An equivalent equilibrium chemistry problem 20

There are two conditions to meet the requirements of the question.

1) The molar ratio of initial SO2 to O2 = 2:1, otherwise, the reaction cannot be complete. So b=;

2) Due to the certain volume of the container, it is necessary to ensure that the volume fraction of each component is consistent with the original situation during equilibrium, and the total number of moles in the container must be consistent with the original during equilibrium, otherwise the pressure will change and cause the equilibrium to move. To ensure that the number of moles is consistent, it is only necessary to ensure that the number of molecules of S or O introduced into the substance is consistent with the original, and now consider the number of molecules of S, because a=, so c=.

The second is based on the effect of pressure on equilibrium according to the conditions of chemical equilibrium. I forgot what it was, and I looked it up myself.

First, there are two questions.

1 The concept of equilibrium.

2 Judgment of equivalence equilibrium.

In the problem, for the reversible reaction of the change of gas coefficient before and after the reaction, only (convert the given amount to one side according to the equation) that the initial amount of the person is equal, is the equivalent equilibrium, and the volume fraction (the quantity fraction of the substance) of each component of the equivalent equilibrium is equal.

The initial amount in the first equilibrium is 2mol SO2, and in the second equilibrium the SO2 b c is unknown.

How can you make the initial quantities equal in two equilibria? Think for yourself.

These two cases belong to the equivalence equilibrium, and the equivalence equilibrium is proportional.

Under constant temperature and constant capacity.

As long as the amount of the initial substance of each substance is equal, the same equilibrium can be established. All the corresponding equilibrium quantities of the two equilibrium (including the rate of forward and reverse reactions, the quantity fraction of the substances of each component, the mass concentration of the substance, the volume fraction of the gas, the mass fraction, etc.) are completely equal. A similar equilibrium can be established as long as the ratio of the initial quantities of each substance is equal.

That is, the relationship between the two balances is an equal relationship. The quantity fraction, gas volume fraction, mass fraction, and conversion rate of each reactant of each component in the two equilibrium correspond to equal; The rate of forward and reverse reactions in the two equilibria, the amount of the substance in equilibrium and the concentration of the substance in equilibrium are proportional.

In these two cases, although the amount of the initial substance of each substance is not equal, it is an equal volume reaction, and the pressure does not work, so it is an equivalent equilibrium.

That's the case, this reaction should be an equal volume reaction.

It is also a container with a fixed volume.

The temperature does not change. So the condition of equilibrium is satisfied.

In fact, you can also think of it as the k-value is constant.

First case: k=a2 (

The second case: k=(

You will find that the two ks have equal values.

Equal volume reduction.

Resulting in equal proportions.

There are generally three situations to do this kind of problem, and this one happens to be the constant temperature volume, so as long as the proportion is the same.

At a certain temperature, the sign that the reaction A2(g)+B2(G)2AB(g) reaches the equilibrium state is ( ).

A N mol of A2 per unit of time is generated while N mol of Ab is generated

b The total pressure in the container does not change with time.

c 2 mol of AB is generated per unit of time, and n mol of B2 is generated at the same time

d N mol of A2 is generated per unit of time, and n mol of B2 is generated at the same time

e The ratio of the amount of A2, B2 and AB in the container is 1 1 2.

f The state of the container when c(a2) = c(b2) = c(ab).

Analysis] The fundamental sign of the chemical equilibrium state is v positive = v inverse, the concentration or mass fraction and volume fraction of each component in the equilibrium mixture remain unchanged, as long as one of the above two points is present, it must be an equilibrium state, otherwise, it cannot be determined as an equilibrium state.

Item A, according to the ratio of the stoichiometric numbers in the chemical equation, the generation of N MoLab must consume molA2, that is, A2 consumes mol per unit of time, and at the same time generates N Molab, which obviously does not reach the equilibrium state.

