High School Chemistry Balance Problems High School Chemistry Balance

The first is that there is an equal balance between the two that belongs to constant temperature and constant capacity.

It is required that the amount of service should be equal, so x must be 1 to achieve.

The second situation is that the two belong to constant temperature and constant pressure, as long as the ratio of the service amount is equal.

Because the container is closed, the volume of gas before and after the reaction is required to be equal, so x is equal to 4

According to the title, it can be obtained: , and, in the chemical equilibrium and 3mol a, 1mol b is equivalent, that is, x=1, and if the pressure does not change after the equilibrium is reached, it means that the coefficients of each substance on both sides of the chemical equation should be equal. This means x=4

x=2 is known from the equation a:b=3:1

and the volume fraction of c is a, then the ratio of the three is.

i.e. 3:1:2x=2

Do you understand equivalence balance? It's rarely talked about in books.

This is a difficult part of the exam.

Under constant capacitance, if it is an equal reaction, it is equivalent, that is, the reactants or products are all converted to one side, and the proportion of the first addition is OK. The same goes for constant pressure.

If it's not an equivalence reaction, it's equivalent, i.e., it's the same after conversion and the first time it was added.

Think about it for yourself. Make it yourself to understand.

It should be easy, if you want to get 120 in chemistry in the 3+2 exam, it should be easy to make it, and work your own. If not, you don't need to learn this.

C should be chosen. This problem requires the use of chemical equilibrium, Le Chatri's principle. I'll explain it here.

First, let's look at the characteristics of the reaction: an exothermic reaction with the same volume of gas. Secondly, when equilibrium is reached in the two containers of A and B, the degree of reaction in the two containers is the same. The concentration of hydrogen iodide is the same.

Now look at option a, while increasing the temperature, according to Le Chatri's principle, the reaction moves in the direction of absorbing heat, and the hydrogen iodide concentration decreases by the same amount at the same time.

Option b, add inert gas, since the reaction is a reaction in which the volume of the gas does not change, the pressure becomes larger, and there is no effect.

Option C, A lowers the temperature, according to Le Chatri's principle, the reaction proceeds in the direction of exothermy, so the reaction in A proceeds in the positive direction, so the concentration of hydrogen iodide in A is greater than that of B.

Option d, add reactants, according to Le Chatri's principle, the reaction proceeds in the positive direction, so the hydrogen iodide concentration increases the same.

To sum up, the answer is C.

In the reaction with the participation of gas, under the condition of constant capacity, inert gas is added, the pressure in the container increases, and the equilibrium moves in the direction of decreasing the amount of gas.

c This reaction is an exothermic reaction, and it is a reversible reaction, it is impossible to react completely, then at the beginning, A is lower than B, and lowering the temperature of A will make the reaction proceed in the direction of exothermic...

The two containers with constant temperature and constant capacity, and then look at the equation, it can be seen that the two are equivalent. 1a False. b Due to the constant capacitance, it does not affect the partial pressure of the reactants, but the total pressure is stronger. c Pair. d is also equivalent.

In this question, the balance is judged by the rate, and two things should be noted:

1. Whether it is a reactant or a product, it is okay, because the rate can be expressed by reactants or products, but there must be one positive and one negative.

2. It must meet the ratio of coefficients.

i.e., n2:h2:nh3=1:3:2

The options are all one positive and one negative, so look at the second point.

a, 3v positive (n2) = v inverse (h2) - i.e., n2:h2 = 1:3, conform, right.

b, v positive (n2) = v inverse (nh3) - false, not compliant.

c, 2v positive (h2) = 3v inverse (nh3) - i.e., h2: nh3 = 3:2, conform, right.

d,v positive (n2) = 3v inverse (h2) --n2:h2 = 3:1, false.

Namely, ac

When the equilibrium state is reached, the rate of the forward and reverse reactions is equal.

