A chemical equilibrium problem that non experts can t do

n2(g)+3h2(g) 2nh3(g) (where denotes the reversible sign).

Reverse reaction. Filling with NH3 increases the concentration of NH3 in a closed container, and comparatively decreases the concentration of N2 and H2. According to the principle of chemical equilibrium reaction, in this case, in order to restore the equilibrium, a reverse reaction takes place, consuming some NH3 and claiming some N2, H2 in order to make its concentration content relatively equilibrium.

Enlarge. In the reverse reaction, 2 molecules NH3 can generate 1 molecule N2 and 3 molecules H2, that is, 2 gas molecules generate 4 gas molecules. The volume of the container is constant, so the pressure naturally increases.

Enlarge. Filling, in the same way as the previous question, the equilibrium will move in the direction of the reverse reaction. At this time, the temperature and pressure are constant, then the gas expands and increases in volume.

No change. The percentage content of NH3 does not change before and after the reaction. For specific reasons, please refer to "Answer Supplement".

Answer: The first question is still empty: the volume fraction of ammonia in the gas mixture will increase.

The reaction is carried out with the same volume, and due to the addition of NH3, the overall pressure of the closed vessel increases, and the increase in pressure causes the equilibrium to shift slightly in the direction of the positive reaction (because the positive reaction can turn 4 molecular gases into 2 molecular gases, reducing the volume and relieving the pressure of the closed container). Therefore, in such a case, NH3 will increase.

The second question is still empty: the volume fraction of ammonia in the gas mixture will increase. With constant temperature and pressure, the reflection reacts in the opposite direction to the original volume percentage.

For the first question, you might think of it this way: imagine that the temperature and pressure are the same, and the reaction is reversed to equilibrium, and the volume increases while the components maintain their original volume percentage. Then the volume is compressed to its original size, and in this process, due to the increase of pressure, the reaction begins to proceed in the positive direction, forming NH3, and thus, the volume percentage (i.e., volume fraction) of NH3 increases.

It's a really hard question. I don't know if it's right or not. For your reference.!...

Step 1: Reverse reaction. Enlarge. Enlarge.

No. 2: Enlargement. No change. No change.

Question 1Because the temperature does not change. That is, k does not change.

The first time it was balanced, it was always kWhen the equilibrium is reached again, we set the equilibrium constant to k1Obviously, if not moved.

k1>k..So we want K1=K, which is to reduce the concentration of NH3, so. I had to move in the direction of the reverse reaction.!

And because pv=rnt!!p is the pressure and v is the volume. r is a constant, n is the amount of gaseous matter inside the container and t is the temperature.

No change. Added NH3...And after joining, it moves in reverse.

Definitely more than that. It's that n has increased. So...

The pressure has increased...(Or.) Let's assume that the pressure is constant, and the volume will increase if it increases.

Since the title says that the volume remains the same. We pressed it down again. At this time, the pressure is relatively large).

Because the volume does not change. Anyway. The coefficient of NH3 is 2

How it also wants to move in the direction of less coefficients. So with NH3 added, it is also mostly NH3 and does not decompose. (Or.)

Let's assume it's not moving, put the pressure on the same as this. The volume fraction does not change, but when we press the volume back, the equilibrium moves forward and the n of NH3 becomes larger. So the volume fraction is also larger!

Question 2: Also based on. pv=rnt..,p.No change. n has increased. So the volume has also increased!

Because the pressure is constant. Added is NH3...NH3 is broken down into the same. So his volume fraction will not change.

The pressure does not change. No matter how you change the amount of NH3. After balancing again. None of his volume scores will change.

1.Reverse reaction, enlarged.

Since the reaction is N2+3H2=2NH3, although the pressure increases after adding NH3, the reverse reaction is still carried out due to the increase in the concentration of NH3. However, it does not compensate for the increased pressure, so the pressure increases.

2.Grow, stay the same.

Although NH3 is added, the reaction proceeds in the opposite direction, and the volume of the mixture increases at equilibrium because the temperature and pressure remain the same. Since the reaction reaches equilibrium without any change in pressure and temperature, the concentration of NH3 is also constant.

