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Mashutka [201]
3 years ago
5

The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism

s for this reaction are: Mechanism A (1) H2(g)+I2(g)−→k12HI(g)(one-step reaction) Mechanism B (1) I2(g)⥫⥬=k−1k12I(g)(fast, equilibrium) (2) H2(g)+2I(g)−→k22HI(g) (slow) Mechanism C (1) I2(g)⥫⥬=k−1k12I(g)(fast, equilibrium) (2) I(g)+H2(g)−→k2HI(g)+H(g) (slow) (3) H(g)+I(g)−→k3HI(g) (fast) Which of these mechanisms are consistent with the observed rate law? mechanism A mechanism C mechanism B In 1967, J. H. Sullivan showed that this reaction was dramatically catalyzed by light when the energy of the light was sufficient to break the I−I bond in an I2 molecule. Which mechanism or mechanisms are consistent with both the rate law and this additional observation? mechanism A mechanism C mechanism B
Chemistry
1 answer:
MatroZZZ [7]3 years ago
5 0

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

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The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.

<h3>How do we calculate the partial pressure of gas?</h3>

Partial pressure of particular gas will be calculated as:

p = nP, where

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  • n is the mole fraction which can be calculated as:
  • n = moles of gas / total moles of gas

Moles will be calculated as:

  • n = W/M, where
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Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole

Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole

Mole fraction of hydrogen = 1 / (1+0.5) = 0.6

Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm

Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.

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An individual is hospitalized and the initial blood work indicates high levels of hco3- in the blood and a ph of 7. 47. This wou
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An individual is hospitalized and the initial blood work indicates high levels of HCO_{3} ^{-} in the blood and a pH of 7. 47. This would indicate the individual probably has compensated respiratory acidosis.

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1 year ago
Select the correct answer. Given: 2Al + 6HCl → 2AlCl3 + 3H2 If the chemical reaction produces 129 grams of AlCl3, how many grams
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The reaction produces 2.93 g H₂.

M_r:                        133.34  2.016

        2Al + 6HCl → 2AlCl₃ + 3H₂

<em>Moles of AlCl₃</em> = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃

<em>Moles of H₂</em> = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂

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7 0
3 years ago
Nitrogen and water react to form nitrogen monoxide and hydrogen, like this: N2(g) + 2H2O(g) → 2NO(g) +2H2(g) Also, a chemist fin
Thepotemich [5.8K]

Answer:

Kc for this reaction is 0.43

Explanation:

This is the equilibrium:

N₂(g) + 2H₂O(g) → 2NO(g) +2H₂(g)

And we have all the concentration at equilibrium:

N₂: 0.25M

H₂ : 1.3M

NO: 0.33M

H₂: 1.2M

They are ok, because they are in MOLARITY. (mol/L)

Let's make the expression for Kc

Kc = ( [NO]² . [H₂]² ) / ([N₂] . [H₂O]²)

Kc = (0.33² . 1.2²) / (0.25 . 1.2²)

Kc = 0.4356

In two significant digits. 0.43

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3 years ago
The molar mass of hydrogen molecules is 2g/mol and the molar mass of oxygen molecules is 32g/mol. What is the ratio of the avera
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Answer:

1

Explanation:

For an ideal gas, the average kinetic energy is given by:

Ek = (3/2)*n*R*T

Where n is the number of moles, R is the gas constant (8.31 J/mol*K), and T the temperature. The gases have the same number of moles, and the same temperature, so they will have the same average kinetic energy:

Ek = (3/2)*1*8.31*300

Ek =3739.5 J

So, the ratio between then is 1.

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