Chemical equilibrium questions, chemical equilibrium questions

Analysis: Equation: Co(G) +H2O(G) = H2(G) +CO2(G).

Before reaction: 1mol---1mol---0---0 equilibrium:

Co-transfer: equilibrium constant. k =([h2]^1*[co2]^1)/([co]^1*[h2o]^1)

Charge H2O(G) again, and the equilibrium constant does not change because of the same temperature and pressure.

There is no essential difference between a recharge and a recharge.

Equation: Co(g) + H2O(G) = H2(G) +CO2(G) Before Reaction:

At equilibrium: (1-x)mol-(

Co-transfer: xmol---xmol--- xmol--- xmol equilibrium constant. k =([h2]^1*[co2]^1)/([co]^1*[h2o]^1)

x*x）/[（1-x）（

x^2 = - x^2)

b^2-4ac = （ 4* =

x1 =[-b+ (b 2-4ac)] 2a = [ =x2 =[-b- (b 2-4ac)] 2a = [ =x1 is not on topic, discard.

So CO2 concentration =

The process is as follows:

In the first equilibrium, the amount of H2 generated is the same as that of CO2, is, and the amount of Co2 is the same as that of H2 in the second equilibrium is Xmol, then the amount of H2 is the same as that of XmolCo, and the amount of H2O is .

The column equation yields: x 2 [(1-x)(.]

Solve x and get it.

Analysis: The key is to know that under the condition of constant temperature, the reaction equilibrium constant is unchanged.

1) According to the relationship between the stoichiometric numbers, AMOLC is generated and the A of AMOL is consumed, so N(A remains) = 1-Amol

2) From the chemical formula of the first question, it can be seen that when the reactant is 3mol, 3amol c is generated

3))(x+1 2) 1=(2+1 2) 1=3a a,x=2 y=x+(1-3a)=3 3a d.

If the amount of 3a 1, b is less than 2mol; If, the amount of substance in b is equal to 2mol;

If the amount of 3a 1,b is greater than 2mol

4) Add another 3molc to the equilibrium mixture of (3), and reach equilibrium again and only add and be completely equivalent to the beginning when it is reached, satisfying , h=; The quantity fraction of the substance of c is.

5) B Because the volume of the container in (5) the problem does not change, and the volume of the container in the problem (1) decreases, the pressure in the container of the problem (5) is less than the pressure in the container of the problem (1), which is conducive to the reverse reaction.

44+12=2*28

Mean molar mass) = m total n total.

Left 44 1 = 44g mol

On the right, 2*28 2=28g mol

The reaction proceeds in the positive direction, and the average molar mass of the gas mixture is ignored by the solid, so it is reduced, do you understand?

The CO in the reducing gas will decompose by 2CO CO2+C at a certain temperature. At 400 CO2 is stable, while at 1000 there is only volume) CO2

Brother, I used to be in the science chemistry class, shame on me, I returned it to the teacher, and I went to check the formula, and the above should be fine.

1. Basic solution (personal thought):

Assuming that at the beginning of the reaction, the gaseous mass is as follows, and the temperature is raised, and the reaction proceeds to the right. Then:

co2（g）+c(s）=2co（g)

Starting amount: 1 2 reaction volume: x 2x equilibrium amount:

1-x 2+2xm total = (1-x) * 44 + (2 + 2x) * 28 = 100 + 12xn total = (1-x ) + 2 + 2 x ) = 3 + x (flat) = m total n total = (100 + 12x) (3 + x) = 12 + 64 (3 + x).

So x increases (i.e., the CO2 of the reaction increases), flat) decreases.

2. Special solution:

Assuming that there is only CO2(g)1mol at the time of reaction, the average molar mass of the gas is 44g·mol, and the reaction proceeds in the right direction when the temperature is raised, and when the reaction is complete, it is all co(g), and the average molar mass of the gas is 28g·mol

Therefore, (flat) = 44-28 range class change, so it is reduced.

d is related to the original gas (mixed) average formula.

(1) In fact, it is equivalent to filling 2molSO3 in two containers, and then container A has a constant pressure and container B has a constant capacity, due to the decomposition of SO3, the pressure in container A becomes larger, while B remains unchanged; Since the volume of container B becomes larger, and the whole mass does not change, the density of A is large.

2) The answer to the second question is very clear.

It's not good that since the starting volume is the same, and the amount of substances with different amounts of matter added to it has the same temperature and pressure, it's simply not possible.

(1) Equal, equal (with extreme transformation).

2) Less than, SO3 reaction, pressure increases, reaction in the direction of positive reaction, less than.

1. Because the volume is variable, this is a problem of equal equilibrium.

So the two options A and B are definitely balanced and do not move. For example, doubling can be understood as two containers of the same volume that reach the same equilibrium 4, 2, 4 on top of each other, so that the equilibrium does not move.

C option, you can add 1mol to understand that adding 1,,1 is added according to the ratio of 2:1:2 at the time of equilibrium, at this time the balance does not move, and the B balance added again moves to the right.