1.Reverse reaction, enlargement, enlargement.

2.Grow, don't change, don't change.

The equation for reversible reaction n2 + 3h2 <-2nh3 is listed first, from which it can be seen that the forward reaction is the process of getting smaller in volume.

When the temperature and volume remain unchanged, recharge.

Since the temperature and volume remain the same, and gas is added, the pressure naturally increases.

And since NH3 is added, the reaction moves in the opposite direction of the reaction.

However, since the pressure is higher than before, this is equivalent to a continuous introduction of gas in a container with a constant volume of N2 and H2 at a ratio of 1:3.

The equilibrium will move towards the smaller side, so the mass fraction and volume fraction of NH3 will increase.

Note that here is the percentage of the total).

When the temperature and pressure are constant, recharge.

This means that the volume of the container is variable, and the gas increases, and the volume naturally increases.

Since the temperature and pressure remain unchanged, the mass ratio and volume ratio of each gas in the gas will not change.

Therefore, the percentage content and volume fraction of NH3 remain unchanged.

Note that although the total amount has increased, the proportion between the various gases remains the same) To sum up, you should carefully analyze this kind of problem verbatim, and distinguish between the total amount and the percentage.

1.Reverse reaction, enlargement, enlargement.

2.Grow, don't change, don't change.

Reverse reaction. According to Le Chatlet's theorem, the reaction in the opposite direction of change decreases the change.

Enlarge. Although the reaction to change in the opposite direction can only reduce the change, not completely.

Enlarge. Because of the unequal reaction, the pressure does not change, i.e., the equilibrium does not move. It can be imagined as a combination of two independent variable containers, each balanced.

No change. The pressure does not change, that is, the balance does not move.

Answer addendum: the first question, (although the reverse direction of the change can only reduce the change, not completely, ammonia must have **).

The second question, (because of the unequal reaction, the pressure does not change, that is, the balance does not move) may not be as detailed as the upstairs, but the two sentences in parentheses are the final idea of doing equivalent equilibrium.

1 The equilibrium will increase in the opposite direction, and the pressure will increase, and the NH3 volume fraction will increase.

2 Increase NH3 white content Increase NH3 volume fraction.

The solution is solved using chemical equilibrium and equivalent equilibrium

First of all, the 2s02+02====2SO3 reaction is a reaction with reduced volume.

A and B are both 2molso molo2 at the beginning, so no matter how they react in the process, the substances in them can be regarded as 2molso molo2, so when you connect A and B (assuming this is a C container), it is equivalent to filling a constant pressure container with 4mol so Molo2, and A is also a constant pressure container with the same temperature, so A and C are an equivalent equilibrium

Therefore: the volume of vessel C at equilibrium (i.e., the total volume of A and B) is 2 times that of A, i.e., and the volume of B is constant at 1L, so the final volume of A is.

After the two containers of AB are connected, the temperature and pressure when the equilibrium is re-reached are the same as when A is balanced alone, so the degree of reaction is also the same, 1L of container A becomes, decreases, and the total volume of communication with B is 2L, which decreases, so the volume of container A is.

a, it is impossible for a reversible reaction to react completely, then there are two limits to the reversible reaction in a:

1. It does not react, and the mass of the gas is 3mol; 2. Complete reaction, the mass of the gas is 2mol. According to Afugadro's law, at the beginning 3mol of gas occupies 6 blocks, so 2mol accounts for 4 blocks. Therefore, the volume of A is (4,6), i.e., the separator k eventually stays between 0 and 2 to the left of the 0 scale.

In the same way, "the separator k finally stays at 1 on the left" means that the gas volume after the reaction is 5 blocks, that is, the mass of matter is.

Let the mass of a participating in the reaction be 2x, so 2-2x+1-x+2x=, x=, then the conversion rate of a in a is 50%.

But for B, it's different. If there is no HE, A and B are equivalent, but B has less pressure than A, then the equilibrium of 2c(g) = reversible = 2a(g) b(g) shifts to the right. Therefore, its conversion rate ratio is greater than 50%. B false.

c, if the diaphragm k ends up on the left near 2, and B in B ends up on the right side with a scale of 4, then B has a volume of 6However, the reaction is reducing the volume of the gas, so the volume of B must be less than 6, that is, f in B must be less than wrong.

