Now look at the segment of the graph between the two data points marked with red squares. Describe how the boiling point and mel
2 answers:
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Answer:
a) ΔS is negative
b) ΔS is positive
c) ΔS is positive
d) ΔS is negative
Explanation:
Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.
The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.
S(solid) < S(liquid) < S(gas)
The entropy change for a given reaction is:
a)
Here the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.
Since S(product) < S(reactant), based on equation 1, ΔS will be negative.
b)
Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.
Since S(product) > S(reactant), based on equation 1, ΔS will be positive.
c)
Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants
Since S(product) > S(reactant), based on equation 1, ΔS will be positive.
d)
Here one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the product which is in the solid phase i.e. lower entropy.
Since S(product) < S(reactant), based on equation 1, ΔS will be negative.
Hence the order of the transition in order of increasing frequency of the photon absorbed or emitted will be : d < a < c < b
×
²/
²
where,
= energy of orbit
n = number of orbit
Z = atomic number
Energy of n = 1 in an hydrogen atom:
₁ = -13.6× 1²/1² = -13.6eV
Energy of n = 2 in an hydrogen atom:
₂ = -13.6 × 1²/2² = -3.40eV
Energy of n = 3 in an hydrogen atom:
₃ = -13.6× 1²/3² = -1.51eV
Energy of n = 4 in an hydrogen atom:
₄ = -13.6× 1²/4² = -0.85eV
Energy of n = 5 in an hydrogen atom:
₃ = -13.6× 1²/ 5² = -0.54eV
a) n = 2 to n = 4 (absorption)
Δ₁ = E₄ - E₂ = -0.85- (- 3.40) = 2.55eV
b) n = 2 to n = 1 (emission)
Δ E₂ = E₁ - E₂ = -13.6 - (- 3.40) = -10.2eV
Negative sign indicates that emission will take place.
c) n = 2 to n = 5 (absorption)
ΔE₃ = E₅ - E₂ = -0.54 -( -3.40) = -2.85eV
d) n = 4 to n = 3 (emission)
ΔE₄ = E₃ - E₄ = -1.51 - (- 0.85) = - 0.66eV
Negative sign indicates that emission will take place.
According to Planck's equation, higher the frequency of the wave higher will be the energy:
h = Planck's constant
frequency of the wave
So, the increasing order of magnitude of the energy difference :
₄< E₁ <E₃ <E₂
The H atom electron transitions in order of increasing frequency of the photon absorbed or emitted will be d < a < c < b
: d < a < c < b
To learn more about transitions visit the link:
brainly.com/question/28304182?referrer=searchResults
The question is incomplete , complete question is:
Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
(a) n = 2 to n = 4
(b) n = 2 to n = 1
(c) n = 2 to n = 5
(d) n = 4 to n = 3
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