Chemistry, reversible reactions, equilibrium, concentration relations, some questions, thank you

1. In fact, the most useful in the solution are two conservations--- conservation of charge and conservation of material.

The so-called conservation of protons can be deduced from the above two conservations, but for some systems, the conservation of protons is more convenient to understand, and for others, it will be more difficult to understand.

Therefore, it is completely possible to ignore the conservation of protons and grasp the two most basic conservation relationships.

2. Conservation of electric charge.

nh4^+ h+ = cl- +oh-

Since it is neutral, i.e., h+ = oh-

So, cl- = NH4+

1) Yes, because the temperature is the same, it is the equilibrium constant of the first one.

2) A is correct, because the volume before and after the reaction is unchanged, and the second set of data is equivalent to 2 times the first set of data, which is equivalent to pressurization, but the equilibrium does not move, so the H2 volume fraction does not change.

B error, the heat released is twice as much, i.e.

c error, the second pressure is strong, the rate is fast.

The first question, I don't know, in fact, I will substitute this formula in high school.

In the second question, C(H+)+C(NH4+)=C(Cl-)+C(OH-) can be known from the conservation of charge

Because it is neutral, then c(h+)=c(oh-) is about c(nh4+)=c(cl-) on both sides

The third question, one, right, the temperature does not change, the equilibrium constant does not change.

Two, a, the second reaction can be seen as two first reactions after reaching equilibrium, then grouping them into a container, and since the sum of the coefficients before and after is the same, the increase in pressure caused by the increase in the amount of matter does not cause the equilibrium to move, so the volume fraction is the same.

B, right. c, right.

The division of the first question into two processes is a little better.

In the second question, if the solution is neutral, the hydrogen ion is equal to the hydroxide, and the solution is electrically neutral, so the chloride ion is equal to the ammonium.

Question 3: Because the reaction is a reaction with constant volume, the compression equilibrium will not move, so the volume fraction is equal to q=2 h and the rate is not equal to the second fast.

Categories: Education, Science, >> Learning Aid.

Problem description: For example, the hail finch has learned that a+b=c (reversible reaction), k=, and asks what is the equilibrium constant of c=a+b?

What is the relationship between its age-eaters?

Analysis: The equilibrium constant source modulus of c=a+b is k'=

The equilibrium constants of the forward and reverse reactions are reciprocal to each other.

If the volume of this process is fixed, it is obvious that the concentration is unchanged means that the amount of the substance is unchanged, and the reaction must be equilibrium If the volume is not fixed, when the volume changes (such as compression by external force), if the concentration is constant, there is only one case, that is, the volume before and after the reaction is unchanged.

For example, A+3B=2C+2D, when the volume compression is 1 2, the concentration becomes 2 times, and the concentration quotient (C2*D2) (A*B3) remains unchanged, and the equilibrium does not move.

Therefore, it can be concluded that as long as the concentration remains unchanged, equilibrium must be reached.

a. The concentration of each component no longer changes, and the reversible reaction reaches equilibrium, so a is correct;

b. The forward and reverse reaction rates are equal, but not 0, which is dynamic equilibrium, so B is wrong;

c. The concentration of reactants is less than the concentration of the products, which is not constant, so C is wrong;

d. It is a dynamic equilibrium, and the forward and reverse rates are greater than 0, so d is wrong; So choose A

1 Equilibrium constant.

k is only affected by temperature, and has nothing to do with the change in the concentration of any reactant or product, nor with the change of pressure and width; Since the catalyst changes the forward and reverse reaction rates to the same extent.

Therefore, the equilibrium constant is not affected by the catalyst.

2 Any reversible reaction, when the temperature remains unchanged and other conditions affecting the chemical equilibrium are changed, even if the equilibrium shifts, the k-value.

No change. 3. When other conditions remain constant, if the forward reaction is an endothermic reaction, K increases (or decreases) due to the equilibrium shifting to the direction of the positive (or reverse) reaction when the temperature is increased (or decreased); If the positive reaction is an exothermic reaction, k decreases (or increases) due to the shift of the deficit equilibrium to the reverse (or positive) reaction direction when the temperature is increased (or decreased); Therefore, the equilibrium constant may increase or decrease as the temperature increases, but it does not change.

Of course, it doesn't have to be, it may increase, it may not change, it may decrease, depending on the situation, it is very complicated!

When the amount of reactants changes, if there is only one reactant at constant temperature and constant capacity (1), such as AA (G) = BB (G) + CC (G) increases the amount of A, and the equilibrium moves to the direction of positive reaction, the change of the conversion rate of reactant A is related to the stoichiometric number, and the conversion rate may increase or decrease.

Less or no change. If the value of the numerator and denominator increases according to the conversion rate formula, the value of the conversion rate cannot be determined, and the compression principle can explain the changes of these three conversion rates. If a=b+c, the balance does not move, and the conversion rate remains unchanged.

Equivalence equilibrium at this point. If A>B+C, the equilibrium shifts to the direction of positive reaction, and the conversion rate of reactants increases. The equilibrium and the original equilibrium are no longer equivalent.

If AC+D, the conversion rate of AB increases. The equilibrium and the original equilibrium are no longer equivalent. ）

The premise is to know the rate.

Signs and judgments that reversible reactions have reached equilibrium.

In a reversible reaction under certain conditions, when the rate of the forward reaction is equal to the rate of the reverse reaction, the state in which the amount and concentration of the reactants and the products of the substances do not change, which is called the state of chemical equilibrium. Its features are:

1) "Inverse": The object of chemical equilibrium research is reversible reactions.

2) "Equal": The essence of chemical equilibrium is that the positive and reverse reaction rates are equal, i.e., v (positive) v (inverse).

3) "Moving": v (positive) v (inverse) ≠0

4) "Fixed": In the equilibrium system, the concentration, mass fraction and volume fraction of each component remain certain (but not necessarily equal) and do not change with time.

5) "Change": The chemical equilibrium is the equilibrium under certain conditions, and if the external conditions change, the chemical equilibrium may move by fractions.

6) "Same": Under the premise that the external conditions remain unchanged, no matter what path the reversible reaction takes, that is, whether it starts from the positive reaction or the reverse reaction, the final equilibrium state is the same, that is, the same equilibrium state.

The signs and judgment methods of reversible reactions reaching equilibrium are as follows:

Take ma(g) + nb(g) pc(g) + qd(g) as an example:

The doubling of the concentration of reactants can be equivalent to doubling its volume is to compress it to half of the original, but the premise of considering this problem is that the state of reactants and products, if they are all gases, they will be compared with each other according to the coefficient before the reactant and the coefficient before the product.

It is regarded as "0" because the solid or liquid does not affect the pressure, in this case, if the concentration of reactants is doubled relative to the product, and the change of entropy and temperature caused by different chemical reactions is not considered, a certain amount of reactants are added to the original container, so that the concentration of the amount of his substance increases, the positive reaction rate will become larger, and the reaction will be on the line, and the concentration of reactants and products will increase during equilibrium. In contrast, the concentration of each group will increase at equilibrium, compared with the original equilibrium.

Hope thank you -