A question on chemical equilibrium 50

Naturally, the noble gas does not participate in the reaction, so the volume does not become larger. The other two have substances involved in the reaction, so they definitely can't be chosen. As for what you said about the volume becoming larger, I don't quite understand, if you add O2, the volume should be reduced, because if the equation is balanced, it should be 2SO2 + O2 = 2SO3, 3 volume becomes 2 volume, the volume decreases, and the reaction moves in the positive direction, SO3 not only increases the amount of matter, but also the air volume becomes smaller, and the concentration must increase.

If it is A, consider that after SO3 is inserted, the equilibrium does not move, but the amount of the substance increases, and the volume increases, but the concentration is uncertain. I still think it's C, and I'm sorry why the answer is A. But I've had a problem with the wrong answer in high school before, so you can ask the teacher again to see if it's my analysis.

Because the O2 molecules are spaced between large and the temperature is unchanged, the filled O2 is easy to compress in the container, so it can be omitted. Consider the volume.

Dizzy, upstairs is all wrong. The equilibrium concentration is the proportion, not the specific concentration, so a is correct because the ratio does not change. c Obvious error. With piston, the volume changed and the balance shifted. B is to add SO2 2mol.

If the temperature volume is constant, the same equilibrium can be obtained if the material is the same twice.

If the temperature pressure is constant, the material added twice is proportional.

The first addition is 2mol SO2 and 1MOLO2, which is equivalent to 2mol SO3, so adding any end of SO3 can ensure that the same equilibrium A is right and B is wrong, and C is equivalent to reducing the concentration, reflecting a shift to the left.

In my opinion, the questioner got the question wrong and should start with "2mol SO2" and 1MOLO2 instead of "2mol SO3" and 1MOLO2, otherwise all three options are incorrect. Killed! If corrected, the answer is naturally A, and the concentration of SO3 gas remains unchanged when the equilibrium is reached no matter how much A is added, because it is equivalent to the original equilibrium.

Pick B. The reaction is an endothermic reaction, which is based on the principle of Le Chatre, the temperature increases, the equilibrium moves in a positive direction, and the mass fraction of a decreases, which is inconsistent with the trend of the curve.

A: If P1 is less than P2, according to the reaction equation, the equilibrium should move in the direction of the positive reaction, that is to say, the mass fraction of A will decrease, and the P1 curve is above the P2 curve, which means that the pressure is strong but the mass fraction of A is high, which is obviously contradictory to the facts, so A is wrong!

There are a few concepts to understand first:

1. The rate of chemical reaction refers to the speed of a chemical reaction at a certain moment;

The concentration of 2N2 is the amount of N2 contained in a unit volume, that is, the amount of N2 divided by the volume of the container;

3H2 conversion, the ratio of the amount of H2 consumed by the reaction to the amount of hydrogen originally present;

The volume fraction of 4H2, the ratio of the volume of H2 in the container to the sum of the volumes of all gases.

The first concept is the amount of representation of the state of the reaction at the beginning of the reaction or at a moment in progress. The value of this amount is related to the course of the reaction.

The 2nd, 3rd, and 4th are the quantities of state representations when the reaction reaches equilibrium, and their values are not related to the process, but only to the final temperature, volume, pressure, etc.

Based on this, it can be assumed that at the beginning, the volume of container B is twice that of container A, and the amount of gas flushed into it remains unchanged. At this time, compare the two containers of A and B. b，c,d

The quantities in should all be equal. Now compress container B to the volume of container A. Rule.

b. The concentration of N2 in container B will increase because the volume becomes smaller;

The conversion rate of C. H2 will become larger, and in the case of volume reduction, the balance of N2+3H2=3NH3 will move to the right;

d. The volume fraction of H2 will become smaller because the amount of H2 will decrease during the right shift of equilibrium. [The numerator in VH2 (VH2+VN2+VNH3) becomes smaller, and the value of the fraction becomes smaller of course).

PV = NRT (gas equation of state, p: pressure v: volume n: quantity of matter r: constant t: temperature (k)).

This equation is very useful and it is recommended that you use it proficiently.

a. From the question, it can be seen that the amount of reactants in B is large, so the concentration is large and the pressure is also large, so the reaction rate is fast.

B. N2 and H2 are added to B, and the concentration of N2 must be larger.

c. For A N2 + 3H2 2NH3, B adds N2 and H2, which is equivalent to pressurization, and the conversion rate of the balance right shift H2 is high.

d. In fact, B is equivalent to mixing two A into a container of the same volume for reaction (that is, pressurization).

It can be seen that the volume fraction of H2 when B (that is, 2 A) and A reach equilibrium is the same, but B is also equivalent to pressurization, and the equilibrium shifts to the right, so the H2 conversion is higher and less, so the volume fraction of H2 of the mixed gas at equilibrium should be: A is greater than B.

B answer: Because the chemical reaction rate is still going on after equilibrium, and B has a large mass, its reaction is fast, so the concentration is large.

If the container volume is the same, the pressure of container B is large, and the balance of N2+3H2=2NH3 will move to the right compared with A, so the conversion rate of N2 and H2 is greater than that of A. But C=Mv, the volume is unchanged, and the amount of N2 and H2 in B is larger than that of A, so the concentration is larger.

Option: Since the ratio of the amount of the reacted substances is 1:::1, a:b:

c:d=2:1:

3:2, which is the simplest ratio, so even if it is equal to 4, it must be divided by the term v = the amount of the reactant divided by the time. Since Z is a solid, it has no effect only if the sum of the left and right gas coefficients is equal, so Y is a solid or liquid, and X, W must be gases.

TermD is an inverse reaction due to the fact that the temperature rises and reacts in the direction of endothermy, and the amount of matter is equal, so the reverse reaction is endothermic.