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Liula [17]
3 years ago
13

Given the following proposed mechanism, predict the rate law for the overall reaction. A2 + 2 B → 2 AB (overall reaction) Mechan

ism A2 ⇌ 2 A fast A + B → AB slow
Chemistry
1 answer:
nlexa [21]3 years ago
5 0

Answer:

The rate of the over all reaction is ;

R=K[A_2]^{1/2}[B]

Explanation:

Step 1 : A_2\rightleftharpoons 2 A fast

Step 2  : A + B\rightarrow AB slow

Equilibrium constant of the reaction in step 1:

K_1=\frac{[A]^2}{[A_2]}....[1]

Overall reaction:

A_2 + 2 B \rightarrow 2 AB

When there is a chemical reaction which taking place in more than 1 step than the rate of the over all reaction is determined by the slowest step occurring during that process;

Here step 2 is slow step, so the rate of the reaction will be;

R=k[A][B]..[2]

Putting value of [A] from [1] in [2]:

R=k\times \sqrt{K_1\times [A_2]}\times [B]

K=k\times (K_1)^{1/2}

K = rate constant of the reaction

The rate of the over all reaction is ;

R=K[A_2]^{1/2}[B]

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Masja [62]
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6 0
3 years ago
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Helga [31]

Answer:

<em> 14, 508J/K</em>

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

<em>m = mass of substance (g) = 0.1184g</em>

1 mole of Mg - 24g

<em>n</em> moles - 0.1184g

<em>n = 0.0049 moles.</em>

Also, q = m × c × ΔT

<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>

<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>

<em>= 14, 508 J/K/kg</em>

ΔT=  (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>

<em>= 1,7117.7472 J °C-1 g-1</em>

<em />

<em>∴ ΔHrxn = q/n</em>

<em>=1,7117.7472  ÷ 0.1184 </em>

<em>= 14, 508J/K</em>

6 0
3 years ago
Freezing point<br> Where would you place this property in the Venn diagram?
Marat540 [252]
For the Venn diagram, freezing water should be in the physical category (A)
7 0
3 years ago
Read 2 more answers
Actuary Tong has to study for two actuarial exams: Exam P and Exam FM. The amount of study time that Actuary Tong will spend on
Georgia [21]

The Probability of spending at least half an hour on exactly one subject is mathematically given as

P(1)= 0.50

<h3>What is the Probability of spending at least half an hour on exactly one  subject </h3>

Question Parameters:

The amount of study time that Actuary Tong will spend on each exam in a day follows a continuous random variable that ranges from 0 to 1 hour.

Generally, the equation for the Probability of spending at least half an hour on a subject is mathematically given as

P(A)= (1 – 0.5)/(1 – 0)

P(A)= 0.5

Hence, the Probability of spending at least half an hour on exactly one subject is mathematically given as

P(1)=2​​​​​​C1 * 0.5 * 0.5

P(1)= 0.50

For more information on Probability

brainly.com/question/795909

6 0
2 years ago
I NEED HELP PLEASE, THANKS! :)
krok68 [10]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
2 years ago
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