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Helen [10]
3 years ago
12

The equilibrium-constant expression is used to describe the concentration of reactants and products for a reaction in dynamic eq

uilibrium. For ideal gases and ideal solutions in homogeneous equilibria, where all reactants and products are in the same phase, the extent to which a particular chemical reaction proceeds to products is given by the equilibrium equation.
aA+bB⇌cC+dD , K=[C]c[D]d[A]a[B]b

where K is the equilibrium constant and the right-hand side of the equation is known as the equilibrium-constant expression.

The concentration of each product raised to its coefficient is divided by the concentration of each reagent raised to its coefficient according the the balanced chemical equation. Therefore, the higher the concentration of products, the larger the value of K will be.

Required:
Identify the proper form of the equilibrium-constant expression for the equation.
N2(g)+O2(g)⇌2NO(g)
Chemistry
1 answer:
Lunna [17]3 years ago
8 0

Answer: K=\frac{[NO]^2}{[N_2]^1[O_2]^1}

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K

For the given chemical reaction:

N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

The expression for K is written as:

K=\frac{[NO]^2}{[N_2]^1[O_2]^1}

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The nucleus of 8Be, which consists of four protons and four neutrons, is very unstable and spontaneously breaks into two alpha p
Mariana [72]

Answer: The given statement is false.

Explanation:

When there is more difference in the ratio or number of protons and neutrons then nucleus of the atom becomes unstable in nature. This unstability is caused due to greater repulsion between the like charges of sub-atomic particles.

As a result, this force of repulsion becomes greater than the binding energy. And, this force is known as the weak force because it is unable to bind the neutrons and protons together.

The proton and neutron ratio for smaller elements is 1:1 and for higher elements it has to be 1:5. Since, ^{8}Be is a smaller element with 4 protons and 4 neutrons. Hence, the proton and neutron ratio is 1:1.

Therefore, ^{8}Be is stable in nature.

Thus, we can conclude that the statement nucleus of 8Be, which consists of four protons and four neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of two protons and two neutrons), is false.

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3 years ago
Elaborate on how the isotopes and their relative abundances affect the average atomic mass of an element.
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What are chemical symbols and chemical formulas used for
lozanna [386]

Answer:

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Explanation:

4 0
3 years ago
Which of the following are true statements about equilibrium systems?For the following reaction at equilibrium:2 H2(g) + O2(g) ?
VMariaS [17]

These are five questions about equilibrium systems each with its complete answer.

<u>Question 1</u><u>.</u> For the following reaction at equilibrium:

2 H₂(g) + O₂(g) ⇄ 2 H2O(g),  the equilibrium will shift to the left if the volume is doubled?

Answer: TRUE

Explanation:

When a force disturbs a chemical <em>equillibrium</em>, the system will shift toward the direction that <em>reduces the effect</em>. This is Le Chatelier's principle.

As per Bolye's law, at constant temperature, the volume and the pressure of a fixed amount of gas are inversely related.

Also, the pressure of the system is directly related to the number of particles (atoms or molecules). Hence, more molecules, more pressure; less molecules, less pressure.

Now, you can reason in this way: if the volume of the given system is doubled, then the pressure is lowered, and the system will try to alleviate this disturbance by shifting the reaction to the side that produces more molecules, to restore the pressure.  Because on the left side three molecules can be produced from the reaction of two molecules of H₂O on the rihgt, <em>the system will shift to the left</em>. And this proves the truth of the statement.

<u>Question 2</u>. For the following reaction at equilibrium:

H₂(g) + F₂(g) ⇄  2HF(g), removing H₂ will decrease the amount of F₂ present once equilibrium is reestablished.

Answer: FALSE.

Explanation:

Note that, since the temperature and other conditions have not changed, the equilibrium constant, Ke, has not changed. And, for the given equilibrium, Ke is given by the following equation.

  • Ke = [ H₂] [F₂] / [HF]²

Hence, to keep Ke unchanged, when removing H₂, the amount of F₂ present once equilibrium is reestablished will have to increase.

This is the opposite of the stated on the question, so the statement is false.

<u>Question 3.</u> Increasing the temperature of an exothermic reaction shifts the equilibrium position to the right.

Answer: FALSE.

Explanation:

You can write an <em>exothermic equlibrium</em> placing heat as a product on the right side of the equation; in this way:

  • A + B ⇄ C + D + heat

There, treating the heat as another product, you can reason that increasing the temperature, which is equivalent to supplying heat, will shift the equilibrium to the left side to consume heat, instead to the proposed by the statement. So, this is a false statement.

<u>Question 4</u>. For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), adding more CaCO₃ will shift the equilibrium to the right.

Answer: TRUE.

Explanation:

CaCO₃(g) is the only reactant of the forward reaction.

Adding more CaCO₃ may be seen as a disturbance against which the system will act by consuming it and producing more CaO and CO₂.

So, the forward reation will be favored and you conclude that <em>adding more CaCO₃ will shift the equilibrium to the right.</em>

<u>Question 5.</u> For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), increasing the total pressure by adding Ar(g) will have no effect on the equilibrium position.

Answer: TRUE.

Explanation:

In accordance to Le Chatelier's principle, increasing the pressure should be addresed by the equilibrium by shifting to the side where such pressure increase could be released.

That is possible when the number of molecules of gases on both sides are different: the equilibrium will shift to the side where more molecules less molecules are produced.

But, when the stoichiometry of the reaction shows the same number of molecules on both sides, which is the case in the given equilibrium, increasiing (or decreasing) the pressure will have no effect on the equilibrium position. Then, the answer is true.

8 0
3 years ago
A solution of rubbing alcohol is 78.5 % (v/v isopropanol in water. how many milliliters of isopropanol are in a 90.8 ml sample o
poizon [28]
.785 x 90.8mL = 71.3 mL
4 0
3 years ago
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