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3 years ago

the reaction below shows a system in equilibrium how would an increase in the concentration of chemicals affect this reaction?

Chemistry

1 answer:

Goryan [66]3 years ago

6 0

Answer:

Option B. The equilibrium of the reaction would shift to the right.

Explanation:

According to Le Chatelier's principle, when an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium position will shift in order to neutralize the effect.

Now, let us answer the question above.

From the question given above, the equation is:

A + B <===> C + D

Increasing the concentration of A will shift the equilibrium position to the right.

This is true because increasing the concentration of A means having more reactants in the system. So the reactants react to produce more product thereby shifting the equilibrium position to the right.

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Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

3 years ago

Energía necesaria para arrancar un electrón de un átomo neutro

zalisa [80]

La energía de ionización o potencial de ionización, es la energía necesaria para extraer un electrón de un átomo neutro. Por ejemplo, cuando un átomo de cloro neutro en forma gaseosa recoge un electrón para formar un ion Cl-, libera una energía de 349 kJ/mol o 3,6 eV/átomo.

Nomas ponlo En tus palabras

5 0

3 years ago

Metallic behavior is generally associated with elements with partially filled p orbitals. elements with unpaired electrons. elem

Mila [183]

Answer:

Elements with low ionization energies.

Explanation:

The ionization energy of an atom reffers to the amount of energy that is required to remove an electron from the gaseous form of that atom or ion.

The greater the ionization energy, the more difficult it is to remove an electron. The ionization energybis one of the indicator that shows the reactivity of an element. Elements with a low ionization energy such as metals are usually reffered to as a reducing agents and form cations, this give metals the tendency to

give away their valence electrons when bonding, whereas non-metals tend to take electrons.

Metallic elements have different properties such as shiny, heat and electricity conductivity . They are malleable and ductile Some metals, such as sodium, are soft and can be cut with a knife. while some are very hard such as iron.

4 0

2 years ago

Why is NX3 Unstable ? ...?

exis [7]

It is because of the small size and high electronegativity of nitrogen.
Nitrogen has very small size as compared to the halogens, which have much larger sizes. Due to this, they can not remain bonded to the nitrogen atom and hence are highly unstable.

3 0

3 years ago

Draw the addition product formed when one equivalent of hcl reacts with the following diene.

pashok25 [27]

Answer:

Major Product = 4-chloro-4-methylcyclohex-1-ene

Explanation:

Alkene are the class of organic compounds which contain one or more double bonds between two carbon atoms. Alkenes are considered most reactive among the unsaturated hydrocarbons and they undergo addition reactions due to high electron density around the double bonds.

In given question it is written that we are provided with one equivalent of HCl while, our compound contains two double bonds (diene) so in selected starting material the HCl will be added across (hydrohalogenation reaction) the substituted double bond because it will give a more stable carbocation (tertiary carbocation) during the reaction course. Hence, as shown in reaction scheme 4-chloro-4-methylcyclohex-1-ene will be the major product.

5 0

2 years ago