How to solve problems related to chemical equilibrium

1. First of all, it can be seen that what occurs in container A is the reverse reaction of the reaction in container B, and the final equilibrium state is the same, and C is equivalent to doubling the concentration of NH3 in container B (that is, doubling the concentration of each substance in container A) and reaching the equilibrium state late.

A. If the pressure in the two containers is always the same, under the same conditions of temperature and volume, there should be the same conversion rate of the two, and the final equilibrium state should be 2C1=C3, but because the pressure in A is significantly less than the pressure in C (the reaction in A is carried out in the direction of volume becoming smaller and pressure becoming smaller, and C is the opposite), so relative to C, the reaction of A moves in the direction of the reverse reaction, and the concentration of NH3 decreases, which should be 2C1P3, which is wrong.

d. The reaction conditions of the system are good, and the actual value of the conversion rate of the three cannot be determined, and the sum of the conversion rate cannot be said to be less than 1, which is wrong.

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 should be chosen, and when water vapor is added to the equilibrium system, the equilibrium shifts positively, and the amount of CO2 increases, so C is wrong. Increase the concentration of water vapor, the conversion rate of water vapor should be reduced, the conversion rate of water vapor in the original equilibrium is still 60% after adding water vapor, then the water vapor should react, and the generated CO2 is also, but because the conversion rate of water vapor should be reduced, the generated CO2 should be less than, so choose A.

It must be somewhere in between.

From Figure A, it can be seen that the initial concentration of A is, the equilibrium concentration is, and the conversion concentration is, and the initial concentration of C is, the equilibrium concentration is, and the conversion concentration is, and the ratio of the stoichiometric numbers of A to C in the reaction is 3:2.

As can be seen from Figure B, since T3 and T4 are catalysts, the T4 and T5 stages should be depressurized. The rate of the forward and reverse reactions of depressurization decreases to the same extent and is equal, and the equilibrium does not move, indicating that the reaction is a gas volume constant reaction, and the reaction is 3a = b + 2c (reversible).

A The starting concentration is the amount of the substance, and the volume is 4L.

a b cc initial x

c The amount of this behavior change).

At the end of c, the amount of change of each substance before and after the reaction is conserved, and the c= of b can be obtained. Therefore, it is concluded that b is the product.

Do you still use it later?

As can be seen from Figure A.

A is the reactant, C is the product, and the ratio of the coefficient is 3:2, and according to Figure B, T4-T5 is the pressure change stage, and after changing the pressure, the positive reaction rate and the reverse reaction rate decrease at the same time and are equal, indicating that the pressure change does not affect the equilibrium state.

Therefore the reactant and product coefficients are equal.

So b is the product, and the ratio of the coefficients is 3:1:2

From Figure A, it can be determined that A reactants, C products and their reaction coefficient ratio A: C = 3:2;

In Figure B, the change of reaction rate with time (and only one condition is changed in each of the four stages, and the conditions are different each time), in which the addition of catalyst has no effect on the reaction equilibrium, only changes the reaction rate; Changes in temperature and concentration inevitably affect the chemical equilibrium, and T2 T3 and T5 T6 change temperature and concentration respectively.

The remaining T4 T5 is to change the pressure conditions, the reaction rate is reduced but the equilibrium is not destroyed, indicating that this reaction is an equal body reaction, in the system ABC is a gas, so B is the product, the reaction is 3A-B+2C (reversible).

b The initial amount is.

Hope it helps.

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.