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Studentka2010 [4]
3 years ago
7

Compare and contrast the Ionization energies of groups 1&2 (Alkali and Alkaline metals) with the Ionization energies of grou

ps 16&17 (Oxygen family and Halogens)
Chemistry
1 answer:
irinina [24]3 years ago
3 0

Answer:

Ionization energy is the energy required to remove an electron from a specific atom. It is measured in kJ/mol, which is an energy unit, much like calories. The ionization energies associated with some elements are described in the Table 1. For any given atom, the outermost valence electrons will have lower ionization energies than the inner-shell kernel electrons. As more electrons are added to a nucleus, the outer electrons become shielded from the nucleus by the inner shell electrons. This is called electron shielding .

Explanation:

a little summary

Ionization energy refers to the amount of energy needed to remove an electron from an atom.

Ionization energy decreases as we go down a group.

Ionization energy increases from left to right across the periodic table.

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The value of Kp for the reaction NO(g) 1 1 2 O2(g) 4 NO2(g) is 1.5 3 106 at 25°C. At equilibrium, what is the ratio of PNO2 to P
expeople1 [14]

Answer : The ratio of p_{NO_2} to p_{NO} is, 6.87\times 10^5

Solution :  Given,

K_p=1.5\times 10^6

p_{O_2} = 0.21 atm

The given equilibrium reaction is,

NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)

The expression of K_p will be,

K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}

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

1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}

\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}

\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5

Therefore, the ratio of p_{NO_2} to p_{NO} is, 6.87\times 10^5

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3 years ago
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