Week 1 Second Quarer Spontaneous Change Entropy and Free Energy PDF

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General Chemistry 2 Quarter 2 –Lesson 1: Spontaneous Change, Entropy, and Free Energy Prepared by: Mr. Ronualdo P. Esteban (Faculty, QCSHS-SHS)

Spontaneity of a Process In this lesson, we will consider the differences between two types of changes that occurs in a system: Changes that occur spontaneously and those that occur by force. In doing so, we will gain an insight and understanding as to why some systems are naturally inclined to change in one direction under certain conditions and how relatively quickly or slowly that natural change proceeds. We will also gain insight on how the spontaneity of a process will affect the distribution of energy and matter within the system.

Spontaneous and Nonspontaneous Processes When a process or change whether physical or chemical occurs naturally under certain conditions it is considered as spontaneous process while a nonspontaneous process or change will not occur or take place unless it is driven by a continual input of energy from an external source. See illustration below.

Figure1: Examples of spontaneous and nonspontaneous process (Dear illustrators, kindly draw an illustration like the ones shown above. Thank you!)

Predicting Spontaneity of a Process based on Entropy Entropy (S) reliably predicts the spontaneity of a process. It is a measure of randomness or disorder.

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Figure 2. Melting of Ice Source: https://bit.ly/35sgOsG

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In the above figure, when ice melts, it becomes more disordered and less structured due to the heat transfer in the system. Factors That Affect the Amount of Entropy in a System: 1. Entropy increases when temperature increases. 2. Entropy increases when a gas expands in a system. 3. Entropy increases when a solid becomes a liquid and when it becomes a gas. 4. Entropy increases when there is an increase in the number of gas molecules in a chemical reaction. 5. Entropy increases when solid dissolves in water to form ions in solution.

ΔS and Spontaneity Table 1. Interplay of ∆Ssys and ∆Ssurr in Determining the sign of ∆Suniv Sign of Entropy Changes ∆Ssys ∆Ssurr Process Spontaneous? ∆Suniv + + + Yes No (reaction will occur in opposite direction) + Yes, if ∆Ssys has a larger magnitude than ∆Ssurr ? + ? Yes, if ∆Ssurr has a larger magnitude than ∆Ssys Calculating Entropy Change: The change in entropy in a chemical reaction is given by the sum of the entropies of the products minus the sum of the entropies of the reactants. The equation must ne balanced and the coefficients of each component must be taken into account. ΔS0=∑nS0(products)−∑ mS0(reactants) Where:

n and m refer to the coefficients

Sample Problems: 1. Calculate the change in entropy associated with the Haber process for the production of ammonia from nitrogen and hydrogen gas. N2(g)+3H2(g)⇌2NH3(g) 2

At 298K as a standard temperature: (Refer to the table for the Thermodynamic Data for the values) S0(NH3) = 192.5 J/mol K S0(H2) = 130.6 J/mol K S0(N2) = 191.5 J/mol K Solution: From the balanced equation we can write the equation for ΔS0 (the change in the standard molar entropy for the reaction): ΔS0 = 2*S0(NH3) - [S0(N2) + (3*S0(H2))] ΔS0 = 2*192.5 - [191.5 + (3*130.6)] ΔS0 = -198.3 J/mol K Based on the calculated ΔS0 which is negative, it shows that there is a decrease in entropy (decrease in disorder) due to a decrease in the number of gas molecules. 2. Calculate ΔS° for the combustion reaction of liquid isooctane with O2(g) to give CO2(g) and H2O(g) at 298 K. Solution: The balanced chemical equation for the complete combustion of isooctane (C8H18) is as follows: C8H18(l)+252O2(g)→8CO2(g)+9H2O(g) (balanced equation) Use this equation to calculate the: ΔS∘rxn=∑mS∘(products)−∑nS∘(reactants) =[8S∘(CO2)+9S0(H2O)]−[S∘(C8H18)+252S∘(O2)] ={[8 mol CO2×213.8J/(mol⋅K)]+[9 mol H2O×188.8J/(mol⋅K)]} −{[1 mol C8H18×329.3J/(mol⋅K)]+[252 mol O2×205.2 J/(mol⋅K)]} =515.3J/K Based on the calculated ΔS∘rxn which is positive, show that combustion reaction in which one large hydrocarbon is converted to many molecules of the gaseous products. Comparing So values Substances with a greater freedom of motion (or number of possible ways to move) have a greater absolute entropy. Example:

I2(g) (So = 261 J/K.mol) and 3

Example:

I2(s) (So = 117 J/K.mol) CH4(g) (So = 186 J/K.mol) and C2H6(g) (So = 230 J/K.mol)

Second Law of Thermodynamics It shows the relationship between entropy and spontaneity. It states that, “all spontaneous changes cause an increase in the entropy of the universe”. ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 Second law of thermodynamics is very important because it tells us about entropy that, ‘entropy dictates whether or not a process or a reaction is going to be spontaneous’ Table 2. The Second Law of Thermodynamics ΔSuniv>0

spontaneous nonspontaneous (spontaneous in opposite direction) reversible (system is at equilibrium)

