
4.3 a/b/c/d/e/f: Intro to Entropy
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Chemistry, Science
•
11th - 12th Grade
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Easy
Erin Hannan
Used 22+ times
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11 Slides • 3 Questions
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b |
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c |
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d |
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f |
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S, unit = J/K
a measure of disorder
the universe tends over time towards disorder/chaotic (2nd Law of Thermodynamics)
measure of how many arrangements with the same energy
we measure entropy change, not entropy directly
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As temp increases, entropy increases
States can predict (i.e. solid = least entropy)
More particles within a system increases energy
When a solute dissolves in a solvent, its entropy increases
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Determine the change in entropy of the chemical reaction:
CaO(s) + H2O(l) --> Ca(OH)2(s)
- there are 2 mol of reactant particles, and 1 mol of product particles = less disordered = ΔS = negative
- decrease in amount of liquids, and increase in amount of solids = less disordered = ΔS = negative
Therefore, overall the system has ΔS = negative
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spontaneous = occurs without any additional energy
when a spontaneous process occurs, there is an increase in entropy of the universe
we determine whether something is spontaneous by 2 factors: change in enthalpy and change in entropy
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We are given standard entropy data
Similar to heats of formation
ΔS° = ΣS° (products) - ΣS° (reactants)
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quantity that helps us determine whether a reaction will be spontaneous
ΔG
The change in GFE is: ΔG = ΔH-TΔS
Combining the effects of enthalpy, entropy and temperature
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if ΔG is < 0 then reaction is spontaneous (exergonic)
if ΔG is > 0 then reaction is non-spontaneous (endergonic)
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State known values
Sub into formula
Answer the q.
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GFE formula can be used to determine the minimum temperature at which reactions switch from spontaneous to not spontaneous
sub in everything but T, and have ΔG = 0 (where reaction is neither spontaneous or NS), and will give you the point of low/high temp
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Multiple Choice
Given the change of phase:
CO2(g) —> CO2(s)
As CO2(g) changes to CO2(s), the entropy of the system
decreases
increases
remains the same
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Multiple Choice
1. Given the following information, calculate ΔG0 for the reaction below at 250C: (K = 0C + 273)
SnCl4(l) + 2 H20(l) → SnO2(S) + 4HCl(g),
ΔH0=133.0 kJ and ΔS0 =401.5 J/K
-252.6 kJ
-13.4 kJ
13.4 kJ
252.6 kJ
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