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dexar [7]
2 years ago
5

What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm? (c = 3.00 × 108 m/s; h = 6

.63 × 10–34 J • s; NA = 6.022 × 1023 moles–1)
Chemistry
1 answer:
ASHA 777 [7]2 years ago
4 0

<u>Answer:</u> The energy of a mole of photons is 2.46\times 10^{5}J

<u>Explanation:</u>

The relation between energy and wavelength of light is given by Planck's equation, which is:

E=\frac{N_Ahc}{\lambda}

where,  

E = energy

h = Planck's constant  = 6.626\times 10^{-34}Js

c = speed of light  = 3.0\times 10^8m/s

N_A =  Avogadro's number = 6.022\times 10^{23}

\lambda = wavelength of photon = 486 nm = 486\times 10^{-9}m     (Conversion factor:  1m=10^9nm )

Putting values in above equation, we get:  

E=\frac{6.022\times 10^{23}\times 6.626\times 10^{-34}Js\times 3.0\times 10^8m/s}{486\times 10^{-9}m}\\\\E=2.46\times 10^{5}J/mol

Hence, the energy of a mole of photons is 2.46\times 10^{5}J

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Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

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Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

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Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

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ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

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K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

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