ΔSuniv − ΔHsys If the process is exothermic, a negative sign is used. The ΔSsurr is a positive quantity and ΔHsys is negative which indicates an increase in entropy. In an endothermic process, ΔHsys is positive and the negative sign ensures that the entropy of the surroundings ΔHsurr decreases. Calculating the Entropy Changes in the Surroundings, ΔSsur ∆Ssurr= ∆H T Where: ∆H =enthalpy change T=temperature Sample Problem: Calculate the ΔSsys and ΔSsurr to the synthesis of ammonia: Is the reaction spontaneous at 25C? N2 (g) + 3H2 (g) —> 2NH3 (g)

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ΔH°rxn = - 92.6 kJ/mol

Calculating the ΔSsur: ΔSsurr

= −ΔHsys/T = − (− 92.6 kJ/mol) (1 kJ/1000 J)/298 K

ΔSsurr = 311 J/K ·mol Calculating the ΔSsys: N2 (g) + 3 H2 (g) —> 2NH3 (g) From the table, S°(J/K· mol): 192

131 193

ΔS° = ΣnS° (products) − ΣmS° (reactants) = [(2) So NH3] – [(1) So N2 + (3) So H2] = [ (2) (193) ] – [ (1) (192) + (3) (131) ] ΔS° = −199 J/K· mol Determining Spontaneity of Reaction Using ΔSuniv: ΔSuniv = ΔSsys + ΔSsurr = −199 J/K· mol + 311 J/K ·mol ΔSuniv = 112 J/K ·mol We can predict that the reaction at 25C is spontaneous since the calculated the ΔS univ is positive. Spontaneity on Temperature If a process releases heat (ΔH is negative) and has an increase in entropy (ΔS is positive), it will always be spontaneous. Various Possible Combinations of ∆H and ∆S for a Process and the Resulting Dependence of Spontaneity on Temperature Table 3. Predicting Spontaneity based on Combinations of ∆H and ∆S Result Case ∆S positive, ∆H negative ∆S positive, ∆H positive ∆S positive, ∆H negative

Spontaneous at all temperatures Spontaneous at high temperatures (where exothermicity is relatively unimportant) Spontaneous at low temperatures (where exothermicity is dominant)

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∆S negative, ∆H positive

Process not spontaneous at any temperature (reverse process is spontaneous at all temperatures)

Gibbs Free Energy (G) It is a thermodynamic quantity that combines entropy, temperature; and heat flow to predict the spontaneity of a process. The change in Gibbs Free Energy at constant temperature and pressure is defines as ∆G=∆H-T∆S. Spontaneity and ΔG If ΔG is negative, the process is spontaneous (and the reverse process is nonspontaneous). If ΔG is positive, the process is non-spontaneous, and the reverse process is spontaneous. If ΔG = 0, the system is at equilibrium.

ΔG (ΔG0) (ΔG CO2(g) ΔGo = -394 kJ Calculate ΔGo for the reaction Cdiamond(s) ---> Cgraphite(s) Cdiamond(s) + O2(g) ---> CO2(g)

ΔGo = -397 kJ

CO2(g) ---> Cgraphite(s) + O2(g) Cdiamond(s) ---> Cgraphite(s)

ΔGo

ΔGo = +394 kJ = -3 kJ

Diamond is kinetically stable, but thermodynamically unstable.

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Thermodynamic Data Table at 298 K Enthalpy (ΔfHo)

Gibbs Free Energy ΔfGo

ΔfHo(HCl(g))

-92.31 kJ mol-1

ΔfGo(NO(g))

86.6 kJ mol-1

ΔfHo(CO2(g))

-393.5 kJ mol-1

ΔfGo(NH3(g))

-16.0 kJ mol-1

ΔfHo(NO(g))

90.29 kJ mol-1

ΔfGo(CuO(s))

-130 kJ mol-1

ΔfHo(NH3(g))

-45.9 kJ mol-1

ΔfGo(NO2(g))

51 kJ mol-1

ΔfHo(CuO(s))

-157.3 kJ mol-1

ΔfGo(C6H6(l))

130 kJ mol-1

ΔfHo(NO2(g))

33.2 kJ mol-1

ΔfGo(Fe2O3(s))

-742 kJ mol-1

ΔfHo(H2O(l))

-286 kJ mol-1

ΔfGo(H2NCH2COOH (s))

-369 kJ mol-1

ΔfHo(C8H18(g)) -208 kJ mol-1

ΔfGo(H2O(l))

-237 kJ mol-1

ΔfHo(HI(g))

27.0 kJ mol-1

ΔfGo(CH4(g))

-51 kJ mol-1

ΔfHo(I2(g))

62.0 kJ mol-1

ΔfGo(HI(g))

1.7 kJ mol-1

ΔfGo(I2(g))

19.4 kJ mol-1

Entropy So So(HCl(g))

186.8 J K-1 mol-1

So(H2(g))

130.6 J K-1 mol-1

So(Cl2(g))

223.0 J K-1 mol-1

So(CO2(g))

213.7 J K-1 mol-1

So(C(graphite)) 5.686 J K-1 mol-1 So(O2(g))

205.0 J K-1 mol-1

So(NO(g))

210.65 J K-1 mol-1

So(NO2(g))

239.9 J K-1 mol-1

So(N2(g))

191.5 J K-1 mol-1

So(NH3(g))

193 J K-1 mol-1

So(CuO(s))

42.63 J K-1 mol-1

So(Cu(s))

33.1 J K-1 mol-1

So(H2O(l))

70 J K-1 mol-1

So(C8H18(g))

467 J K-1 mol-1

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