Entropy Flashcards
What is the second law of thermodynamics?
Spontaneous changes result in an increase in disorder or entropy
What does the second law of thermodynamics determine?
- Whether a physical or chemical change is likely to happen at a particular temperature
- Whether redox reactions will take place
- The position of equilibrum
What is standard molar entropy?
The entropy per mole of substance under standard conditions. Usually given the units J mol^-1 K^-1, and usually given per mole of atoms, not molecules
Diamond has a lower value for standard molar entropy than lead, why is this?
The carbon atoms in diamond are held in place by strong, highly directional covalent bonds, but the metallic bonding in lead is not directional so atoms can vibrate more freely and share out their energy in more ways than carbon atoms in diamond
Lead has a higher standard molar entropy than iron, why is this?
In hard solids, the atoms have rigid structures with stiff bonds and the thermal motion of the atoms is very restricted so entropy of a hard solid like iron is lower than lead which is soft. In soft solids, there is more thermal disorder and entropy is larger
Why is the standard molar entropy of liquids generally higher than solids?
Atoms or molecules are freer to move and there are more ways of distributing the particles and energy so there is more disorder.
Why is the standard molar entropy of gases generally higher than liquids?
The atoms or molecules are free to move but are also very widely spaced, so there are even more ways of distributing the particles and energy so the disorder is even greater
Why are more complicated molecules higher in entropy than simpler molecules?
More energy levels are available in a complicated molecule, so they can vibrate, rotate and arrange themselves in more ways
What is the equation for ΔSsystem which concerns standard molar entropies of reactants and products?
ΔSsys = ΣSproducts - ΣSreactants
What is the equation for ΔStotal which concerns ΔSsys and ΔSsurr?
ΔStotal = ΔSsys + ΔSsurr
ΔSsurr is positive for which type of reaction?
Exothermic reactions
ΔSsurr is negative for which type of reaction?
Endothermic reactions
What effect does the temperature of the surroundings have on the value for increase in entropy?
Entropy increase is small because the molecules have high entropy and the area is already in chaotic motion.
The opposite is true if the surroundings are cold.
The entropy change in the surroundings caused by transfer of heat depends on the value of the heat change and is also inversely proportional to the temperature of the surroundings
What is the equation which concerns ΔSsurr, ΔH and temperature?
ΔSsurr = -ΔH / T
Temperature measured in K, enthalpy has to be in Jmol^-1
What is the equation which concerns ΔStotal, ΔSsys, ΔH and T?
ΔStotal = ΔSsys - (ΔH/T)
For a spontaneous reaction, is ΔStotal positive or negative?
Positive
What happens if ΔStotal is zero?
It is in equilibrium/balanced - becomes very sensitive to temperature and could go either way
What is the definition of a spontaneous reaction?
A spontaneous reaction is one for which the total entropy change is positive
Mathematically, why do most exothermic reactions tend to go at about room temperature?
-ΔH/T is much larger and more positive than ΔSsys, so ΔStotal is positive
Under what conditions (to do with the equation for entropy) can an endothermic reaction be feasible?
If the increase in the entropy of the system is greater than the decrease of the entropy of the surroundings
Mathematically, why might a reaction that doesn’t go at room temperature become feasible as temperature rises?
ΔSsurr decreases in magnitude as T increases
What is the minimum value for ΔStotal for which reactions tend to go to completion?
200 J mol^-1 K^-1
What is the equation for ΔG that concerns ΔH, T and ΔSsys?
ΔG = ΔH - TΔSsys
Does a positive or negative value for ΔG indicate a feasible reaction?
Negative
What happens if ΔG = 0?
The reaction is at equilibrium
What is the equation for ΔG that concerns T and K (equilibrium constant)?
ΔG = -RTlnK
When ΔG becomes more negative than a certain value (approx. -40kJ mol^-1), what happens to K and what does that indicate?
K becomes so large that the reaction has effectively gone to completion
When ΔG becomes more positive than a certain value (approx. 40 kJ mol^-1), what happens to K and what does that indicate?
K becomes so small that equilibrium lies almost entirely to the reactants’ side, so hardly any of the reactants wil actually react
What are the two main reasons that a reaction can be thermodynamically feasible but not take place?
- Kinetic stability - if the activation energy is too high/if reactants are kinetically stable
- Non-standard conditions - ΔG is calculated using values calculated under standard conditions
