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frozen [14]
3 years ago
15

The value of K at constant temperature depends on the amounts of reactants and products that are mixed together initially. A lar

ge value of K means the equilibrium position lies far to the right. For the following reaction: CaCO3(s) ⇌ CaO(s) + CO2(g) the [CaCO3] appears in the denominator of the equilibrium expression. For the following reaction: H2(g) + F2(g) ⇌ 2HF(g) the values of K and Kp are not the same. For a reaction with K >> 1, the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
Chemistry
1 answer:
Verdich [7]3 years ago
8 0

Answer:

a) K = [ CO2(g) ]

⇒ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.

b) Kp = K (RT)∧Δn

⇒ the values of K and Kp are not the same

c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.

Explanation:

a) CaCO3(s) ↔ CaO(s) + CO2(g)

⇒ K = [ CO2(g) ]

∴ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.

b) H2(g) + F2(g) ↔ 2 HF(g)

⇒ K = [ HF(g) ] ² / [ F2(g) ] * [ H2(g) ]

⇒ Kp = PHF² / PF2 * PH2

for ideal gas:

PV = RTn

⇒ P = n/V RT = [ ] RT

⇒ Kp = K (RT)∧Δn

⇒ the values of K and Kp are not the same.

c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.

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Why is true about electron dot diagrams
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Hello! Let's solve this!

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Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves
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This is an incomplete question, here is a complete question.

Manganese commonly occurs in nature as a mineral. The extraction of manganese from the carbonite mineral rhodochrosite, involves a two-step process. In the first step, manganese (II) carbonate and oxygen react to form manganese (IV) oxide and carbon dioxide:

2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)

In the second step, manganese (IV) oxide and aluminum react to form manganese and aluminum oxide:

3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)

Write the net chemical equation for the production of manganese from manganese (II) carbonate, oxygen and aluminum. Be sure your equation is balanced.

Answer : The net chemical equation for the production of manganese is:

6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)

Explanation :

The given two chemical reactions are:

(1) 2MnCO_3(s)+O_2(g)\rightarrow 2MnO_2(s)+2CO_2(g)

(2) 3MnO_2(s)+4Al(s)\rightarrow 3Mn(s)+2Al_2O_3(s)

First we are multiplying reaction 1 by 3, and reaction 2 by 2, we get:

(1)

(2) 6MnO_2(s)+8Al(s)\rightarrow 6Mn(s)+4Al_2O_3(s)

Now we are adding both the reactions, we get the overall chemical reaction.

6MnCO_3(s)+3O_2(g)+6MnO_2(s)+8Al(s)\rightarrow 6MnO_2(s)+6CO_2(g)+6Mn(s)+4Al_2O_3(s)

The  MnO_2 is common on both side, by cancelling it, we get:

The net chemical equation for the production of manganese is:

6MnCO_3(s)+3O_2(g)+8Al(s)\rightarrow 6CO_2(g)+6Mn(s)+4Al_2O_3(s)

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