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oksano4ka [1.4K]
2 years ago
6

There are 1.00 mol each of H2 and I2 in a 2.00 L flask. The Kc for this reaction is 55.3. Determine the equilibrium concentratio

ns of all chemical species at room temperature.
Chemistry
2 answers:
Oksana_A [137]2 years ago
5 0

The equilibrium concentrations of all chemical species at room temperature are given below.

[H2]  = 0.212M

[I2]  = 0.212M

[HI]  = 1.576M

H2(g)  +   I2(g) <-----> 2HI(g)

I              1          1                   0

c            -x        -x                  2x

E            1-x      1-x                 2x

[H2] ⇒ 1-x/1 = 1-x

[I2] ⇒ 1-x/1 = 1-x

[HI] ⇒ 2x

Kc = [HI]^2/[H2][I2]

55.3 = (2x)^2/(1-x)(1-x)

55.3 = (2x/1-x)^2

7.44 = 2x/1-x

7.44*(1-x) = 2x

x  = 0.788

[H2] ⇒ 1-x = 1-0.788 = 0.212M

[I2] ⇒ 1-x = 1-0.788 = 0.212M

[HI] ⇒ 2x = 2*0.788 = 1.576M

In a chemical reaction, while each reactant and the products are in a concentration that does not alternate with time any extra, it is said to be in a state of chemical equilibrium. in this kingdom, the price of forwarding response is the same as the rate of backward response.

Equilibrium is whilst the fee of the ahead reaction equals the charge of the opposite reaction. All reactant and product concentrations are consistent at equilibrium.

We are saying that a chemical is in an equilibrium concentration while the products and reactants do not change as time moves on. In different words, chemical equilibrium or equilibrium awareness is a state when the price of an ahead reaction in a chemical response will become equal to the rate of a backward response.

Learn more about equilibrium here: brainly.com/question/13414142

#SPJ4

Margaret [11]2 years ago
3 0

The equilibrium concentrations of all chemical species at room temperature are given below.

  • [H2]  = 0.212M
  • [I2]  = 0.212M
  • [HI]  = 1.576M

 H2(g)  +   I2(g) <-----> 2HI(g)

I              1          1                   0

c            -x        -x                  2x

E            1-x      1-x                 2x

[H2] ⇒ 1-x/1 = 1-x

[I2] ⇒ 1-x/1 = 1-x

[HI] ⇒ 2x

Kc = [HI]^2/[H2][I2]

55.3 = (2x)^2/(1-x)(1-x)

55.3 = (2x/1-x)^2

7.44 = 2x/1-x

7.44*(1-x) = 2x

x  = 0.788

[H2] ⇒ 1-x = 1-0.788 = 0.212M

[I2] ⇒ 1-x = 1-0.788 = 0.212M

[HI] ⇒ 2x = 2*0.788 = 1.576M

In a chemical reaction, while each reactant and the products are in a concentration that does not alternate with time any extra, it is said to be in a state of chemical equilibrium. in this kingdom, the price of forwarding response is the same as the rate of backward response.

Equilibrium is whilst the fee of the ahead reaction equals the charge of the opposite reaction. All reactant and product concentrations are consistent at equilibrium.

We are saying that a chemical is in an equilibrium concentration while the products and reactants do not change as time moves on. In different words, chemical equilibrium or equilibrium awareness is a state when the price of an ahead reaction in a chemical response will become equal to the rate of a backward response.

Learn more about equilibrium here: brainly.com/question/13414142

#SPJ4

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3 0
3 years ago
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A first-order reaction has a half-life of 16.7 s . How long does it take for the concentration of the reactant in the reaction t
andrew11 [14]

It takes 33.4 s for the concentration of A to fall to one-fourth of its original value.

A <em>half-life</em> is the time it takes for the concentration to fall to half its original value.

Assume the initial concentration is 1.00 mol/L. Then,

\text{1.00 mol/L }\stackrel{\text{1st half-life }}{\longrightarrow}\text{ 0.50 mol/L } \stackrel{\text{ 2nd half-life }}{\longrightarrow}\text{0.25 mol/L}\\

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