the reaction will shift to the left.

If Qc < Kc, the reaction will shift to the right.

[SO2] = 0.20 M

[O2] = 0.60 M

[SO3] = 0.60 M

Qc = [SO3]² / [SO2]² [O2]

Qc = (0.60² )/(0.20² *0.60)

Qc = 15

Qc = Kc, the reaction mixture is in equilibrium.

[SO2] = 0.14 M

[O2] = 0.20 M

[SO3] = 0.40 M

Qc = [SO3]² / [SO2]² [O2]

Qc = (0.40² )/(0.14² *0.20)

Qc = 40.8

Qc > Kc, the reaction will shift to the left.

[SO2] = 0.50 M

[O2] = 0.50 M

[SO3] = 0.10 M

Qc = [SO3]² / [SO2]² [O2]

Qc = (0.10² )/(0.50² *0.50)

Qc = 0.08

Qc < Kc, the reaction will shift to the right

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