Ask for help with questions about chemical equilibrium conversion rates. 20

The addition of NH3 is equivalent to increasing the pressure of the entire container.

When you increase the pressure of the whole container, you will cause the equilibrium to move in the opposite direction, that is, to the left.

As the pressure increases, the rate of the reaction increases in the forward and reverse directions, which is what you call the concentration.

But this is increasing the rate of the reaction, not equilibrium.

It will cause the equilibrium to go in the opposite direction, which is to generate NH3

Conversion rate = converted original generated NH3 that is, NH3 was not converted, so the conversion rate decreased.

To distinguish between equilibrium and velocity, you have to take the equilibrium and velocity problems and compare them to each other.

Look at the notes more often you read the notes the more you will have a new understanding.

Remember 1. Increasing the concentration of one reactant (referring to forward progress) will increase the conversion rate of another reactant.

Self-conversion rate decreases.

2. The conversion rate of pressurized (also refers to forward shift) reactants is increased Of course, the reverse is the opposite.

When you encounter a conversion rate question, you think about 2 questions, and after doing a few questions, you will remember it.

You have to look at the issues separately.

If the pressure increases, the concentration increases, but that's the rate of reaction, not the equilibrium.

It's just a reversible reaction that increases the concentration without taking into account the pressure.

The conversion rate of PCL5 itself decreases.

I hope my answer will make you understand.

You look at how the pressure and concentration on the note affect the equilibrium movement.

There is how it affects the rate.

No matter how much others talk about, it's better to think about it yourself, as long as you figure it out, you'll be fine.

Because of the addition of amolar NH3, it is impossible for this amole substance to be completely transformed, and the pressure of the normal system increases rather than the pressure of the product. If there is a b mol NH3, the equilibrium conversion rate is x a (after the reaction to x), and now, after the reaction, it is converted to y (a + b). The former is greater than the latter.

Hehe: That's a good question.

The first problem is actually like this: because NH3, N2, and H2 are all gases, they are mixed in a container, so when NH3 is added, the pressure becomes larger because the volume does not change, so it moves to the side with the smaller coefficient. Instead of only increasing the pressure of NH3, how can the pressure of a substance be increased only when the gas in a container is mixed?

As for the second reaction, it's the reverse reaction, but I don't know why the answer is a positive reaction, and I don't know why.

But your last question, I remember, is that the main factor is the pressure, in fact, the two are not contradictory, hehe, if the pressure and concentration change at the same time, first pay attention to the pressure, the pressure has changed, but I hope you can ask the teacher.

After all, the teacher is the most professional.

First of all, it is necessary to clarify the essence of pressure and concentration to change the equilibrium, and the essence of the balance is to change the equilibrium through concentration, so there is no contradiction between the two.

This problem can be regarded as the volume of the original container first changes to do isobaric expansion, and then equilibrates and then compresses to the original volume, because it is a reaction of volume change before and after, in order to establish equivalent equilibrium, the reactants must be proportional, and because there is only one reactant, no matter how much a is equal to, it is a proportional change, becoming the original (1+a) times.

Therefore, the equivalent equilibrium is established at this time (at this time, the volume is larger than the original volume, the percentage content of each component remains unchanged, and the next step is the compression process) In the compression process, each concentration becomes larger, and according to the principle of Le Chatelier, the reaction moves in reverse.

Key point: The idea of this question is to find the intermediate equivalent equilibrium and constancy, and then use the principle of Le Chatletle, you are wrong that it is not a balance system at the moment when you just join, and the principle of Le Chatre is not valid!

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For example, if 10 users see the results of a search promotion, 5 of them click on a promotion result and are redirected to the target URL, and then 2 of them have subsequent conversions. Then hail blind, the conversion rate of this promotion result is (2 5) 100%=40%.

The conversion rate is the core of whether the company can make a profit in the end, and improving the conversion rate is the result of the comprehensive operational strength.

Branding. Popularity, credibility, word-of-mouth, and positioning all determine the conversion rate.

Service features: User experience.

