Which of the following enthalpies of reaction would the reaction represented by the diagram have?
1 answer:
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Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
- On addition of reactant at equilibrium shifts the equilibrium in forward direction.
- On addition of product at equilibrium shifts the equilibrium in backward direction.
- On removal of reactant at equilibrium shifts the equilibrium in backward direction.
- On removal of product at equilibrium shifts the equilibrium in forward direction.
Reactants = A , B
Product = C
1. Increase A
On increasing the amount of A at equilibrium will shift the equilibrium in forward or rightward direction.
2. Increase B
On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.
3. Increase C
On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.
4. Decease A
On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.
5. Decease B
On decreasing the amount of B at equilibrium will shift the equilibrium in backward or leftward direction.
6. Decease C
On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.
7. Double A and Halve B
Equilibrium constant of the reaction = K
On doubling A and halving B, equilibrium constant of the reaction = K'
The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.
8. Double both B and C
Equilibrium constant of the reaction = K
On doubling B and C, equilibrium constant of the reaction = K'
K' = 2 K
The value of equilibrium constant K' is double the K, which means that product is increasing which means that equilibrium will shift in backward or leftward direction.
Answer:
3,78 mL of 12,0wt% H₂SO₄
Explanation:
The equilibrium in water is:
H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)
The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] =
Thus, you need to add:
[H⁺] = = <em>5,31x10⁻⁸ M</em>
The total volume of the pool is:
7,00 m × 18,0 m ×1,50 m = 189 m³ ≡ 189000 L
Thus, moles of H⁺ you need to add are:
5,31x10⁻⁸ M × 189000 L = 1,00<em>x10⁻² moles of H⁺</em>
These moles comes from
H₂SO₄ → 2H⁺ +SO₄²⁻
Thus:
1,00x10⁻² moles of H⁺ × = <em>5,00x10⁻³ moles of H₂SO₄</em>
These moles comes from:
5,00x10⁻³ moles of H₂SO₄ × ×
×
= <em>3,78 mL of 12,0wt% H₂SO₄</em>
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I hope it helps!