The same option d can be understood as first decreasing the amount of 1,,1, then the balance does not move, and then the reduced b then the balance moves to the left. So choose C for this question

2. When constant t, v Because the substance is added on the basis of the old equilibrium, the volume remains unchanged at this time, and the pressure will change, so that the equilibrium will move, except for those whose volume remains unchanged before and after the reaction. Therefore, to maintain the equivalence (the percentage content of each component is unchanged) can only be exactly the same as the original equilibrium of the various quantities. And when it is constant t, p, the volume can be changed, as long as the proportion is the same.

For example, in the reaction of 2SO2+O2=2SO3, the initial feeding 2,1,0 reaches a certain equilibrium state.

Then the feeding of 4,2,0 is twice as much as the original, which can be understood as using two containers of the same volume, respectively according to 2,1,0, and the equilibrium state of the two containers after equilibrium is exactly the same. Now let's stack these two containers on top of each other, imagine that after the partition is pulled away, will the volume of the container be doubled? So will the balance move at this point?

No. Then this is the equivalence of constant t and p.

But if it is constant t and v, then after removing the partition, the volume is twice as large, and we still have to compress the container, then the pressure of the container will increase, so what will happen to the balance? Of course, it will move to the right. This is isobaric equivalence and equivalence.

Of course, what you are talking about in your question is a reaction with a change in the volume of the gas before and after the reaction, and if it is a reaction with a volume difference of 0, then no matter what the conditions are, it is equivalent proportionally.

This part of the content is more difficult to understand, if you don't understand, you can communicate online.

3. The four reactions are all gases, then the reactants are decomposed by 8% and the total volume is increased, these four reactions are 2 volumes of gas decomposition, we can think, first decomposed 2 volumes of gas, which is less than 8%, but in the end it increased by 8%, indicating that the generated gas volume first supplemented the previous reduction of 2 volumes, and at the same time there should be 2 more volumes, so that a total of 4 volumes of gas will be generated, then only option A is compliant.

After increasing by 1mol, the amount of substance b is larger than that required for equilibrium, so the reaction shifts to the right, and if 1,,1 is added, the equilibrium does not move.

The final volume of the reaction is 8% more, that is, the reaction product is equivalent to 16% of the original volume, of which 8% fills the volume of the reaction, and 8% is the excess, so the ratio of product to reactant is 2:1

ca.Constant temperature, compression volume, equilibrium does not move, the color of the mixed gas deepens --- pairs, the volume of the gas before and after remains unchanged, pressurized, and the balance does not move, but the concentration of each component increases, so the concentration of colored I2 also increases, and the color deepens.

b.Constant pressure, filled with Hi (G), the positive reaction rate decreases at the beginning--- pair, constant pressure, the vessel has a piston, filled with Hi, the piston will move outward, at this time the concentration of reactants H2 and I2 decreases, that is, the positive rate decreases.

c.Constant volume, increase temperature, decrease of positive reaction rate--- wrong temperature, increase temperature, positive rate and reverse rate are increased, but the amplitude of increase is not the same, the equilibrium moves in the direction of endothermy, indicating that the direction of endothermy increases more.

d.Constant volume, filled with H2, the volume fraction of I2 (g) decreases by --- pairs, constant volume, indicating that the volume of the container remains unchanged, adding H2, the equilibrium moves positively, i2 is converted, and the fraction of i2 decreases.

The first thing to note is that i2 is brown, aObviously, the number of gases on the left and right sides of the equation is 2, and changing the pressure will certainly not affect the equilibrium. But when the volume is compressed, the number of i2s per unit volume increases, so the color becomes darker.

b.Needless to explain, the reaction moves in the direction you react.

c.Since it is an endothermic reaction, the positive reaction speed will inevitably increase when the temperature rises. Wrong option.

d.When H2 is rushed, the positive reaction rate accelerates, and the volume fraction of I2 inevitably decreases. That's right.

The answer is c

c a, the volume difference before and after is 0, the balance is not affected by pressure, the balance does not move, the amount of I2 substance remains unchanged, but due to the decrease in volume, the concentration of I2 increases, and the color deepens.

B. Constant pressure, the amount of H2 and I2 substances remains unchanged, after filling into HI, the volume becomes larger, the concentration of H2 and I2 substances decreases, and the reaction rate decreases at the beginning.

c. Regardless of the temperature increase, the positive reaction rate and the reverse reaction rate are increased, regardless of whether the heat is absorbed or not.

D. Constant volume, filled with H2, the volume fraction of H2 increases, the volume fraction of I2 decreases, the equilibrium is positive, and the volume fraction of I2 is low.

c.As the temperature increases, the rates of both forward and reverse reactions increase.

c When the temperature increases, the rate of forward and reverse reactions must be accelerated.

Option d, calculate the conversion rate of the positive reaction of 60%, if the feeding ratio is the same, starting from the reverse reaction, it is said that the conversion rate of the reverse reaction is 40%.

However, the constant capacity sealing condition and the constant volume are the same, the original feeding is total, generated, and remaining after the reaction, while the D option is fed total, mol, and the feeding should also work hard in the direction of the remainder, so it will increase the number of mol and increase the volume in the direction. So the conversion rate of N2 must be greater than 40%.

You can see it by sending a ** alone.