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1. Reaction properties, k standard value of reaction;

2. Reaction conditions, heating, pressurization, etc.

3. The concentration of the product;

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Problem description: Why for the reaction ma (gas) + nb (gas) Analysis: Assuming that A and B each have a return to 1mol, then after a period of reaction, the amount of both substances decreases, and then 1mol B is added, then the A participating in the reaction is less than 1mol, so the B added later cannot reach the first conversion amount, and the conversion rate is not reduced; And A responds partially, and the conversion rate increases.

It is equivalent to a small amount of A pinning down B's reaction.

If there is only one reactant, there is no pull-in effect. Then the conversion rate of the gas depends on the change in pressure.

For example, 2Hi = H2 + I2, the pressure before and after the reaction is unchanged, and then Hi is added, and the balance does not move, the conversion rate remains unchanged;

PCL5=PCL3+CL2, after the reaction, the pressure becomes larger, then the gas increases and the equilibrium moves in the opposite direction, and then Pcl5 is added, and the conversion rate decreases.

2NO2 N2O4, the pressure becomes smaller after the reaction, the gas increases and the equilibrium moves forward, and NO2 is added, and the conversion rate increases.

The relationship between the direction of chemical equilibrium movement and the conversion rate of reactants is:

1. If the temperature is changed, if the equilibrium shifts to the direction of the positive rapid hunger reaction, the conversion rate of the reactants must increase. On the contrary, if the equilibrium shifts to the opposite reaction direction, the conversion rate of the reactants must be reduced.

2. Change the pressure, if the equilibrium moves to the direction of positive reaction, the conversion rate of reactants must increase. On the contrary, if the equilibrium shifts in the direction of the opposite reaction, the conversion rate of the reactants must be reduced.

The reaction equilibrium of a gas depends on the (gas) volume of reactants and products.

1. The volume of the reactant is 2, the volume of the product is "2+1", that is, 3, the volume of the product is greater than the volume of the reactant, so when the reaction is balanced, the addition of gas will promote the reverse reaction, thereby reducing the conversion rate (your question must be "I added the reactant, how to reduce it", remember, it is the conversion rate rather than the conversion amount, and the addition of reactants must increase the reaction amount and the reaction rate is not necessarily);

In the same way, it is still a problem of conversion rate, increasing SO2 will provide O2 with more opportunities to contact SO2 (to put it more deeply, it is called a relative increase in reaction area), and the conversion rate of O2 will increase. What about SO2: to explain by the reaction rate formula, the amount added to the denominator increases, while the amount of reaction in the molecule increases because there is not enough O2 to react with it (i.e., the reaction area is relatively reduced).

2. This is a common problem, the test point is that N2O4 is solid, so the reaction equilibrium no longer looks at the "volume" of reactants and products (referring to the volume of gas), so no matter which reactant on both sides of the increase = on both sides, the corresponding product will increase.

3. Here Hi and H2 are both gases, and I2 is solid, so the volume of gas before and after the reaction remains unchanged, increasing H2, the relative reaction area of H2 increases, so the conversion rate increases, and the relative reaction area of H2 decreases, so the conversion rate decreases (the same as 1); Because there is no change in the volume of the gas before and after the reaction, the conversion rate remains the same, while the conversion amount of Hi must increase.

In short, we must distinguish the relationship between conversion volume and conversion rate in our consciousness.

If you still have any questions, you can add them.

(1).Constant volume, adding SO3 is equivalent to pressurization. The reversible reaction proceeds in the direction of a decrease in the number of gas molecules. So the SO3 conversion rate decreases.

If the concentration of SO2 increases, the effective collision between O2 and SO2 increases, and the conversion rate of O2 increases. The SO2 and O2 reactions are reversible, and the increased SO2 cannot reach the previous conversion rate. **Turnover rate = total reactants of the reaction).

2) The volume remains unchanged, that is, the volume is fixed. Increasing the NO2 concentration is equivalent to pressurization, and the reversible reaction proceeds in the direction of decreasing the number of gas molecules. As mentioned above, the increased reactants cannot reach the previous reaction rate, but the attraction effect of pressure is great, so the conversion rate of total NO2 becomes larger.

Increasing the concentration of N2O4 is also equivalent to pressurization. The reversible reaction also proceeds in the direction of a decrease in the number of molecules of the gas. So the NO2 conversion rate becomes larger.