Options A and D are reactants on both sides and cannot be stated;

In b and c, it should be noted that the reaction rate is different with different coefficients, because the generation of 2molNH3 consumes 3molH2, so only V positive (H2) = 3 2V inverse (NH3) when the reaction reaches equilibrium.

So choose C

The N2 conversion rate remains the same.

The H2 conversion rate increases.

The conversion rate is the concentration of the conversion divided by the coal conversion, so it depends on the direction of the balance movement.

The reflection will be carried out in a positive way. Because the concentration of N2 increases, then there are more reactants, so the N2 conversion rate becomes smaller and H2 becomes larger when the positive reflection is carried out. Conversion rate = total amount of conversions Because N2 converts, but it converts only a part of what is added, so it becomes smaller.

The law is that whichever substance has more reacts to the lesser side.

25 (a) a chemical reaction 2a(g) b(g)+c(g) in a constant capacity closed container, carried out under three different conditions, in which the experiment is 800, the experiment is 850, the initial concentration of b and c is 0, and the concentration of reactant a (mol·l-1) changes with time (min) as shown in the figure:

1) In the experiment, the average reaction rate of a reaction within 20 min to 40 min was mol·l-1min-1.

2) Experimental Compared to Experiment I, the possible implicit reaction conditions are:

3) According to the comparison of experiments and experiments, it can be inferred that the reaction lowers the temperature and the equilibrium moves in the direction of the "positive" or "inverse") reaction, which is the "exothermic" or "endothermic") reaction.

4) Compared with the experiment, if the initial concentration of a in the experiment is changed to mol·l-1, and other conditions remain unchanged, the time required to reach equilibrium in the experiment (fill in "greater than", "equal to" or "less than").

Analysis, Answer:

Q1: Concentration change divided by time change and average reaction rate:

Q2: As you can see from the diagram, the experiment reaches equilibrium faster than the experiment, that is, the reaction rate becomes faster.

Since the concentration and temperature are constant, it is impossible to cause any change in concentration or temperature, so it is possible that a catalyst is added.

Question 3: As can be seen from the diagram, the temperature of the experiment is higher than that of the experiment, and the experiment reaches the equilibrium point first.

So it can be known that the temperature and height equilibrium move in the direction of the positive reaction, so the direction of the positive reaction is endothermic. Conversely, if the temperature is lowered, the reaction will want to move in the opposite direction.

Q4: From the reaction equation 2a(g) =b(g)+c(g), we can know that if the initial reaction concentration of a is reduced, the reaction rate will be slower, so it will take more time to reach equilibrium than the experiment.

Relative to the experiment, it is equivalent to the experiment after equilibrium, the amount of a is reduced, the equilibrium moves in the direction of the reverse reaction, and the time is consumed after the equilibrium is reached.

I feel less than, based on:

1. Assuming that the reaction starts from the beginning, and there are b, c, then the reaction time is less 2, and there is no b, c reduces the reverse reaction rate, and the equilibrium is faster.

In summary, t (< t(,c)

This problem is really not easy to explain, in fact, the concentration is large, the rate is fast, this situation is compared with the instantaneous rate, and when it comes to the time problem, it is not the instantaneous rate, but the average rate.

Therefore, if this problem is explained from the perspective of activating molecules, with the help of a hypothetical feeding state, i.e., a, b, c, this state is a point that A must pass through when it reaches the equilibrium state, so the time required to assume the feeding state must be shorter than the time required for a to reach the equilibrium state; And direct injection a, the concentration of forward activated molecules is the same as the concentration of positive activated molecules in the hypothetical feeding state, and the concentration of reverse activated molecules is smaller than that of the reverse activated molecules in the hypothetical feeding state, but with the advancement of the reaction, after this moment, the concentration of positive activated molecules in the hypothetical state will be larger than the concentration of positive activated molecules in the direct feeding state, so the time required to reach equilibrium is shorter or the hypothetical state is shorter, so whether the feeding is or not, the time required to reach the equilibrium state is longer than the time required for this hypothetical feeding state, So it's really hard to compare!

