A high school chemistry balance problem. High School Chemistry Balance Issues.

No. 1 fast. It's a question of equivalence. Since only carbon dioxide is a gas in the reactants, the percentage of CO in the product must also be the same.

After the previous consideration, pay attention to the effect of pressure on chemical equilibrium, because the amount of gaseous substances added to No. 2 is 4mol, it is stronger than No. 1 pressure, because the reaction is in the direction of the increase of gas volume, so the degree of equilibrium moving in the opposite direction is larger, so it takes more time to reach the same percentage content. To sum up, the speed of No. 1.

The time to reach the equilibrium of the reaction is related to the temperature and pressure, and the concentration of the reactants, if you want to generate 2 parts of carbon monoxide as mentioned in the title, does it mean 2mol? If so, it must be faster at No. 2, because No. 2 has high air pressure and high concentration of reactants, and it must be 2mol carbon monoxide, but No. 1 cannot generate 2mol carbon monoxide, and this reaction cannot have a 100% conversion rate. If it doesn't mean 2mol, then what does 2 parts of carbon monoxide mean?

The rate is fast, the temperature is the same in the two containers, the volume is the same, the larger the amount of the substance, the greater the concentration, the faster the reaction rate, and the shorter the time required for the same amount of product.

Second, at the same temperature, the latter has a higher concentration, higher pressure, and a faster reaction rate.

Other things being the same as you said, then the second one takes less time because the concentration of carbon dioxide is larger.

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.

NH3 is 2mol, and the conversion rate of NH3 is 20% = > initial amount 2 (1-20%) = > reaction.

Initial O2 = 5mol, remaining.

The change in the concentration of the generation of no = no indicates the average rate of the reaction = (

Equilibrium concentration of O2 =

The reaction reaches equilibrium after 2min under certain conditions, NH3 is 2mol, and the conversion rate of NH3 is 20%, indicating that NH3 has 10mol, so oxygen has 20mol.

1.2mol NH3 can give 2mol of NO, so the change of NO concentration indicates that the average rate of the reaction is: 2mol 2min = 1mol min

2.Consuming 2molNH3, the oxygen consumed, so O2 remains, so the equilibrium concentration =

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.

When N2 is added and moved forward, the H2 conversion rate increases and the N2 conversion rate decreases.

Why does it have little effect on balance? Because equilibrium is a constant, it is mainly related to temperature, so it has little effect on equilibrium.

However, when N2 is added, the instantaneous concentration of N2 increases rapidly, which is an increase in the concentration of reactants for H2, so there will be another part of H2 in equilibrium that will continue to be converted into NH3

So why is N2's conversion rate dropping?

Obviously, if the part of N2 added to the equilibrium state reacts with another part of the H2 in equilibrium state, if it reacts in equal proportion with the one before the addition, the conversion rate of N2 remains unchanged, in fact, the H2 in the equilibrium state can no longer be transformed according to the original proportion, that is, the actual reaction ratio is less than the proportion before the addition, that is, the N2 conversion rate after the addition is lower than before the addition, and the total N2 conversion rate naturally decreases.

In the reaction, NO2 is added, assuming that the part added reacts in the original proportion, but in the actual constant volume system, the pressure still increases, so it will continue to move in the direction of pressure reduction, that is, NO2 will continue to transform until the pressure is reduced to meet the equilibrium, so the total NO2 conversion rate increases.

Add more, move less (balanced moves: weaken you plus changes, but can't cancel them out).

This is the only reactant concerned (adding one reactant is equivalent to adding all the reactants).

The first question, Professor Wang answered, I won't answer.

Second, if you're interested, I suggest you look at what equilibrium is. I don't think you're familiar with the concept of equivalence. You'll know it when you learn about equivalence.

PV=NRT, in a container with constant temperature and volume, the pressure ratio is equal to the ratio of the amount of the substance of the gas, and the ratio of the amount of the substances participating in each substance is equal to the ratio of the stoichiometric number.

When m+n=p+q, that is, the amount of gas before and after the reaction is unchanged, that is, whether it is an equilibrium state or not, the pressure is equal to the initial pressure, and the pressure is a fixed value, which cannot be used as a basis for judgment.

If m+n>p+q is assumed, the pressure decreases in the forward direction of the reaction, increases the pressure in the reverse direction, and the pressure does not change in the equilibrium state, and m+n is vice versa< p+q, it can also be analyzed in the same way.

If there is no equilibrium, the amount of total matter in the vessel will change regardless of whether it is a reaction to the left or the right, because it is a gas, so the pressure will definitely change, so on the other hand, when the pressure is certain, the amount of the total substance will be certain, which is the reaction must be balanced Got it?

This is an all-gas reaction, defined according to equilibrium, as "a state in which the concentrations of the components of reactants and products no longer change." "Because m+n is not equal to p+q, the reaction to the left or right will cause the concentration of the reaction components to change, when the total pressure is unchanged, it means that the reaction is not carried out to the left and right, and the reaction has reached a state of equilibrium... Understand??

ma(g)+nb(g)<==>pc(g)+qd(g)

If m+n=p+q, the pressure is equal in equilibrium or in other states because the amount of gas does not change. And according to the principle of Lechatre equilibrium movement: when m+n≠p+q and isothermal and equal volume, the pressure of the system is increased, the equilibrium moves in the direction of gas decrease, and the system pressure decreases, and the equilibrium moves in the direction of gas increase.

There is an ideal gas equation of state, pv=nrt, p is the pressure, v is the volume, n is the amount of the substance, r is the constant, t is the temperature, isothermal and equal volume, n=pv nrt, the change of the quantity of the system can be reflected by the change of pressure, the pressure is constant, that is, the amount of the substance of each substance in the system is constant, indicating that the equilibrium has been reached.

Hope it helps.