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2 years ago

8

Consider the mechanism. step 1:step 2:overall:2a⟶bb c⟶d2a c⟶dslowfast determine the rate law for the overall reaction, where the

overall rate constant is represented as k.

1 answer:

Ne4ueva [31]2 years ago

3 0

Answer:

Rate: R = k [A]²

Explanation:

The rate law equation for a given complex chemical reaction is given as the product of molar concentrations of the reactant raised to the power their respective partial reaction orders. The sum of the partial reaction orders is equal to the overall order of reaction.

For a complex reaction, the rate law is generally determined by the slowest step, which is known as the rate-determining step.

<u>Given reaction mechanism:</u>

Step 1: 2 A⟶ B, slow step

Step 2: B+ C⟶ D, fast step

Overall: 2 A + C ⟶ D

In this given reaction mechanism, the <em>step 1 is the slow step and thus the rate determining step.</em>

Therefore, the rate law for the given mechanism is:

Rate: R = k [A]²

Here, k is the overall rate constant

Also, the overall order of the reaction = 2

Therefore, the given chemical reaction is said to be second order reaction.

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Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)

Sedbober [7]

Answer:

The yield of the product in gram is $\mathsf{{w_P}=0.26 \ gram}$

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = $2.7 \times 10^{-3} \ moles$

We know that:

Number of moles = mass/molar mass

Then:

$2.7 \times 10^{-3} = \dfrac{ mass}{257.997}$

$mass = 2.7 \times 10^{-3} \times 257.997$

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = $\dfrac{n_P}{n_R}\times 100\%$

where;

$n_P = \dfrac{w_P}{m_P}$ and $n_R = \dfrac{w_R}{m_R}$

Theoretical yield = $\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

Given that the theoretical yield = 100%

Then:

$100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

$\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}$

${w_P}=\dfrac{w_R \times m_P}{m_R}$

where,

$w_P$ = derived weight of the product

$m_P =$ the molecular mass of the derived product

$m_R =$ the molecular mass of the reagent

$w_R$ = weight in a gram of the reagent

${w_P}=\dfrac{w_R \times m_P}{m_R}$

${w_P}=\dfrac{0.697 \times 96.2}{257.997}$

$\mathsf{{w_P}=0.26 \ gram}$

8 0

2 years ago

What is the average kinetic energy and rms speed of N₂ molecules at STP? Compare these values with those of H₂ molecules at STP.

Otrada [13]

The average kinetic energy and rms speed of N₂ molecules at STP is $5.65686 \times 10^{-21}$ and $493 \mathrm{~m} / \mathrm{s}$

Given,

$\begin{aligned}&\mathrm{P}=1.013 \times 10^{5} \mathrm{~Pa} \\&\mathrm{~T}=273.15 \mathrm{~K} \\&\rho=1.25 \mathrm{~kg} / \mathrm{m}^{3}\end{aligned}$

The average kinetic energy of a molecule is given by, $K . E .=\frac{3}{2} k T$ where k is the Boltzmann constant and Tis the absolute temperature of the gas.

$K . E .=\frac{3}{2} \times $1.380649 \times 10^{-23}$ \times 273.15 K$

$K.E=$5.65686 \times10^{-21}$$

The rms speed of $\mathrm{N}_{2}$ molecules is given by

$\begin{aligned}&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \times 1.013 \times 10^{5}}{1.25}}=493 \mathrm{~m} / \mathrm{s}\end{aligned}$

The average kinetic energy of a gas's particles is inversely related to its temperature. As the gas warms, the particles must travel more quickly since their mass is constant.

The average kinetic energy (K) is equal to one half of the mass (m) of each gas molecule times the RMS speed (vrms) squared.

Learn more about average kinetic energy brainly.com/question/1599923

#SPJ4

6 0

1 year ago

For the reaction Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)

mojhsa [17]

Answer : The value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = 151.2 kJ = 151200 J

$\Delta S^o$ = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(151200J)-(328.0K\times 169.4J/K)$

$\Delta G^o=95636.8J=95.6kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = 95636.8 J

R = gas constant = 8.314 J/K.mol

T = temperature = 328.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k$

$k=1.70\times 10^{15}$

Therefore, the value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

3 0

2 years ago

How does energy change form when friction causes the temperature of an object to increase?

Oksanka [162]

I believe the answer would be through the movement of the molecules, the higher the temperature the more the molecules move around and vibrate.

7 0

2 years ago

How would the addition of protons affect the concentration of CH3COOH? How would the addition of OH– affect the amount of CH3COO

Deffense [45]

Answer:

1) increase concentration

2) decrease the amount

3) decrease the concentration

4) it would increase

Explanation: edge 2021

8 0

2 years ago

Read 2 more answers