Item B is described in terms of pressure, because this reaction is a reaction with the same number of gas molecules before and after the reaction, and the pressure is the same when equilibrium is reached or not, so the pressure cannot be used to indicate that the equilibrium state has been reached. On the other hand, if the reaction is not an equal-volume reaction and the total pressure in the vessel does not change with the extension of time (the temperature is constant), the reaction can be considered to be in equilibrium.

C, from the meaning of the title, when 2 mol Ab is generated per unit time, N molA2 and N molB2 are also consumed, so that the A2 consumed and the A2 generated in the same volume of container at the same time are both N mol, indicating that the reaction is in equilibrium.

In terms d, we only know how much A2 and N molB2 are generated per unit time, but we don't know how much A2 and how much B2 are consumed at the same time, so it is not possible to determine whether the equilibrium is reached.

The options e and f are described by the content of the components of the mixture, because when the chemical equilibrium state is reached, the concentration of each substance in the reaction mixture remains unchanged and is not equal, nor is the ratio of their molecular numbers necessarily equal to the ratio of the stoichiometric numbers of each substance in the chemical equation, so e and f are also not in line with the topic.

Answer: c

Topic: T, 3mola (g) and 1molb (g) into a closed container with a volume of 2L (volume unchanged) occurs as follows, 3a (g) + b (g) = 4c (g), the reaction reaches the equilibrium state (temperature unchanged) at 2min, at this time the container is left, keep the temperature and volume unchanged, the amount of the three substances in the original equilibrium mixture is adjusted as follows, the equilibrium can move positively

a.All halved bBoth are doubled cdare reduced.

Analysis: Explanation of DD, the first type: the ratio of the amount of the substance in the equilibrium state ABC is to increase and become the reaction to proceed in the opposite direction and in the same way as to the positive direction.

The second type is !!, when the equilibrium state is reached

At this time, if you do it according to the batch joining method: only overtime, although it is equivalent to the original balance!

But this amount lasts: a b c ; Balance should move in reverse!

Only when the added substance also conforms to the amount ratio of ABC substance is 3:1:1, it does not move!

It's so simple, first of all, this is not the equivalent equilibrium or a simple balance movement problem, the conditions of the three equilibrium systems are constant temperature and constant capacity, directly judged by the equilibrium theory, after adding the amount of reactants, the concentration of reactants is increased, the balance will move to the right, but the conversion rate of reactants will decrease, this is because the conversion rate = the amount of transformation has been compared with the initial amount, the amount of transformation after the balance moves to the right increases, but the starting amount increases more "weakened but not offset", so that the ratio decreases, That is, the conversion rate decreases.

Your misconception is that you don't understand the conversion rate formula and ignore the denominator getting bigger.

In fact, judging whether it is an equivalent equilibrium is very simple, you have not understood the substance, the so-called equivalence refers to the equivalence at the beginning, not the equivalence after equilibrium, and it is compared under the two systems, the judgment method: the percentage content of the equilibrium of the two systems is the same, and the equivalent equilibrium will not involve the moving "two systems on the study of the amount at the beginning".

The principle of balance movement is inseparable from any balance problem and is a tool to solve balance problems.

Quite simply, because there is only one reactant, and increasing the concentration of reactants is equivalent to increasing the pressure, and in the first reaction, the chemical equilibrium shifts to the left with increased pressure, and the conversion rate decreases. In the second reaction, the chemical equilibrium does not move when the pressure is increased, and the conversion rate remains unchanged. In the third reaction, the pressure increases, the chemical equilibrium shifts to the right, and the conversion rate increases.

Le Chatre's principle can be used at any time, but care must be taken in the analysis of conditions.

The volume remains the same, and the respective reactants are added, which is equivalent to pressurizing and increasing the concentration, but the problem is the change of conversion rate, so be more careful.

Assuming that the same reaction is carried out in two systems, and the conversion rate is a% when the equilibrium is reached, the two reaction systems are combined without changing the conditions, and all concentrations are increased, which has no effect on the conversion rate, but is equivalent to pressurization.

