All categories

2 years ago

What is the rate constant of a reaction if rate = 1 x 10-2 (mol/L)/s, [A] is 2 M,

Chemistry

2 answers:

boyakko [2]2 years ago

8 0

The rate constant of a reaction : 8.3 x 10⁻⁴

<h3>Further explanation</h3>

Given

rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M, [B] is 3 M, m = 2, and n = 1

Required

the rate constant

Solution

For aA + bB ⇒ C + D

Reaction rate can be formulated:

$\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}$

the rate constant : k =

$\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}$

Firlakuza [10]2 years ago

8 0

Answer:

8.3x10^10-4

Explanation:

a p e x

You might be interested in

Question:

Elina [12.6K]

H2SO4 + ZN ------- ZNSO4+ H2

(SO4)²The sulphate salt is formed......

Hope it helps

8 0

3 years ago

A reversible reaction is one where A) there are large changes in the net free energy from substrate to product. B) there is litt

laila [671]

Answer:

A reversible reaction is one where B) there is little change in the net free energy between substrate and product.

Explanation:

A reversible reaction is one that reagents are transformed into products and at the same time products are transformed into reagents. That is to say that as the products appear in the reaction, they can react with each other by regenerating the reagents again. It is represented by a double arrow, indicating that the reaction can be carried out both in one direction and the other way around.

At the start of the reaction, there is a large amount of reagents. As time goes by, that amount decreases and speed too.

On the other hand, at the beginning of the reaction there are no products. As the reaction happens, the products are being formed and their speed will increase to match the speed of the reagents. When the rates of products and reagents are equal and constant, it is possible to say that the reaction is in chemical equilibrium. At this point, both reactions continue to happen, but the total concentrations of reagents and products no longer change.

The Gibbs free enthalpy or free energy of a system is a measure of the amount of usable energy (energy that a job can perform) in that system.

When a reaction system is in chemical equilibrium, it is in the lowest possible energy state (it has the lowest possible free energy). If a reaction is not in equilibrium, it will move spontaneously towards it because that allows it to reach a state of lower and more stable energy. Then when the reaction moves towards equilibrium, the free energy of the system decreases more and more.

Finally, a reversible reaction is one where B) there is little change in the net free energy between substrate and product.

4 0

3 years ago

What is the most common workplace for people in the Finance cluster?

Romashka [77]

https://www.onetonline.org/find/career?c=6

8 0

3 years ago

Read 2 more answers

Define motion of particles in matter

Wittaler [7]

Answer:

Explanation:

The states that all of the particles that make up matter are constantly in motion. As a result, all particles in matter have kinetic energy. The kinetic theory of matter helps explain the different states of matter—solid, liquid, and gas. ... Particles do not always move at the same speed.

5 0

3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.