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Gekata [30.6K]
2 years ago
14

The energy difference between the 5d and 6s sublevels in gold accounts for its color. Assuming this energy difference is about 2

.7 eV (electron volt; 1 eV = 1.602 x 10⁻¹⁹ J), explain why gold has a warm yellow color.
Chemistry
1 answer:
sergejj [24]2 years ago
7 0

Answer:

\lambda=459.1\times 10^{-7}\ m = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

Explanation:

Given that:- Energy = 2.7 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.602 × 10⁻¹⁹ J

So, Energy = 2.7\times 1.602\times 10^{-19}\ J=4.33\times 10^{-19}\ J

Considering:-

E=\frac{h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light

So,  

4.33\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

4.33\times \:10^{26}\times \lambda=1.99\times 10^{20}

\lambda=459.1\times 10^{-7}\ m = 459.1 nm

This wavelength corresponds to yellow color and thus gold has warm yellow color.

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The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

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<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

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Base = proton (receiver) acceptor (H + ion)

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From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

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Keywords : acid base reaction, the equilibrium constant

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