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Elena L [17]
3 years ago
11

How to set up the rate expressions for the following mechanism?

Chemistry
1 answer:
Anestetic [448]3 years ago
3 0

Answer:

Explanation:

From the given information:

A → B k₁

B → A k₂

B + C → D k₃

The rate law = \dfrac{d[D]}{dt}=k_3[B][C] --- (1)

\dfrac{d[B]}{dt}=k[A] -k_2[B] -k_3[B][C]

Using steady-state approximation;

\dfrac{d[B]}{dt}=0

k_1[A]-k_2[B]-k_3[B][C] = 0

[B] = \dfrac{k_1[A]}{k_2+k_3[C]}

From equation (1), we have:

\mathbf{\dfrac{d[D]}{dt}= \dfrac{k_3k_1[A][C]}{k_2+k_3[C]}}

when the pressure is high;

k₂ << k₃[C]

\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_3[C]}= k_1A \ \  \text{first order}

k₂ >> k₃[C]

\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_2}= \dfrac{k_1k_3}{k_2}[A][C] \ \  \text{second order}

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The conversion of methyl isonitrile to acetonitrile in the gas phase at 250 °C CH3CN(g) is first order in CH3NC with a rate cons
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Answer: The concentration of CH_3NC will be 1.56\times 10^{-2}M after 416 seconds have passed.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

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k = rate constant = 3.00\times 10^{-3}s^{-1}

t = age of sample = ?

a = let initial amount of the reactant = 5.45\times 10^{-2}M

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t=\frac{2.303}{3.00\times 10^{-3}}\log\frac{5.45\times 10^{-2}}{1.56\times 10^{-2}}

t=416s

The concentration of CH_3NC will be 1.56\times 10^{-2}M after 416 seconds have passed.

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5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------&gt; 1 Mg3(
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Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

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