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Anna11 [10]
4 years ago
15

For the reaction 2 SO 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 SO 3 ( g ) the equilibrium constant is K c = 15 M − 1 at 850 K . Three sets

of concentrations are given for the three gases in the reaction. For each set, predict the direction in which the net reaction will proceed toward equilibrium. Left No net reaction Right (SO2)=0.16M (SO2)=0.20M (SO2)=0.50M (O2)=0.20M (O2)=0.60M (O2)=0.60M (SO3)=0.50M (SO3)=0.60M (SO3)=0.15M
Chemistry
1 answer:
Pachacha [2.7K]4 years ago
6 0

Answer:

2SO₂(g) + O₂(g) ↔ 2SO₃(g)     Kc = 15M⁻¹

The sets are:

(A) (SO₂)=0.16M; (O₂)=0.20M; (SO₃)=0.50M

(B) (SO₂)=0.20M; (O₂)=0.60M; (SO₃)=0.60M

(C) (SO₂)=0.50M; (O₂)=0.60M; (SO₃)=0.15M

The reaction quotient is

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

  • If Q < Kc, then the reaction will proceed towards the right (products)
  • If Q > Kc, then the reaction will proceed towards the left (reactants)
  • If Q = Kc, then we're at equilibrium.

Now we <u>calculate Q for each of the sets</u>:

(A) Q = 0.50² / (0.16²*0.20) = 48.8

Q > Kc. So the reaction will proceed towards the left.

(B) Q = 0.60² / (0.20²*0.60) = 15

Q = Kc. So the reaction is at equilibrium.

(C) Q = 0.15² / (0.50²*0.60) = 0.15

Q < Kc. So the reaction will proceed towards the right.

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The molarity of Na_2S_2O_3 is 0.108 M  

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