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H 2 ( g ) + I 2 ( g ) 2 HI ( g ) K c = 50.0 at 400 °C

The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with [H 2 ] = [I 2 ] = 0.221 M and [HI] = 1.563 M is at equilibrium; for this mixture, Q c = K c = 50.0. If H 2 is introduced into the system so quickly that its concentration doubles before it begins to react (new [H 2 ] = 0.442 M ), the reaction will shift so that a new equilibrium is reached, at which [H 2 ] = 0.374 M , [I 2 ] = 0.153 M , and [HI] = 1.692 M . This gives:

Q c = [ HI ] 2 [ H 2 ] [ I 2 ] = ( 1.692 ) 2 ( 0.374 ) ( 0.153 ) = 50.0 = K c

We have stressed this system by introducing additional H 2 . The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess H 2 , reducing the amount of uncombined I 2 , and forming additional HI.

Effect of change in pressure on equilibrium

Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for K c ) or partial pressure (for K P ). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.

As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Châtelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.

Consider what happens when we increase the pressure on a system in which NO, O 2 , and NO 2 are at equilibrium:

2 NO ( g ) + O 2 ( g ) 2 NO 2 ( g )

The formation of additional amounts of NO 2 decreases the total number of molecules in the system because each time two molecules of NO 2 form, a total of three molecules of NO and O 2 are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of NO 2 into NO and O 2 , which tends to restore the pressure.

Now consider this reaction:

N 2 ( g ) + O 2 ( g ) 2 NO ( g )

Questions & Answers

so is HCl ionic compound
Honest Reply
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scientific study of structure of substances and of the way that they react with other substances
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Fathmat Reply
A measure of the amount of matter contained by a given volume. The ratio of one quantity to that of another quantity.
mass divided by volume i.e. g/cm^3
what's molarity?
Okpaka Reply
the concentration of a substance in solution, expressed as the number moles of solute per litre of solution
Please help me solve this question. A is a solution of 0.995mol/dm cube hydrochloride acid. B was prepared by diluting 10cm cube of a saturated solution of sodium trioxocarbonate (iv) to 100cm cube at room temperature. Assuming that 21.50cm cube of A reacted with 25cm cube of B. Calculate: i. Concentration of solution B in mol/dm cube. ii.Solubility of sodium trioxocarbonate (iv) at room temperature. Equation of the reaction: Na2Co3 +2HCL------> 2NaCL +H2O +CO2.
Mercy Reply
I don't know whether it's ok or not, but the answers I got are: I. 0.428 mol/dm^3 II. 4.54g per 100 g of water
In the first one, I first found out the amount of HCl in mol using moles=concentration x volume. Then I checked the ratio of Na2CO3 to HCl, which is 0.5 to 1. Therefore the moles of Na2CO3 will be half of HCl. Using the amount in moles and the volume as 25 cm^3, I reached my answer!
In the second one, it says that 10 cm^3 has saturated Na2CO3 solution. Using the concentration we found in previous answer, I found out the moles present in 10cm^3. After that, using mass= moles x RFM, I got it's mass. As for the mass of water, we know 1 cm^3 gives 1g, so 10 cm^3 gives 10g.....
Using solubility= mass of solute/mass of solvent x 100, we reach the answer.
Note: we will not use the volume of solution to be 100 cm^3, because then the solution will be dilute.
plz do correct me if I'm wrong!! ☺️
is like the answer is 900
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Traceyo Reply
Write the resonance hybrids of furan and thiophene
Hydrolysis of CH3CH2NO2 with 85% H2SO4 gives? 2/Acetaldehyde is oxidised with potassium dichromate and sulphuric acid gives 3/ When benzyl alcohol is oxidised with KMnO4, the product obtained ? 4/ Benzyl chloride is oxidised with KOH4, the 5/
Hydrolysis of CH3CH2NO2 with 85% H2SO4 gives?
Define reduction in term of loss or gain of oxygen or hydrogen give an example.
CuO + Mg → Cu + MgO removing oxygen is reduction. here Mg is reducing agent(loss of electrons)
reduction >> reduc(+)ion mean (+)ion reduced mean electron gained by (+)ion (+)ion means H(+).
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Practice Key Terms 3

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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