The d,y axis is not possible to represent the amount of a substance in the container B. Because in the beginning, there was no gas in B.

The mind is a little dizzy, I hope you can understand. In fact, the exam is to save time, and if you look at the A right, you will choose it directly. I'll have time to come back and think about it.

Select A and Ca, the volume is increased to 2 times, the pressure decreases, the balance moves to the left, and the value of C(NO2) C(N2O4) increases.

b, the volume is unchanged, the amount of matter that increases NO2, first assumes that the pressure is constant, then the equivalent equilibrium C(NO2) C(N2O4) value is formed as the original equilibrium remains unchanged, and then compressed to the original volume, the pressure increases, the equilibrium moves to the right, and the C(NO2) C(N2O4) value decreases.

C, keep the pressure unchanged, fill N2, the volume increases, which is equivalent to reducing the pressure, the balance moves to the left, and the C(NO2) C(N2O4) value increases.

D, the volume is unchanged, the amount of substances that increase N2O4 is analyzed, and the analysis is the same as B.

b is wrong. Because it is a reversible reaction. Increasing the amount of nitrogen dioxide with the same volume shifts that balance to the right, and you may think that increasing nitrogen dioxide even to the right increases nitrogen dioxide.

The actual situation is that the volume remains the same, and you increase the gas into the system, which is equivalent to increasing the pressure, which is not difficult to judge from the law of four sameness, and the reaction equilibrium shifts to the right by increasing the pressure, that is, just opposite to the question requirement, the C(NO2) C(N2O4) value decreases.

Choose C solution:

Item A is correct, because the sliding of the separator K indirectly reflects the change of the mass of the substances in the system, because the reactants and the products are both gases, and the total amount of reactants and products is not equal (which can be seen from the coefficient of the equation), so the sliding of the separator K indicates that the mass of each substance in the system remains unchanged and the equilibrium is reached.

The amount of matter in which A and B finally reach equilibrium is the same, because the two are equivalent equilibrium, that is, the substance that is converted to the other half of the equation according to the stoichiometric relationship in the equation is equivalent equilibrium if it is the same as the amount directly added to the other half.

A: 2a(g)+b(g).

>2c(g)

2mol1mol

B: 2a(g)+b(g).

>2c(g)

If 2mol is converted, that is, the quantity of substance C becomes 0, which is equivalent to the change of substance C is 2mol, then A is 2NOL and B is 1mol, it is obvious that the two are an equivalent system, and the quantity of substance of substance C is equal, so C is wrong.

D, if A is at the left 1, because it is the equivalent balance B's piston should be at the right 5, but with the addition of 1mol of HE, it is changed to 6, so the final volume of system A should be smaller than that of system B, so term B is okay.

3a（g）+2b（g）==4c（？）2d（？According to the pressure ratio of 5:4, the amount of gaseous substances after the reaction is 6 4 5 =, and according to the total AB after the reaction is 4mol, then C is solid or liquid, and D is gaseous; b is transformed into;

The ratio of the pressure before and after the equilibrium reaction is 5:4 by 3a(g)+2b(g)==4c(l)+2d(g), and the ratio of the amount of matter is equal to the ratio of pressure under the same temperature and volume, that is, n (before): n (flat) = 5:

4. It shows that this is a reaction with the volume of the gas decreasing, increasing the pressure of the system, and the equilibrium shifts to the right, but because the temperature does not change, the chemical equilibrium constant does not change, so c is wrong;

Since C is a liquid or solid, increasing C has no effect on the equilibrium, and the equilibrium conversion rate of B remains unchanged, so C is correct;

In this problem, we can know from the ratio of the pressure before and after is 5:4 that the volume of gas in the reaction is reduced, so there must be one of c d who is not a gas, and according to the three-stage formula, we can know that 3a (g) + 2b (g) = 4c (?).)2d

So to satisfy 5:4, it has to be c as a solid or liquid, and then the rest can come out, and you have to count it yourself.