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Spontaneity & Entropy
Unit 3 - Thermodynamics
Spontaneity
• In chemical terms, a spontaneous reaction is a reaction that occurs on its own.
• Speed is not an issue (that is it can happen fast or slowly).
Spontaneity
• Here are some examples of spontaneous chemical reactions:
Na(s) + ½Cl2(g) → NaCl(s) + 411.2 kJ
H2(g) + ½O2(g) → H2O(g) + 242 kJ
• Note that these are all exothermic. Energy, like boulders, tend to run downhill.
• The lower energy state of the products is more stable, which is why exothermic reactions tend to be spontaneous.
Spontaneity
• Some endothermic reactions are also spontaneous:
Br2(l) + Cl2(g) + 29.3 kJ → 2 BrCl(g)
H2(s) + ½O2 + 6.01 kJ → H2O(l)
• However, the energy needed for these two reactions is very little.
• If a reaction is highly endothermic (a lot of energy is required) the reaction will not be spontaneous.
Stability of Compounds
• The stability of a compound refers to how readily it breaks down.
• Therefore, the reverse of a heat of formation reaction will tell us how readily a compound breaks down.
• That is, we can tell how stable a compound is by looking at the ΔH°f value.
Stability of Compounds – Situation 1
• If ΔHf is large and negative• These compounds give off a lot of energy
when they form. • For example:
• Mg(s) + ½O2 → MgO(s) + 601.7 kJ• The decomposition of MgO(s) requires a lot of
heat to break down into its component elements.• MgO(s) + 601.7 kJ → Mg(s) + ½O2
• Because of this, we say that MgO(s) is stable.
Stability of Compounds – Situation 2
• If ΔHf is small and negative• These compounds give off little energy when
formed. Hence, when they decompose they require little energy to do so.
• For example, the formation of HBr(s) is:• ½ H2(g) + ½ Br2(g) → HBr(g) + 36.4 kJ
• Therefore, the decomposition of HBr(s) is:• HBr(g) + 36.4 kJ → ½ H2(g) + ½ Br2(g)
• Since little energy is required for HBr to break down spontaneously, we call it unstable.
Stability of Compounds – Situation 3
• If ΔHf is positive• If a compound requires heat to form, then its
decomposition will give heat off.• For example:
• C(s) + 2 S(s) + 117.4 kJ → CS2(g)• Reversing this to see the decomposition
reaction:• CS2(g) → C(s) + 2 S(s) + 117.4 kJ
• Since exothermic reactions tend to be spontaneous, we say these types of compounds are also unstable.
Entropy
• Entropy (S) is the measure of randomness in an object. The units for entropy are J/K•mol.
• All substances, whether it be an atom or a compound, contains a certain degree of disorder due to the constant motion of particles.
• Thus, entropy is always a positive value.
Entropy
• The second law of thermodynamics states that spontaneous systems always proceed in the direction of more entropy. This is also known as the law of disorder.
• That is, over time, systems tend to become more random instead of more ordered.
• An analogy for entropy:• Your room at home tends to become more
messy over time as compared to more clean.
Entropy Example
• Entropy can be very hard to explain and often harder to understand. Here are a couple of examples that hopefully will help clarify the topic of entropy.• Imagine you are throwing bricks into a
pile• When you throw the bricks, they
will likely land into a random pile• This is a HIGHLY DISORDERED • This is likely what will happen
because it is HIGH IN ENTROPY
• Imagine you start through bricks into a pile again• When you throw these bricks, they
land it a perfectly stacked pile.• This is HIGHLY ORDERED• This likely WILL NEVER HAPPEN
because it is LOW IN ENTROPY
Entropy Example
• Place a gas the container flask. That flask is connected to an empty flask.
• If you open the gap between the flasks, the gas will spread to fill both containers
• This is MORE DISORDERED so it is HIGHER IN ENTROPY
• If you open the gap, the gas stays only on one side.
• This is MORE ORDERED so it is LOW IN ENTROPY
Can entropy (S) ever be equal to 0?
• The third law of thermodynamics says that as the temperature approaches absolute zero all processes cease and entropy approaches a minimum value (0 J/mol*K).
Can entropy (S) be a negative number?
• Concerning chemical reactions, we will focus on the change in entropy (ΔS):• As a system goes from a state of high disorder
(a gaseous state for example) to a state of less disorder (a liquid state for example) ΔS will be a negative value.
• That is, a negative number indicates more order in the system.
• Likewise, a positive number indicates an increase in randomness.
Predicting Entropy Changes
• The following suggest an increase in entropy:• i. Changes in state:
• a. solid liquid• b. liquid gas• c. solid gas• d. solid or liquid aqueous state (dissolving)
• ii. An increase in the number of moles in the products compared to the reactants. That is, more particles moving about imply more disorder.
• iii. Increase in temperature
Calculating Entropy Changes
• To calculate the change in entropy we will use the same formula as Hess's law. However, we will be using entropy values instead of enthalpy values.
ΔS = ΣSproducts – ΣSreactants
• ΔS values for compounds can be found on thermochemical data sheet.
• ΔS values that are positive indicate less order (spontaneous).
• ΔS values that are negative indicate more order (not spontaneous).
• Note ΔS values must be multiplied by balancing coefficients.
Example:
• Predict whether the following chemical reaction will become more or less random.
• Then, using ΔS values, calculate the change in entropy.
2 NO(g) + O2(g) N2O4(g)