Chemical balance, rush and hurry

The state of chemical equilibrium has the characteristics of inverse, equal, dynamic, definite, and variable.

Inverse: The object of chemical equilibrium study is reversible reactions.

Equal: At equilibrium, the rate of forward and reverse reactions is equal, i.e., v positive = v inverse.

Dynamic: When equilibrium, the reaction is still ongoing, it is dynamic equilibrium, and the reaction is carried out to the maximum.

Constant: When the equilibrium state is reached, the concentration of each component in the reaction mixture remains unchanged, the reaction rate remains unchanged, the conversion rate of reactants remains unchanged, and the content of each component remains unchanged.

Variety: Chemical equilibrium, like all dynamic equilibrium, is conditional, temporary, relative, when the conditions change, the equilibrium state will be destroyed, from equilibrium to disequilibrium, and then a new equilibrium will be established under new conditions.

There are many factors that affect chemical balance.

Such as pressure, temperature, concentration, etc. (Note: The catalyst does not affect the chemical equilibrium, only the reaction rate).

Other things being equal, increasing the concentration of reactants or decreasing the concentration of products can shift the equilibrium in the direction of positive reaction.

Le Chatlet's principle: if you change one of the conditions that affect the equilibrium (concentration, pressure, temperature, etc.), the equilibrium will be able to weaken the change.

Without chemical equilibrium, this question has nothing to do with this. Item B says that the amount of I2 has decreased, think about a question, if the iodine is reduced, where does the point go, and it becomes S2?? Just kidding, if yes, then you can get Lobel.

If you look at the chemical equation carefully, it is reversible, so when it comes to low temperatures, then the reverse reaction will naturally be released to form I2, so the amount of i2 will not decrease. Now look at the third option, TaS2 is a solid, in the high temperature and gaseous I2 reaction to form two gases of substance, due to the air pressure, they flow to the low temperature zone, when they reach the low temperature zone just to reach the conditions for the reverse reaction to occur, then TaS2 solid and gas I2So C is correct.

Under adiabatic conditions, since the positive reaction is an exothermic reaction, as the reaction progresses and the temperature of the system increases, the positive reaction is inhibited, so that the conversion rate of reactants decreases.

Got it?

The amount of iodine remains the same.

In the high temperature zone, the reaction moves forward and is converted into a gaseous tantalum iodide, which fills the entire quartz tube.

Entering the low temperature zone, the reaction moves in reverse, and the gaseous tantalum iodide is converted into solid Tas2 to achieve the purpose of purification.

The amount of I2 is unchanged, Tas2 in the high temperature zone on the right reacts with I2, and after drifting to the low temperature zone, TaS2 and I2 are formed, so the concentration of I2 remains the same, but the impure TaS2 is transported from the right to the low temperature zone and becomes pure.

Select A, adiabatic, which is equivalent to heating up, positive reaction exothermy, so the conversion rate of equilibrium left shift B is reduced, so A is greater than B

(1) The second small void, due to the existence of a reversible reaction 2NO2(G)==N2O4(G), with the reaction B gas decreases, the volume decreases, and the A gas decreases, but the volume remains unchanged, then the medium pressure in A (1) the first small void and B medium pressure are strong, and the reaction rate is accelerated by increasing the pressure, V(A)(2) (A) will increase (equivalent to increasing the pressure), if the same amount of Ne gas is introduced, the conversion rate of NO2 in A will remain unchanged when equilibrium is reached (the filling of irrelevant gas has no effect on the reaction), The conversion rate of NO2 in B will be reduced (the volume of B will be increased by filling with irrelevant gas, which is equivalent to reducing the pressure).

3) Not comparable. At the beginning of the reaction, the volume of A and B is the same, the color is the same, and when the equilibrium is reached, A has more NO2 left, but the volume is large; There is little remaining NO2 in B, and the volume is small. Can't be compared.

4) The mass is conserved before and after the reaction, and the amount of the total substance of the mixed gas is obtained according to n=mmm=, before the reaction, according to the equation 2NO2==N2O4, the reaction is lost, generated, and remained.

Container A is equivalent to a constant volume, and container B is equivalent to a constant pressure [because the volume of the balloon is variable], so NO2 rushes into B and a reaction occurs, and 2NO2 (G) ==N2O4 (G) and the reaction moves in a positive direction.

NO2 rushes into A while the same reaction occurs, according to the Carabolon equation.

pv=nrt, where n is the amount of gaseous substance, decreases.

b In order to ensure that the pressure is constant, v on the right side of the equation must be reduced.

A In order to ensure that the volume is constant, it is equal to the right p must be reduced.

Because p is the same at the beginning, after equilibrium, p(a).< p(b) so when the k2 gas is opened, the gas in b flows into a, and the volume becomes smaller by < p>

The balance shifted to the left

Because hydrochloric acid will react with ammonia gas to form ammonium salt solids, the concentration of ammonia gas is reduced, and the reaction rate is reduced, so the reverse reaction rate is greater than the positive one, so the equilibrium shifts to the left.

1. What are the conditions for the reaction? 2.The above reaction has reached equilibrium, is there anything left in 02?

A is definitely true, and the conversion rate of B is definitely equal when a is added in 3 according to the ratio of chemical coefficients; B is also right, when A, B is added according to 2:1, the conversion rate is equal, in the addition of 1mola affirmation that the conversion rate of B is greater than A, C is wrong, under constant pressure, 3 is equivalent to adding 1molB in 1, so that the balance is shifted to the right, C increases, A decreases; D pair, 2 is equivalent to adding 1molb to 3, and the balance is shifted to the right, so C is chosen. The above is under constant pressure, just look at the coefficient ratio.

