All categories

1 year ago

A reaction is in ____________ kinetics when the reaction rate depends only on enzyme concentration

Chemistry

1 answer:

Anvisha [2.4K]1 year ago

7 0

A reaction is in zero order kinetics when the reaction rate depends only on enzyme concentration.

<h3>What do we mean by chemical kinetics?</h3>

Understanding the speeds of chemical reactions is the focus of the physical chemistry field of chemical kinetics, commonly referred to as reaction kinetics. Thermodynamics, in contrast, deals with the direction in which a process happens but does not, by itself, reveal the pace at which it occurs. Chemical kinetics comprises studies of how experimental circumstances affect a chemical reaction's rate and reveal details about the reaction's mechanism and transition phases, as well as the creation of mathematical models that can also characterize a chemical reaction's features.

To know more about enzymes:

brainly.com/question/14953274

#SPJ4

You might be interested in

WHAT IS THE CHEMICAL NAME FOR ALKYNE HAVING

madreJ [45]

Answer:

Ee your house 6r2f5r56rrrr6gjyf

Explanation:

4 0

2 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2

Gekata [30.6K]

The image is not given in the question, it is attached below:

Answer: The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

Explanation:

In the given image:

Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms

The combination of 1 black and 2 red spheres will represent carbon dioxide $(CO_2)$ compound

The combination of 2 blue spheres will represent nitrogen molecule $(N_2)$

The combination of 1 blue and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

The combination of 1 black and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

We are given:

Given moles of NO = 6 moles

Given moles of CO = 4 moles

For the given chemical equation:

$2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)$

By stoichiometry of the reaction:

If 2 moles of CO reacts with 2 moles of NO

So, 4 moles of CO will react with = $\frac{2}{2}\times 4=4mol$ of NO

As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, CO is considered a limiting reagent because it limits the formation of the product.

Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

3 0

3 years ago

Which one is more acidic, Chloroacetic acid or Bromoacetic acid? And why?

Mama L [17]

Chloroacetic acid is stronger.

This is because it contains (more electronegative) chlorine atoms in place of (less electronegative) hydrogen atoms.

3 0

3 years ago

The following data were collected for the rate of disappearance of NO in the reaction 2NO(g)+O2(g)→2NO2(g)::

Anit [1.1K]

Answer:

a) The rate law is: v = k[NO]² [O₂]

b) The units are: M⁻² s⁻¹

c) The average value of the constant is: 7.11 x 10³ M⁻² s⁻¹

d) The rate of disappearance of NO is 0.8 M/s

e) The rate of disappearance of O₂ is 0.4 M/s

Explanation:

The experimental rates obtained can be expressed as follows:

v1 = k ([NO]₁)ᵃ ([O₂]₁)ᵇ = 1.41 x 10⁻² M/s

v2 = k ([NO]₂)ᵃ ([O₂]₂)ᵇ = 5.64 x 10⁻² M/s

v3 = k ([NO]₃)ᵃ ([O₂]₃)ᵇ = 1.13 x 10⁻¹ M/s

where:

k = rate constant

[NO]₁ = concentration of NO in experiment 1

[NO]₂ = concentration of NO in experiment 2

[NO]₃ = concentration of NO in experiment 3

[O₂]₁ = concentration of O₂ in experiment 1

[O₂]₂ = concentration of O₂ in experiment 2

[O₂]₃ = concentration of O₂ in experiment 3

a and b = order of the reaction for each reactive respectively.

We can see these equivalences:

[NO]₂ = 2[NO]₁

[O₂]₂ = [O₂]₁

[NO]₃ = [NO]₂

[O₂]₃ = 2[O₂]₂

So, v2 can be written in terms of the concentrations used in experiment 1 replacing [NO]₂ for 2[NO]₁ and [O₂]₂ by [O₂]₁ :

v2 = k (2 [NO]₁)ᵃ ([O₂]₁)ᵇ

If we rationalize v2/v1, we will have:

v2/v1 = k *2ᵃ * ([NO]₁)ᵃ * ([O₂]₁)ᵇ / k * ([NO]₁)ᵃ * ([O₂]₁)ᵇ (the exponent "a" has been distributed)

v2/v1 = 2ᵃ

ln(v2/v1) = a ln2

ln(v2/v1) / ln 2 = a

a = 2

(Please review the logarithmic properties if neccesary)

In the same way, we can find b using the data from experiment 2 and 3 and writting v3 in terms of the concentrations used in experiment 2:

v3/v2 = k ([NO]₂)² * 2ᵇ * ([O₂]₁)ᵇ / k * ([NO]₂)² * ([O₂]₂)ᵇ

v3/v2 = 2ᵇ

ln(v3/v2) = b ln 2

ln(v3/v2) / ln 2 = b

b = 1

Then, the rate law for the reaction is:

v = k[NO]² [O₂]

Since the unit of v is M/s and the product of the concentrations will give a unit of M³, the units of k are:

M/s = k * M³

M/s * M⁻³ = k

M⁻² s⁻¹ = k

To obtain the value of k, we can solve this equation for every experiment:

k = v / [NO]² [O₂]

for experiment 1:

k = 1.41 x 10⁻² M/s / (0.0126 M)² * 0.0125 M = 7.11 x 10³ M⁻² s⁻¹

for experiment 2:

k = 7.11 x 10³ M⁻² s⁻¹

for experiment 3:

k = 7.12 x 10³ M⁻² s⁻¹

The average value of k is then:

(7.11 + 7.11 + 7.12) x 10³ M⁻² s⁻¹ / 3 = 7.11 x 10³ M⁻² s⁻¹

The rate of the reaction when [NO] = 0.0750 M and [O2] =0.0100 M is:

v = k [NO]² [O₂]

The rate of the reaction in terms of the disappearance of NO can be written this way:

v = 1/2(Δ [NO] / Δt) (it is divided by 2 because of the stoichiometric coefficient of NO)

where (Δ [NO] / Δt) is the rate of disappearance of NO.

Then, calculating v with the data provided by the problem:

v = 7.11 x 10³ M⁻² s⁻¹ * (0.0750M)² * 0.0100M = 0.4 M/s

Then, the rate of disappearance of NO will be:

2v = Δ [NO] / Δt = 0.8 M/s

The rate of disappearance of O₂ has to be half the rate of disappearance of NO because two moles of NO react with one of O₂. Then Δ [O₂] / Δt = 0.4 M/s

With calculations:

v = Δ [O₂] / Δt = 0.4 M/s (since the stoichiometric coefficient is 1, the rate of disappearance of O₂ equals the rate of the reaction).

3 0

3 years ago