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lisabon 2012 [21]
3 years ago
5

The reaction between nitrogen and oxygen is given below: 2 N2(g) + O2(g) 2 N2O(g) We therefore know that which of the following

reactions can also occur? 2 N2O(g) 2 N2(g) + O2(g) N2(g) + 2 O2(g) 2 NO2(g) 2 NO2(g) N2(g) + 2 O2(g) None of the Above
Chemistry
1 answer:
Ostrovityanka [42]3 years ago
3 0

Answer:

  • a)  2N₂O(g) → 2N₂(g)  + O₂(g)

Explanation:

Arrange the equations in the proper way for better understanding.

T<em>he reaction between nitrogen and oxygen is given below:</em>

<em />

  • <em>2N₂(g) + O₂(g) → 2N₂O(g)</em>

<em />

<em>We therefore know that which of the following reactions can also occur?</em>

<em />

  • <em>a)  2N₂O(g) → 2N₂(g)  + O₂(g)</em>
  • <em>b)  N₂(g) + 2O₂(g) → 2NO₂(g)</em>
  • <em>c)  2NO₂(g) → N₂(g) + 2O₂(g)</em>
  • <em>d) None of the Above</em>

<h2>Solution</h2>

Notice that the first equation,  a) 2N₂O(g) → 2N₂(g)  + O₂(g), is the reverse of the original equation, 2N₂(g) + O₂(g) → 2N₂O(g).

The reactions in gaseous phase are reversible reactions that can be driven to one or other direction by modifying the conditions of temperature or pressure.

Thus, the equilibrium equation would be:

  • 2N₂(g) + O₂(g) ⇄ 2N₂O(g)

Which shows that both the forward and the reverse reactions occur.

Whether one or the other are favored would depend on the temperature and pressure: high temperatures would favor the reaction that consumes more heat (the endothermic reaction) and high pressures would favor the reaction that consumes more moles.

Thus, by knowing that one of the reactions can occur you can conclude that the reverse reaction can also occur.

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When we stand on the floor, we apply a force on the floor surface in the downward direction and in return the floor also exerts an upward and equal force on us.

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Newton' third law is vey famous and it states that for each and every action, there applies an equal but opposite reaction. Thus the action force and the reaction force always acts on pairs. But they does not contribute to the motion of the object.

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Decide which of the following statements are True and which are False about equilibrium systems:A large value of K means the equ
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Answer:

a. True

b. False

c. True

d.  False

e. False

Explanation:

A. (true) The equilibrium constant K is defined as

\frac{Products}{reagents}

In any case  

aA +Bb ⇌ Cd +dD

where K is:

K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}

A large value on K means that the concentration of products is bigger than the concentrations of reagents, so the forward reaction is favored, and the equilibrium lies to the right.

B. (False) When we work with gases, we use partial pressure to make calculations in the equilibrium, so we estimate Kp as:

Kp= \frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}

Using the ideal gas law, we can get a relationship between K and Kp  

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Kp= \frac{[C]^{c}*(RT)^{c}[D]^{d}*(RT)^{d}}{[A]^{a}*(RT)^{a}[B]^{b}*(RT)^{b}}

Reorganizing the equation:

Kp= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}*\frac{(RT)^{c+d}}{(RT)^{a+b}}

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Delta n = 2-1-1=0

Kp= K*(RT)^{0}

So Kp=K in this case.

C. (true) The value of K just depends on the temperature that’s why changing the among of products won’t have any effect on its value.  

D. (false) as we can see this reaction involve a heterogeneous system with solids and gases. For convention the concentration for solids and liquids can be considered constant during the reaction that’s why they’re not include in the calculation for the equilibrium constant. Taking this into account the expression for the equilibrium for this reaction is:

CaCO_{3}(s)---equilibrium----CaO(s) + CO_{2}(g)

K= [CO_{2}]

So we can see that [CaCO_{3}] is not include in the expression.  

E. (False) The equilibrium is defined as the point where the rate of the forward reaction is the same to the rate of the reverse reaction. The value of K is telling you which reaction is favored but the rate of both reactions is the same in this point. (see picture)  

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