So I can only tell you that the concentration is large and the rate is fast, and the comparison is the instantaneous rate, and if it is a matter of time, it is the average rate, and with the help of the hypothetical state, if the conclusion is one large and one small, it can still be compared, but if the two major are obtained, it is not comparable. If you look at it from the perspective of average rate, the average rate = concentration change time, and the average rate ratio is larger; The concentration also varies greatly, and the length of time between the two cannot be compared.

I also turned to my math teacher to see if I could solve this problem with derivatives, but it still didn't work, and it could be a communication problem; Perhaps there is a need for a reintegration between chemical theory and mathematical methods.

I hope it helps.

In fact, add a reference reaction and you will understand.

1.Experiment I: 1mola, reaction volume of 1 L

2.Reference reaction: , reaction volume.

3.New experiment: , reaction volume 1 l

Obviously, 1 and 2 are a balanced reaction with exactly the same effect. It takes the same amount of time from the beginning to the time it reaches equilibrium. From your question, you can conclude that you have a good score in chemistry, and I believe you will understand, so I will not waste time here to explain.

Comparison between 2 and 3: the starting concentration is the same, the reaction volume of 3 increases, the concentration decreases, the pressure decreases, and the reaction speed slows down. It takes longer to reach equilibrium.

3 vs. 1: The time it takes to reach equilibrium increases.

The conversion of 2 3 CO to CO2 indicates that there is 2 3 mol CO2 produced when 1 molco and 1 molH2O (g) are added and 1 mol co is added to the equivalent of molh2.

If only 1molCo molH2 is added in case 2, 2 3molCO2 will remain, because the total volume before and after the reaction is unchanged, so the volume fraction of CO2 is 2 3 divided by 2= Now 1molH2O is added to shift the reaction to the right and the CO2 conversion rate is increased.

Only d is greater than d, so choose d

It's not counted.

If it is an equivalent equilibrium, then the volume fraction of CO2 is , and the equivalent equilibrium that matches the conditions in a fixed volume is 1molCo and 2molH2O, and there are 1molH2O reactants relative to 1molHC2O in the original question, then the reaction proceeds in the positive direction on the original basis, so the volume fraction of CO2 is greater than in this option is .

Without the volume when filling 4mol a and 1mol b, this problem cannot be solved. If the volume is 10l, the answer is.

Now assume that the volume is known and is 10 liters.

The solution is as follows: by filling 4mol a and 1mol b, the concentration of d at equilibrium is , and the volume is 10l, then the equilibrium coefficient of the reaction is [d] [a] = (

When filling c and d, the equilibrium is reached, and d is to consider whether the equilibrium is restricted by c at the same time, and this question does not need to be considered, because there is still c remaining in the system at the time of equilibrium).

For the concentration of d to be, the volume is required. (The change in volume does not affect the movement of the equilibrium).

From 4a(g) b(s) =3c(s) 4d(g): 1. The volume of gas remains unchanged before and after the reaction; From (in the initial container to refill C, D, the temperature remains the same) knows: 2, the temperature of the two cases does not change, the balance does not move, the conversion rate is the same, so the same relationship.

Can be shortened.

For the following equilibrium systems, the equilibrium shifts in the direction of the reverse reaction when the conditions are changed.

a c+CO2 2CO (with soda lime added),Soda lime consumes CO2 and moves in reverse of equilibrium

B Ca(OH)2Ca2+ +2OH- (HCl),Consumes oh- to balance the positive movement

C H2S H+ + Hs- (NaOH added),Consumes H+ to balance positive movement

D 3NO2+H2O 2HNO3+NO (air penetrated).I consume の and move the balance positively

Select A for this question

a) Soda lime consumes reactant CO

b) HCl depletes the product OH

c) NaOH consumes the product H

d) The air (o in ) consumes the product no