Answer:
A. Increasing the temperature will favor forward reaction and more CaCo3 formed.
B. More CaCo3 will be formed.
C. CaCo3 will decrease and more react ants formed.
D. Less CaCo3 will be formed.
E. Iridium is a catalyst so there is no effect
Explanation:
A. Temperature will increase because it's an endothermic reaction.
B. Adding Cao will favor forward reaction and more CaCo3 formed.
C. Removing methane, more react ants are formed and CaCo3 decreases.
D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.
The landscape region of Long Island is the Atlantic coastal plain.
Answer:
Answers are in the explanation.
Explanation:
<em>Given concentrations are:</em>
- <em>SO₂ = 0.20M O₂ = 0.60M SO₃ = 0.60M</em>
- <em>SO₂ = 0.14M O₂ = 0.10M SO₃ = 0.40M </em>
- <em>And SO₂ = 0.90M O₂ = 0.50M SO₃ = 0.10M</em>
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In the reaction:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
Kc is defined as:
Kc = 15 = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are equilbrium concentrations.</em>
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Also, you can define Q (Reaction quotient) as:
Q = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are ACTUAL concentrations.</em>
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If Q > Kc, the reaction will shift to the left until Q = Kc;
If Q < Kc, the reaction will shift to the right until Q = Kc
If Q = Kc, there is no net reaction because reaction would be en equilibrium.
Replacing with given concentrations:
- Q = [0.60M]² / [0.60M] [0.20M]² = 15; Q = Kc → No net reaction
- Q = [0.40M]² / [0.10M] [0.14M]² = 82; Q > Kc, → Reaction will shift to the left
- Q = [0.10M]² / [0.50M] [0.90M]² = 0.015; Q < Kc → Reaction will shift to the right
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