It is easier to choose BCA and B.

The conversion rate is the ratio of the amount of reactions that are likely to occur.

The conversion rate of a is equal to n(a), i.e., 1-n(c), n(a), and b's conversion rate is equal to n(b), i.e., 1-n(c) 2n(b), and the numbers are substituted into the above equation respectively.

The conversion rate of a in balance 3 is 1-n(c) 6 The conversion rate of b in balance 3 is 1-n(c) (2*3), which is 1-n(c) 6

So item A is correct.

The conversion rate of a in balance 1 is 1-n(c) 3 The conversion rate of b in balance 1 is 1-n(c) (2*1), which is 1-n(c) 2

It is easy to conclude that the n(c) 3c option uses the chemical reaction equilibrium constant k

The above three groups have the same equilibrium temperature and the same k.

k=n(c) 2 So 2=n(b)*k, which is the size of n(b) when comparing equilibrium, it is obvious that equilibrium 3 is greater than equilibrium 1, and as for term d, I haven't thought of how to explain it, so I thought of sending it again.

Choose C, it is best to use the limit method to try this kind of problem, which is commonly used in product and horizontal problems

BC bar! What is the state of the generated c? It's been too long since I've been exposed to chemistry and I don't remember!

Observe the equation Jianzhou 3h2 + n2 = 2nh3 and its coefficient ratio is 4:2

It shows that the pressure becomes larger, and the balance shifts to the right, and the first explanation holds.

As for the second explanation, it can be understood that if the concentration increases, it is said that although the concentration is moving to the left, the pressure in the container gradually increases as it moves, and the balance moves to the right.

So to sum up, the balance shifts to the right!

It's all gas. Only 4mol of gas produces 2mol of gas, and the total concentration decreases significantly, and the pressure increases.

The second case can be regarded as the premise of the first one, and then put in 2 moles of ammonia, at this time, due to the increase in pressure, the reaction can not be as complete as in the first type (that is, the 2 moles of ammonia put in later are not as much as the first 2 moles of ammonia), so the overall balance moves, relative to the first one, the second equilibrium shifts to the right.

As the concentration of the product increases, so does the concentration of reactants! ...... both left and rightI had to look at the pressurized ......

I don't agree with the upstairs statement, because it is clearly mentioned in the stem that it is all gaseous (e.g. does water have to be liquid?). Obviously, it depends on the stem), and the addition obviously affects the balance.

This kind of reactant is a kind, there are many kinds of products (or a variety of reactants, a kind of products), when the reactants (or products) are added, the container can be expanded equivalently to keep the pressure unchanged, then the balance is not moved, and then the volume is reduced to the original, then the pressure becomes larger, and the balance naturally moves in the direction of volume reduction, so the conversion rate is reduced. As the answer is.

This kind of problem can be solved by equivalence balance, that is, the above method can be solved.

There is another category of more troublesome problems. It is the dimerization of nitrogen dioxide into nitrogen tetroxide.

At the moment when nitrous tetroxide is added, the equilibrium will move in the opposite direction due to the sudden increase in concentration, but according to the equivalent equilibrium, the final result is a shift in the direction of the positive reaction, and the same can be said for phosphorus pentachloride.

Equivalence is a good way to do this, and you need to be flexible.

Of course, when the equilibrium constant can be quantitatively calculated, this problem can also be solved by using the concentration quotient.

Off topic: The imposed factor is greater than the balance factor.

In high school, just remember this, for example, the dimerization of nitrogen dioxide is nitrogen tetroxide, which expands in volume, and although the equilibrium shifts in the direction of nitrogen dioxide, it looks lighter in color.

In the synthesis of ammonia, nitrogen is added, the balance is moved positively, and the H2 conversion rate is larger, but the N2 conversion rate is reduced, because the impact of adding it is large (the actual production of excessive N2 is to increase the H2 conversion rate, because H2 is relatively expensive).

Hope it helps!

Yes, reactant conversion is the ratio of the amount of reactant that has been transformed to the amount of reactant that was before the reaction.

Although you increase the concentration of reactants, according to Le Chartrete's principle, the amount of the transformed reactant increases, but don't forget that the total amount of reactant also increases, generally speaking, the amount of the converted reactant is always less than the amount of the reactant you originally added, so the conversion rate decreases.

As for your answer, I'm sorry, I beg to differ.

The conversion rate of phosphorus pentachloride will be reduced because after the addition of phosphorus pentachloride according to Le Chatelier's principle (change one of the conditions that affect the chemical equilibrium such as: concentration, temperature, pressure. The chemical equilibrium will move in a direction that will weaken this change.

The result of the move: it can only weaken the change of external conditions, but cannot completely offset the impact of the change of external conditions. The equilibrium will move in the direction of the decrease in phosphorus pentachloride.

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.

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.

The first question uses the extreme value method, assuming that the reaction is completely reactive and completely non-reactive, and the relative molecular mass is calculated.

The second question is done in a three-stage form.

Let the reaction amol, the average relative molecular mass is equal to the total mass divided by the total number of moles m = m n

So bring in m = + 2*2 - 12a) ( +4 - a ).

78 - 12a）/( -a )

Mention 12 = 12 (-a).

a )