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pishuonlain [190]
2 years ago
8

Does anyone have answers of a worksheet called “ Intro to Graphs” Chemquest 2?

Chemistry
1 answer:
r-ruslan [8.4K]2 years ago
8 0

Answer:

no

Explanation:

Because I have lost and I dont know where is the worksheet

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the particles of solids move but very slowly.

the particles of liquids move moderately fast

the particles of gas move very fast.

The reasons for this movement is the space the particles are together. Since there is not move space between solids, the particles move slowly, water there is moderate space, and air there is a lot of space.

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1.A telescope is a tool that helps people observe distant objects. In the simplest telescope design, incoming light hits a lens
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3 years ago
Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)
Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g)              K₁= ?                       (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g)                 K₂= 3.61                 (2)

2 A(g) + D(g)  ⇄ C(g)             K₃= 7.19                 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3)  algebraically  to get our desired equilibrium (1).

We are allowed to reverse  reactions, in that case we take the reciprocal of K as our new K' ; we can also  add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this  in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3)  and we want D as a reactant . That  suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B  as product in our  equilibrium (1) . Finally we would add (2) and (3) to get  (1) which is our final  goal.

2C(g)             ⇄  2A(g) + 2B(g)  K₂´= ( 1/ 3.61 )²  

                                   ₊

2 A(g) + D(g)  ⇄     C(g)               K₃ = 7.19  

<u>                                                                                    </u>

C(g) + D(g)     ⇄    2B(g)       K₁ = ( 1/ 3.61 )²   x  7.19

                                             K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves  gases.

To compute the last part lets setup the following mnemonic  ICE table to determine the quantities at equilibrium:

pressure (atm)        C             D           B

initial                     1.64          1.64         0

change                    -x             -x        +2x

equilibrium          1.64-x         1.64-       2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)²  where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x  we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

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