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natali 33 [55]
3 years ago
8

Rank the following elements by electron affinity 1) fluorine 2) selenium 3) arsenic 4) potassum 5) argon

Chemistry
1 answer:
denis23 [38]3 years ago
7 0

Answer:

Fluorine > Selenium > Arsenic > Potassium > Argon

Explanation:

Electron affinity describes the ability or readiness or tendency of an atom to gain an electron.

The higher the value, the higher the tendency. Electron affinity depends on the on the nuclear charge and atomic radius. When nuclear charge is more, electron affinity is high, when atomic radius increases electron affinity reduces.

Noble gases such as Helium, Neon, and Argon would have 0 affinity for electrons because of their stable electronic configuration. From the list, Ar is the least in terms of electron affinity.

Potassium is a metal with large electropositivity which describes the tendency of an atom to lose electrons. Potassium would readily lose electrons instead of gaining.

Between Arsenic and Selenium: Arsenic belongs to group V and Selenium group VI. The two elements both belong to period IV on the periodic table. Across a period, electron affinity increases due to increase in nuclear charge. Therefore, Selenium would have a greater electron affinity compared to Arsenic.

Fluorine has the highest electron affinity of all. It needs just an electron to complete its octet.

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Answer:

The value of the equilibrium constant for reaction asked is 1.24\times 10^{-7}.

Explanation:

CO_2(g)\rightleftharpoons C(s)+O_2(g)

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2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)..[1]

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2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)..[2]

K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}

CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)..[3]

K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}

[1] +  [2] + [3]

2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)

 ( on adding the equilibrium constant will get multiplied with each other)

K=K_1\times K_2\times K_3

K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}

K=1.53\times 10^{-14}

K=\frac{[C]^2[O_2]^2}{[CO_2]^2}

On comparing the K and K_{goal}:

K^2=K_{goal}

K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}

The value of the equilibrium constant for reaction asked is 1.24\times 10^{-7}.

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