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allsm [11]
3 years ago
8

If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?

Chemistry
1 answer:
Fiesta28 [93]3 years ago
6 0
The given sentence is part of a longer question.

I found this question with the same sentence. So, I will help you using this question:

For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>).  If </span>Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs?


(a) Q = K<span>;   The reaction </span>is at equilibrium.
(b) Q < K<span>;   The reaction </span>will proceed to the right.
(c) Q > K<span>;   The reaction </span>will proceed to the left.

The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left, since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
Explanation:

Kp is the equilibrium constant in term of the partial pressures of the gases.

Q is the reaction quotient. It is a measure of the progress of a chemical reaction.

The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.

At equilibrium both Kp and Q are equal. Q = Kp

If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,

If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.

Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.

Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
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Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide
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Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

       According to question

                                2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\

Since 'NO' is the limiting reagent according to this ratio.

According to equation

         2 moles NO reacts to form 2 moles NO₂

So,  0.1855 moles NO give  = 0.1855 moles of NO₂

            Mass of 1 mole NO₂ = 46 g/mole

            Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

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4 years ago
An organic fertiliser in which plants are ploughed back into the soil​
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It was called MANURE

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3 years ago
A 1.26 m aqueous solution of an ionic compound with the formula mx2 has a boiling point of 101.63 ∘c. part a calculate the van't
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<span>Answer is: Van't Hoff factor (i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT =i · Kb · b.
Kb - </span><span>molal boiling point elevation constant</span><span> is 0.512°C/m.
b -  molality, moles of solute per kilogram of solvent.
b = 1.26 m.
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i = 1.63°C ÷ (0.512°C/m · 1.26 m).

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Answer:

12 Ethene gas, CH is completely burned in excess oxygen to form carbon dioxide and water

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CH 30, - 200, 2H,0

The table shows the bond energies involved in the reaction

bond

bond energy

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614

413

C-C

CHH

0 0

СО

495

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467

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A-954 kJ/mol

B-1010 kJ/mol

C-1313 kJ/mol

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Explanation:

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