Respuesta :
Answer:
a) Q<Kc the reaction will shift to the right
b) Q=Kc the reaction is at the equilibrium
c) Q>Kc the reaction will shift to the left
Explanation:
Step 1: Data given
temperature = 850 K
Kc = 15
When Q=K, the system is at equilibrium. Since the system is already at equilibrium, there will no shift to left or right happen.
When Q>K, it means we have more products than reactants. As reaction, the amount of products will decrease, and produce more reactants. in other words, the reaction will shift to the left.
When Q<K, it means we have more reactants than products. As reaction, the amount of reactants will decrease, and will produce more products. In other words, the reaction will shift to the right.
Step 2: The balanced equation
2SO2(g)+O2(g)⇌2SO3(g) Kc=15
Step 3: Calculate Q
Q = [SO3]²/[O2][SO2]²
a. [SO2] = 0.60 M, [O2] = 0.70 M, [SO3] = 0.16 M
Q = 0.16²/0.70*0.60²
Q =0.10
Q << Kc
we have more reactants than products. As reaction, the amount of reactants will decrease, and will produce more products. In other words, the reaction will shift to the right.
b. [SO2] = 0.20 M, [O2] = 0.60 M, [SO3] = 0.60 M
Q = 0.60²/0.60*0.20²
Q = 15
Q=K, the system is at equilibrium. Since the system is already at equilibrium, there will no shift to left or right happen.
c. [SO2] = 0.10 M, [O2] = 0.20 M, [SO3] = 0.40 M
Q = 0.40²/0.20*0.10²
Q = 80
Q>K, it means we have more products than reactants. As reaction, the amount of products will decrease, and produce more reactants. in other words, the reaction will shift to the left.
For the equation 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) Kc=15,
- When the concentrations are a. [SO₂] = 0.60 M, [O₂] = 0.70 M, [SO₃] = 0.16 M, the reaction will proceed to the right to attain the equilibrium.
- When the concentrations are b. [SO₂] = 0.20 M, [O₂] = 0.60 M, [SO₃] = 0.60 M, there will be no net reaction.
- When the concentrations are c. [SO₂] = 0.10 M, [O₂] = 0.20 M, [SO₃] = 0.40 M, the reaction will proceed to the left to attain the equilibrium.
Let's consider the following reaction at equilibrium.
2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) Kc=15
Given the concentrations of the 3 gases, we can predict in which direction the net reaction will proceed toward equilibrium calculating the reaction quotient (Q).
- If Q < Kc, the reaction will proceed to the right to attain the equilibrium.
- If Q = Kc, there will be no net reaction.
- If Q > Kc, the reaction will proceed to the left to attain the equilibrium.
a. [SO₂] = 0.60 M, [O₂] = 0.70 M, [SO₃] = 0.16 M
[tex]Q = \frac{[SO_3]^{2} }{[SO_2]^{2}[O_2] } = \frac{0.16^{2} }{0.60^{2}\times 0.70 } = 0.10[/tex]
Since Q < Kc, the reaction will proceed to the right to attain the equilibrium.
b. [SO₂] = 0.20 M, [O₂] = 0.60 M, [SO₃] = 0.60 M
[tex]Q = \frac{[SO_3]^{2} }{[SO_2]^{2}[O_2] } = \frac{0.60^{2} }{0.20^{2}\times 0.60 } = 15[/tex]
Since Q = Kc, there will be no net reaction.
c. [SO₂] = 0.10 M, [O₂] = 0.20 M, [SO₃] = 0.40 M
[tex]Q = \frac{[SO_3]^{2} }{[SO_2]^{2}[O_2] } = \frac{0.40^{2} }{0.10^{2}\times 0.20 } = 80[/tex]
Since Q > Kc, the reaction will proceed to the left to attain the equilibrium.
For the equation 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) Kc=15,
- When the concentrations are a. [SO₂] = 0.60 M, [O₂] = 0.70 M, [SO₃] = 0.16 M, the reaction will proceed to the right to attain the equilibrium.
- When the concentrations are b. [SO₂] = 0.20 M, [O₂] = 0.60 M, [SO₃] = 0.60 M, there will be no net reaction.
- When the concentrations are c. [SO₂] = 0.10 M, [O₂] = 0.20 M, [SO₃] = 0.40 M, the reaction will proceed to the left to attain the equilibrium.
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