Question

Can someone explain to me how Electron affinity and ionization energy works?

Answer 1

ionization energ is the energy required to remove an electron from atom or ion in the gas phase. Electron affinity is need when an atom in the gas phase acquires an electron.

Answer 2

does it help? or I did not answer the question

Answer 3

So for example: \[\sf Ca~ (g) \rightarrow Ca^+ (g) ~ + ~ e^{-}\] This is Ionization energy, right? The `+` indicates we have taken away an electron? To show that we write it as \(\sf +~e^{-}\)?

Answer 4

Yes

Answer 5

Awesome :)

Then the reverse (electron affinity) would work as ...\[\sf Ca^+ (g) +~ e^- \rightarrow Ca~(g) \]

Answer 6

yep

Answer 7

Just reversing it I'm guessing... and this is hypothetical :D

Answer 8

Awesome!

Answer 9

good night

Answer 10

I've got a test today on enthalpy changes, etc... trying to understand all the types of enthalpy atm

Answer 11

Goodnight!

Answer 12

Oh! May Iask one last question?

Answer 13

If you're looking for bond energy... and the formation gives you half of that amount, your total reaction would be multiplied by how many MOLES of the atom there actually are, right?

Answer 14

Give me example

Answer 15

I'll pull out my example here..

I'm looking for the bond energy of \(\sf O_2\), but my formation equation is \[\Delta H = 2Na(s)~ +~ \frac{1}{2}O_2 (g) \rightarrow ~ Na_2O(s)\]

Answer 16

This is not all the parts to it, but according to this, the energy I would find here would only be half the amount.. so I would have to work backwards to realize that \[\sf O_2 \rightarrow 2O\] right?

Answer 17

1/2 mean half mole of the oxygen molecule. Oxygen is diatomic molecule that is why it write O2

Answer 18

Half of the mole has nothing to do with half energy.

Answer 19

it just balance equation

Answer 20

\[\sf Na(s) \rightarrow Na(g)~~~\Delta H = 107\]\[\sf Na(g) \rightarrow Na^+ +e^-~~~\Delta H =496 \]\[\sf O(g) +e^- \rightarrow O^{-}(g) ~~ \Delta H = -141\]\[\sf O^- +e^- \rightarrow O^{2-}(g) ~~ \Delta H = 878\]\[\sf 2Na(s) + \frac{1}{2}O_2(g) \rightarrow Na_2O ~~ \Delta H = -416\]\[\sf 2Na^+ O^{2-} \rightarrow Na_2O(s) ~~ \Delta H = -2608\]

Answer 21

And here i am told to find bond energy of \(O_2\)

Answer 22

(i forgot some of the phases, quickly wrote it out)

Answer 23

So \(\sf \Delta H = 107 +496 + (-141) +878 + (-2608) + \Delta H_{BE} = -416\)?

Answer 24

The answer choices were:

a. 356

b. 498

c. 426

d. 249

e. 852

Answer 25

When I solved for this, I got 249 , but the answer was double of it, 498

Answer 26

I wanted to know why it was doubled..

Answer 27

NaO2. Sodium requires 2 oxygen that is why doubled

Answer 28

one oxygen is required only 249, but energy Na requires to make Nao2 is twice as much

Answer 29

Hmm...

Answer 30

Oh, if we only required half it would be something like \(NaO\) right?

Answer 31

look at O-2. there are two negative ( 2 electron).

Answer 32

That's when we would get 249.

Answer 33

yes

Answer 34

Yeah I see that

Answer 35

I mean, two EAs.

Answer 36

we are good now. I need to go to sleep

Answer 37

Oh ok, thank you very much!

Answer 38

I took chemistry long time ago.....that is all I can do.

Answer 39

It was helpful :)

Answer 40

I love chemistry. please come ask me in the future.

Answer 41

Willdo!

Answer 42

I also can help you with math , up to pre-cal

Answer 43

:) Ok.

Have a good night :D

Answer 44

electron affinity is given in units of \(\sf \color{red}{\frac{J}{mol}}\) or kJ as well, in which case, some atom accepts an electron, or in otherwords, the likelihood that it will accept that electron.

Ionization, as it's name implies, with "\(ion\)" in general relates to formation of cations, because you're removing the electron.