So for the first reaction to the left, decrease, the second to the same, the third to the right, to the right, to increase.

Similar questions can be considered like this:

Suppose the same amount of reactants is added to another identical vessel and equilibrium is reached at the same temperature, it is obvious that the equilibrium state of the two vessels is exactly the same, and the pressure, concentration, conversion rate, etc. of each substance are also the same.

Then merge, except for the accident that the volume is twice as large as the original, everything else is the same.

What is done now is to reduce the volume to the original volume.

1. Compression, balance shift left, so the conversion rate is reduced.

2. Compression, the balance does not move, so the conversion rate remains the same.

3. Compression, balance shift to the right, so the conversion rate increases.

1. Compression, balance shift left, so the conversion rate is reduced.

2. Compression, the balance does not move, so the conversion rate remains the same.

3. Compression, balance shift to the right, so the conversion rate increases.

The amount of product increases, but the conversion rate decreases.

2a(g) +b(g) =3c(g) +d(s) start: 2 1

Equilibrium: Since the volume is 1L, the concentrations of A, B, and C are ., respectively2（mol/l）k= ÷ =2。

a) 2a(g) +b(g) = 3c(g) + d(s), if c is at equilibrium, the amount of its substance is.

Start: 1 Balance: Can't, unreasonable).

b) 2a(g) +b(g) = 3c(g) +d(s) start: 3

Equilibrium: When the equilibrium is reached, the concentrations of a, b, and c are consistent with the equilibrium mentioned in the title, and k can be seen to be consistent, indicating that the same equilibrium can be reached).

c) 2a(g) +b(g) = 3c(g) +d(s) start: 2 1 0 1

Equilibrium: When the equilibrium is reached, the concentrations of a, b, and c are consistent with the equilibrium mentioned in the title, and k can be seen to be consistent, indicating that the same equilibrium can be reached).

d) 2a(g) +b(g) = 3c(g) +d(s) start: 0

Balance: Can't, unreasonable).

3C+, C: 3C+2D, D: 3C+, it turned out to be 3C+1D

Equivalence Balance··· It should be 3:1...

It is noted that the conditions are not the same, so the equilibrium constant of the reaction is not the same.

The H2 and BR2 added earlier were 1mol and 2mol, respectively, and then XMOLhbr survived

So as long as the matter added later is converted to H2 and Br2 and the ratio of the two is 1:2, the same balance can be maintained (I don't know if you can understand this sentence).

Therefore, after conversion, H2 is A+C 2mol and BR is B+C 2mol2(A+C 2)=B+C 2, and 4A+C=2B is obtained, so 1 is correct Option 2: There is no need for the amount of substance to be the same as the original, as long as the proportion of the initial material is the same.

Option 4: Corresponding to 1molH2 is xmolHBR corresponding to A+C 2molH2 is (A+C 2) XMOL HBR is known earlier as B=2A+C, so A+C 2 is not equal to (A+B+C) 2, so 4 is wrong.

So pick A

In a constant temperature and constant volume closed container, 1 and 2 are actually equivalent, because their equilibrium point is the same (i.e. k), so they equilibrate v and the same percentage of each component, so ac is right.

The reaction and reaction are actually equivalent equilibrium, which should be equal in volume of gas before and after them, and proportional to the cutting and feeding.

a.From the definition of equivalent equilibrium, it is known that the two equivalent equilibrium components have the same percentage content. So the percentages are the same. Right.

b.Wrong. c.

Average relative molecular weight = total mass of the total amount of matter. By the law of conservation of mass, the mass of the gas before and after is unchanged, and the volume of the gas before and after this reaction is equal, by Avergadro's law The quantity of the substance does not change, so the average relative molecular mass does not change. The average relative molecular mass is the same.

d. The total mass remains unchanged, and the total volume remains unchanged (for the volume of the container). So AC is chosen for equal density