High School Chemistry The following reaction occurs by mixing the quantities of A and B of the same

Solution: The concentration of the substance c (c) = vt = according to the ratio of the stoichiometric number = the ratio of the amount concentration of the substance.

Then x: 2=x=2v(d)=c(d) t=, or the ratio of stoichiometric numbers = rate.

i.e. x:2=: x=2].

Let the concentration of the substance starting with a be y, then the concentration of the substance starting with b is also y3a(g) +b(g) = 2c(g)+2d(g) starting y y 0 0

Reaction for 5min

The concentration of y=a at the end of 5 min is: c(a)=

v（b）=c(b)/t=

I hope mine will help you in your studies!

The amount of substance in b is less, you know. The amount of the substance of b is reduced by the unit and the time is 5min, so it is divided by 5min. The rest will be counted by yourself, and you won't ask again.

According to the chemical equation, the molar ratio of reactants A and B is 3:1. Therefore, if we take the total number of moles of reactants A and B as 4N, then after the reaction is complete, the number of moles of A is 3N, the number of moles of B is n, the number of moles of product C is n, and the number of moles of product D is 2N.

According to the ideal gas equation of state PV = NRT, the partial pressure of reactants A, B, Product C, Product D in a 2 L vessel can be expressed as:

pa = 3nrt / 2v)

pb = nrt / 2v)

pc = nrt / 2v)

pd = 2nrt 2v) = nrt v brings pa and pb into the equilibrium constant expression of the reaction, yielding:

kc = pc] xc[d] 2 a] 3[b] due to the mixing of the masses of a and b and so on, so [a] = b], substitution of the roll balance and into the above equation, obtain:

kc = c] 3[d] 2 a] 4 denote the molar numbers of c, d, and a by n, and substitute them into the upper staring formula, and get:

kc = n 1 4n] 4 = 1 256 Therefore, mixing an equal amount of A and B in a 2 L airtight container results in products C and D, which have an equilibrium constant of 1 256 kc.

(1) Concentration of A at the end of 5 min = mol l

2）v（b）=

3) x=2 analysis: let the amount of substances a and b be n5min to generate n(d)=

3a(g) +b(g) = xc(g)+2d(g) start: n n 0 0

Change: After 5min:

Because c(a):c(b)=3:5

Then (:5 solves: n=

So the concentration of (1) a at the end of 5 min = ( mol l(2) v(b) =

3）v（c）=

Solution: x=2

Hope it helps.

The concentration of d is then in a 2L container to produce a total of 1mol of d and molar AB, respectively.

Then according to the average reaction rate of C is mol (l·min) 0 0 in a container of 2L in 5 minutes, the amount of substance that generates C is 1 mol of D at the same time, and 1 mol of C is also generated, so the coefficient of Cd is the same, that is, x=2(1) from the title C(A): C(B)=3:5, let the amount of the original Ab substance be n to generate 1 mold needs to consume and :

5 So n=3

The concentration of a at 5 min is ( = .

The amount of substance starting with AB is 3 moles.

2）v（b）=

3) At the same time as 1 mol d is generated, 1 molar c is also generated, so the coefficient of cd is the same, i.e. x = 2.

The rate at which d is pushed out from the concentration after 5 min is the same as that of c, so x=2;Then calculate the consumption of AB according to the ratio, and then obtain the remaining AB after 5min.

c *4min=

3a(g)+ b(g)=== xc(g)+ 2d(g)

Start a a 0 0

Change 4 min after a-a-

c(a):c(b)=3:5 =( a- )a- )a=

a a- = -

Meter-to-number ratio = rate ratio = varying concentration ratio.

x:2= : x =2

v(b):v(c)=1：2 v(b): 2 v(b)= mol/(l*min)

Average reaction rate of d = average reaction rate, so average reaction rate of x = average reaction rate of 2b = average reaction rate of c 2=, 3a(g) +b(g)===2c(g) +2d(g), c-3x c-x 2x 2x2x=x=

c = the concentration of the amount of a substance =

Mixing the quantities of substances A and B in a closed container of 2 liters should be the same amount of substances, right?

The concentration of D measured after 2L with 5mln is, C(A):C(B)=3:5, and the average reaction rate of C is.

Calculate d 1mol,c at this time

Therefore, the coefficient is the same, and the value of x is 2

Find: (1) the concentration of A at this time and the amount of substances A and B put into the container before the reaction starts, and (2) the average reaction rate of B? (3) What is the value of x?

-3a(g) +b(g)=2c(g)+2d(g), starting a....A change 1mol1mol

Later(c(a):c(b)=(:(5

a=3mol

1) The concentration of a at this time =

Before the start of the reaction, the amount of substance A and B was put into the container 3mol2), the average reaction rate of B.

3a(g)+b(g) = xc(g)+2d(g) The amount of substance in D after 5 minutes = mol·l-1*2l=1mol There is a reaction formula to know, and the reactant consumed at this time.

a consumption = 3 2= b consumption = 1 2 =

Let the amount of the original reactant a=b=y

According to the meaning of the question, (y-a consumption) :(y-b consumption) = 3:5 solution to obtain y=3mol

At this time, the concentration of a [a] = (y-a consumption) 2l =

The second question is to find the value of x

The concentration of c at 5 minutes [c] = mol·l-1·min-1*5min = has the relationship between c and d, and it can be seen that [c]:[d]=x:2 is brought into [c]=[d]=, and x=2 is obtained

1) A concentration at the end of 5min =

2）v（b）=

3) x=2 analysis: let the amount of a and b of the initial isomatter He Huishan be n5min and then generate n(d)=

3a(g) +b(g) = xc(g)+2d(g) start: n n 0 0

Change: After 5min:

Change to c(a): c(b) = 3:5

Then (:5 solves: n=

So (Zen skin answer 1) the concentration of a at the end of 5 min = (

2）v（b）=

3）v（c）=

Solution: x=2

Hope it helps.

Solution: The concentration of d is, so the amount of the substance of d is 1mol Let the amount of the substance of a and b be nmol, 3a(g) +b(g)=xc(g)+2d(g), start n mol n mol 0 0

Participate in the reaction culture grip 1mol

After 4 min ( 1mol

c(a):c(b)=( : 2l =3:5, so the amount of n=3a at the end of 4min is: (n-1,5)mol=, and the amount concentration of the substance is .

The concentration of d is , the reaction time is 4min, the reaction rate of d is , the reaction rate of c is that according to the same reaction, the ratio of the reaction rate of each substance is equal to the ratio of the number of measurements, so x=2;

The concentration of d is, so the amount of the substance of d is 1mol

Let the amount of substances of a and b be nmol, 3a(g)+b(g)=xc(g)+2d(g), and n mol n mol 0 0 before the start of the bond

Reaction 1mol

5min （ 1mol

The amount of a substance at the end of 5min is: (n-1,5)mol=, and the concentration of the substance is =;

The concentration of d is, the reaction time is 5min, the reaction rate of d is , the reaction rate of c is that according to the same reaction, the ratio of the reaction rate of each substance is equal to the ratio of the number of measurements, so x=2;

The reaction rate of c is that according to the ratio of the reaction rate of each substance in the same reaction is equal to the ratio of the number of measurements, so the rate of counter-belief of b is;

So the answer is:; 2；；