Chemical equilibrium 2SO2 O2 2SO3 conversion rate problem in the constant temperature and volume con

2SO2 + O2 = 2SO3, when 2 Mo SO2, 1 Mo O2, when the equilibrium is reached, the SO2 conversion rate is A%, 2 Mo SO2, 1 Mo O2 is re-equilibrium SO2 conversion rate is B, after re-equilibrium is equivalent to 4 Mo SO2, 2 Mo O2 at the beginning, and the equilibrium SO2 conversion rate is B%, if the volume of this container is V.

Then after the equilibrium of 4 Mo SO2 and 2 Mo O2 is filled into a 2V container, the equilibrium is equivalent to that of 2 Mo SO2 and 1 Mo O2 in the container with volume V. That is, the 2V container is filled with 4 mos of SO2, and after the equilibrium of 2 mos of O2, the SO2 conversion rate is a%.

The 2V container is filled with 4 mos of SO2, and after the equilibrium of 2 mos of O2, the compressed volume is V. Since the equilibrium is to reduce the volume and compress the volume, this equilibrium is shifted to the right, that is, the SO2 conversion rate will increase! And the conversion rate is b%.

That is, b% > a%.

and your answer is not consistent.

First of all, assuming that the volume of container 1 is V, and it is filled with 2 moles of SO2 and 1 mole of O2, when the equilibrium is reached, the SO2 conversion rate is a%.Then assume that the container 2 with a volume of 2V is filled with 4 Mo SO2 and 2 Mo O2, and the two equilibrium are equivalent. At this time, the SO2 conversion rate is a%.

Then compress the container 2 to the volume V, increase the pressure, the equilibrium moves in the direction of volume reduction, and the conversion rate of SO2 is B%, B% >A% again

In the constant temperature and volume vessel, the reactants or products are rushed in proportionally, which is equivalent to pressurizing the system, and the conversion rate increases for the positive movement of the reaction equilibrium.

Pick B. The amount of SO2 in equilibrium is 20*(1-80%)=4mol

For reactions with volume changes before and after the reaction, if you want to achieve equilibrium, the volume fraction of each component is the same as the original, and the SO3 needs to be converted into the original ratio, so 20mol SO3 is required.

The SO2 requirement for equilibrium is 16mol

2so3<*****>2so2+o2

Stoichiometric number 2 2 1

Starting 20 0 0

Variation 4 4 2

Balance 16 4 2

The conversion rate of SO3 is 4 20*100%=20%.

Original question: A 10mol and 10% B 20mol and 20%.

c 20mol and 40% d 30mol and 80%.

Answer: B According to the equivalent equilibrium, the conversion rate of 20mol and 20mol is equivalent to 80% for 20mol and 10mol respectively at the beginning, indicating that the amount of SO2 participating in the reaction is 16mol, then 16 mol SO3 is generated. That is, when the starting substance is SO3, the amount of species participating in the reaction is 20mol-16 mol="4"mol, so the conversion rate is 20%.

Detailed analysis: The amount of SO2 in equilibrium is 20*(1-80%)=4mol

When the equilibrium is reached, the volume fraction of each component is the same as the original, and the original proportion of SO3 needs to be converted into 20molSO3

The SO2 requirement for equilibrium is 16mol

Stoichiometric number 2 2 1

Starting 20 0 0

Variation 4 4 2

Balance 16 4 2

The conversion rate of SO3 is 4 20*100%=20%.

So choose B.

Option B is an equivalence balance issue.

Adopt a one-sided calculation. Assuming that SO3 is completely converted to SO2 and O2, we can get and then the maximum value of SO2 is O2 and the maximum value of O2 is (which is actually not reached)Assuming that SO2 and O2 are completely converted to SO3, we can get it, then the maximum value of SO3 is the range of the remaining SO2 The range of O2 The range of SO3 is true and b.

c s is conserved, and the sum is equal to wrong.

When d is converted to O2, we can get the answer that so that the remaining SO3 is O2 and the remaining D pair AD

1. Increasing C does not affect the reaction equilibrium, and the concentration remains unchanged.

2. Decreasing the volume is equivalent to increasing the pressure, and the increase in CO is more obvious, that is, the equilibrium moves in the direction of the reverse reaction; CO2 concentration increases.

3. The introduction of nitrogen does not affect the concentration of reactants, the equilibrium does not move, and the concentration remains unchanged.

4. N2 makes the concentration of reactants decrease, and the CO decreases more obviously, so the equilibrium moves to the positive reaction direction; Because the volume increases, the amplitude of the volume change is greater than the increase in the CO2 concentration due to the equilibrium shift, so the CO2 concentration decreases overall.

Found the original question: the concentrations at a certain time are and respectively.

Option B assumes that the reaction group remainder is completely to the left. At most, there is SO2 and O2, which should be less than this value, so A is wrong;

When SO2 is, as long as the SO3 is converted, and the remaining SO3, so it is possible;

Obviously, according to the conservation of sulfur, the total amount of SO2 and SO3 is maintained at a certain amount, so C is wrong;

Assuming that the reaction collapses and rolls to the right to fully react, there is a maximum of SO3, which should actually be less than this value, so D is wrong.

k only changes with temperature, and the temperature does not change k.

When oxygen is added, the oxygen concentration increases, and the denominator in the concentration quotient increases, so q becomes smaller q

T does not change, so K does not change, the reaction of adding O2 proceeds positively, Q increases, and D is selected

I found the answer to the original question: the concentrations at a certain time are and respectively.

Option B assumes that the reaction is completely to the left. At most, o2 is, and the actual value should be less than that, so a is wrong;

When SO2 is, as long as the SO3 of the collapse roll is converted, and the SO3 of the remaining group L remains, so it is possible;

Obviously, according to the conservation of sulfur, the total amount of SO2 and SO3 is maintained at a certain amount, so C is wrong;

Assuming that the reaction is completely to the right, at most there is, but the actual value should be less than that, so d is wrong.

The essence of the equilibrium movement of the chemical anti-forest cover should be to weaken the reverse of this change!

In this problem, SO3 is added, that is, SO3 is added, and in order to reduce this change, the balance is shifted to the left, that is, in the direction in which SO2 is generated. However, the result of the chemical equilibrium shift does not completely prevent the increase of SO3, which is only a little resistant, such as adding 5 units of SO3, the balance moves, so that a small part of the SO3 decomposes, and finally reaches another equilibrium or the amount of SO3 increases a lot, and the amount of S02 increases a little, volume fraction = the content of a substance The total content of the substance, the total content increases by 5, that is, the denominator increases greatly, and the SO2 content increases a little, that is, the molecule increases less, So although it increases, the proportion decreases.

There should be a lot of illustrations of the chemical balance you learned in high school, and it is recommended that you ponder it carefully and grasp the essence of the problem.