1. Spontaneity & Entropy Unit 3 - Thermodynamics

2. 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).

3. Spontaneity Here are some examples of spontaneous chemical reactions: Na(s) + ½Cl 2 (g) → NaCl(s) + 411.2 kJ H 2 (g) + ½O 2 (g) → H 2 O(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.

4. Spontaneity Some endothermic reactions are also spontaneous: Br 2 (l) + Cl 2 (g) + 29.3 kJ → 2 BrCl(g) H 2 (s) + ½O 2 + 6.01 kJ → H 2 O(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.

5. 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.

6. 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) + ½O 2 → 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) + ½O 2 Because of this, we say that MgO(s) is stable.

7. 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 H 2 (g) + ½ Br 2 (g) → HBr(g) + 36.4 kJ Therefore, the decomposition of HBr(s) is: HBr(g) + 36.4 kJ → ½ H 2 (g) + ½ Br 2 (g) Since little energy is required for HBr to break down spontaneously, we call it unstable.

8. 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 → CS 2 (g) Reversing this to see the decomposition reaction: CS 2 (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.

9. Entropy Entropy (S) is the measure of randomness in an object. The units for entropy are J/Kmol. 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.

10. 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.

11. 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

12. 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

13. 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).

14. 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.

15. 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

16. 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.

17. 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) + O 2 (g)  N 2 O 4 (g)