3), Ibid. (1), the second question. The increase in Hi conversion rate does not change, because it is a reaction with the same number of gas molecules. So increasing hi has no effect on reaction speed.

That's all I can say, I hope it can be understood and helpful to you.

1 Because this reaction is a reaction in which the gas participates, your reaction volume is 2 volumes, and after the reaction is 3 volumes, when you increase the concentration of sulfur trioxide, it is equivalent to the pressure of your reaction system increases, and when the pressure increases, the reaction proceeds in the direction of decreasing pressure, so for reversible reactions, the conversion rate of sulfur dioxide decreases.

2 Because they are all gases, when the concentration of reactants increases, the concentration of the system increases, and the reaction proceeds in the direction of volume reduction, so the concentration of nitrogen dioxide increases, the conversion rate of nitrogen dioxide increases, and vice versa.

3 This reaction is a reaction with constant volume.

When the concentration of hydrogen increases, it means that there is an excess of hydrogen, so the conversion rate of iodine vapor increases, and the concentration of hydrogen iodide increases, which is equivalent to the reaction volume of the whole system does not change, so the conversion rate does not change.

The process of increasing the concentration is equivalent to increasing the pressure in the direction of the reaction, and the direction of the reaction is judged according to the direction of the change in pressure.

This kind of problem is mainly considered from the volume change, and if you understand this knowledge point thoroughly, you can basically get points for this piece of knowledge.

1. For the reversible reaction of reactants with a variety of substances, the conversion rate of this substance will decrease if one substance is added. This is because, the conversion rate = the amount of the transformed substance The total amount of the substance, now this substance is added, so the amount of the transformed substance increases, but the total amount of the substance increases more, so the ratio decreases, that is, the conversion rate decreases (this is only qualitative analysis, it is difficult to quantitatively analyze).

Increasing the SO2 concentration reduces the SO2 conversion rate (analyzed above); The increase in the conversion rate of O2 is due to the fact that after the addition of SO2, the equilibrium shifts in a positive direction, so the amount of substances converted by O2 increases, while the total amount of substances in O2 remains unchanged, so the conversion rate increases.

22NO2(G) = reversible = N2O4

This reaction is special because there is only one reactant, which does not satisfy the situation mentioned in question 1. A different method of analysis is to be employed. (Set Intermediate Method).

To increase the concentration of NO2, we can first set the intermediate state, that is, perform an isobaric transformation (increase in the volume of the container), so that the conversion rate of NO2 does not change (the conversion rate of isobaric change does not change), and then reduce the volume of the container, for this reaction, decreasing the volume of the container will make the equilibrium move in a positive direction, so NO2 will continue to convert to N2O4, so the conversion rate of NO2 increases.

The same is true for increasing the concentration of N2O4 and increasing the conversion rate of No2.

3 This reaction is a special case of problem 2 (the most special case of a reversible reaction, there is only one reactant, and the volume of gas before and after the reaction is unchanged).

It is still possible to use the intermediate state method, but the volume of the intermediate container is the same as the volume of the final container, so the conversion rate remains the same.

If you still have any questions, you can hi me.

Because after adding one of the reactants, the reaction proceeds in the direction of reducing this reactant, that is, to the right, so the other reactants are also reduced, the products are increased, and the conversion rate is increased.

For this reactant, the conversion rate of the newly added part is lower than that of the reaction before the addition is lower, so the overall conversion rate is reduced.

After adding a reactant, the chemical equilibrium moves to the direction of the positive reaction, and the conversion rate of the rest of the substances increases, but due to the principle of Le Chattelier, the proportion of the increased part of the reaction is smaller than the proportion of the original part of the reaction due to the increase in the initial amount of the added substance, so in general, the conversion rate of the increased substance is reduced.

For example, a + b --c + d

Let's say a conversion rate is 50% 2 2 0 0

A b 2mol each, then transform 1 1 1 1

a + b --c + d

Add 1mol of a at this point, then, 3 2 0 0

The most ideal state is 1mol conversion rate conversion.

Also 50%, then.

Then, we know that this is a chemical equilibrium, and the added 1mol cannot be completely converted, so the conversion rate of A is less than 50%, so even if BCD is not converted, it is at least more than the original 1mol, so the conversion rate will also increase.