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jok3333 [9.3K]
2 years ago
9

Why are pure liquids and solids not included in the reaction quotient or the equilibrium constant expression for a given heterog

eneous reaction
Chemistry
1 answer:
AfilCa [17]2 years ago
3 0

Pure liquids and solids at the equilibrium stage:

Pure solids and pure liquids are not included in expressions for the equilibrium constant of heterogeneous equilibria since their concentrations are constant across time. Therefore, including it in our equilibrium expression is not significant.

Equilibrium expression:

When a certain chemical reaction is in equilibrium, the rates of the forward and backward reactions are equal, which means that there is no net change in the concentrations of the reactants and products and that the reaction mixture's composition does not change.

The concentration of solid and liquid:

Solid or liquid concentration,

= No. of Moles/ Volume in L

= Mass/(Volume × molar mass)

= Density/ Molar mass

Pure solids and liquids have constant densities and molar masses at a constant temperature.

Because of this, they have constant molar concentrations, which means that they are not considered in the equilibrium constant for a heterogeneous reaction.

Learn more about the pure solids, liquids, and heterogenous mixture here,

brainly.com/question/1811148

#SPJ4

 

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Determine the value of the equilibrium constant, Kgoal, for the reaction
hodyreva [135]

Answer:

Part A

K_{goal}= 3.22\times 10 ^{-66}

Part B

K_{goal}= 2.26\times10^{-21}

Explanation:

Part A:

The given equation is not balanced. So, initially let's balance the equation by taking 24 moles of each of the reagent and NO.

24N₂ + 24H₂O ⇌ 24NO + 12N₂H₄

Now, simplify the equation by dividing both sides by 12. The final balanced equation is the following

2N₂ + 2H₂O ⇌ 2NO + N₂H₄

The above-balanced equation can be solved algebraically to obtain the required Kgoal value.

Adding given equations 1, 2 and 3 we obtain the required equation.

When the equations are added, the equilibrium constants of each equation are multiplied. Mathematically it can be represented as,

K_1 \times K_2 \times K_3 = K_{goal}

K_{goal}= 4.10 \times 10^{-31} \times 7.40 \times 10^{-26} \times 1.06 \times 10^{-10}

K_{goal}= 3.22\times 10 ^{-66}

Part B:

The required equation is balanced, Now

Let.

P₄(s)+6Cl₂(g) ⇌ 4PCl₃(g), K₁=2.00×10¹⁹ ------------------------------------ (a)

PCl₅(g) ⇌ PCl₃(g)+Cl₂(g), K₂=1.13×10⁻² --------------------------------------- (b)

By multiplying equation 2 by 4 and subtracting equation 1 from it, we get

4PCl₅(g) ⇌ P₄(s)+10Cl₂(g)

The Kgoal for the above equation is the product of four times K₂ and inverse K₁ according to the applied operation. Mathematically,

K_{goal}= 4K_2 \times \frac{1}{K_1}

K_{goal}= 4(1.13\times10^{-2}) \times \frac{1}{2.00\times10^{19}}

K_{goal}= 2.26\times10^{-21}

5 0
3 years ago
PROJECT: THE SCIENTIFIC METHOD
aleksley [76]

Answer:

how many it is you should give only one question

8 0
3 years ago
Read 2 more answers
A chemical equation is balanced when the number of each
erik [133]

Answer:

element

Explanation:

4 0
3 years ago
What is the oxidation state Ca(NO3)2
mylen [45]
Oxidation state is a number that is assigned to an element in a chemical combination. The number represents the number of electrons that an atom can gain, loses, or share when chemically bonding with an atom of another element. In this case the oxidation state or the oxidation number of Ca(NO3)2 is Zero, since its an electrically neutral compound. Thus the sum of oxidation number of calcium, nitrogen and oxygen is zero.
8 0
3 years ago
determine the maximum amount of NaN03 that was produced during the experiment. Explain how you determined the amount
Angelina_Jolie [31]

Answer:

9 moles of NaNO3 is obtained

Explanation:

The balanced chemical reaction equation for the reaction is;

Al(NO3)3 + 3NaCl-------> 3NaNO3 + AlCl3

Now, we have to determine the limiting reactant. The limiting reactant yields the least amount of NaNO3.

1 mole of Al(NO3)3 yields 3 moles of NaNO3

4 moles of Al(NO3)3 yields 4 * 3/1 = 12 moles of NaNO3

Also,

3 moles of NaCl yields 3 moles of NaNO3

9 moles of NaCl yields 9 * 3/3 = 9 moles of NaNO3

Hence, NaCl  is the limiting reactant and 9 moles of NaNO3 is obtained.

7 0
3 years ago
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