Basic Definitions and Equations
Entropy
Based on the number of ways a system can be arranged, represents the disorder of a system
Third Law of Thermodynamics? State and what does it imply
entropy of a perfect crystal at absolute zero (0 K) is zero, shows that when only a single micro-state is available the entropy is zero, So when S = klnW, W = 1 and S = 0
Write the balanced chemical equation for the combustion of benzene, C6H6(l), to give carbon dioxide and water vapor. Would you expect ΔS to be positive or negative in this process?
C6H6(𝑙)+7.5O2(𝑔)⟶3H2O(𝑔)+6CO2(𝑔)
There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and ΔS is positive.
First Law of Thermodynamics
Change in internal energy is the sum of the heat and the work
A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process.
high to low concentration, energy is released from system to surroundings
Consider the decomposition of red mercury(II) oxide under standard state conditions.
2HgO(𝑠,red)⟶2Hg(𝑙)+O2(𝑔)
(a) Is the decomposition spontaneous under standard state conditions?
(b) Above what temperature does the reaction become spontaneous?
Standard Delta G_HgO 298: -58.5 kJ/mol
Delta G: Calculation:
Del G = Products - Reactants
= 2(Hg +O_2) - 2(HgO)
= 0 - (2)(-58.5)
= 117.0 kJ/mol
Non Spontaneous bc Del G positive
Second Law of Thermodynamics
A spontaneous process is when the change in entropy is greater than 0
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.
Consider the following reaction at 298 K:
N2O4(𝑔)⇌2NO2(𝑔)
𝐾_𝑃=0.142
What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.
Δ𝐺∘=−𝑅𝑇ln𝐾=4.84 kJ/mol
At equilibrium Q=K, and Delta G = 0
Gibbs Free Energy
Can be used to predict the spontaneity of a process at a constant temperature and pressure, difference between energy produced in a process and the energy lost to the surroundings... this difference is "free" to do work
At room temperature, the entropy of the halogens increases from I2 to Br2 to Cl2. Explain.
The masses of these molecules (I>Br>Cl) would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I2 is a solid, Br2 is a liquid, and Cl2 is a gas. Another way to think about this is that Cl2 will have an easier time distributing itself easily because it is less massive.
By calculating ΔSuniv at each temperature, determine if the melting of 1 mole of NaCl(s) is spontaneous at 500 °C and at 700 °C.
𝑆∘NaCl(𝑠)=72.11J/mol·K
𝑆∘NaCl(𝑙)=95.06J/mol *K
Δ𝐻∘fusion=27.95 kJ/mol
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
NaCl(s) --> NaCl (l)
Del (S_uni)=Del(S_sys) + Del(S_surr)
= Del(S_sys) + q_surr/T
At 500:
Del (S_uni) = (95-72) + (-27*10^3 J/mol)/(500+273) = roughly -13
At 700: Also negative number roughly -5.8
Neither is spontaneous
What is a state function? What is a standard quantity mean?
State function: value only depends on the initial and final states (why we can use a final minus initial instead of integrating)
Standard: change in some quantity for 1 mol in its standard state at 1 atm and 298.15 K
Fill in what is Del G at high T and low T, and when spontaneous
Del S and Del H +:
Del S and Del H -:
Del S +, Del H -:
Del S-, Del H +:
Del S and Del H +: Del G neg at high T, Del G pos at low T, spont at high T
Del S and Del H -: Del G pos at high T, Del G neg at low T, spont at low T
Del S +, Del H -: Del G neg at all temp, always spont
Del S-, Del H +: Del G pos at all temp, never spont
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
G6P⇌F6PΔ𝐺°=1.7 kJ
(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate ΔG when the concentrations of G6P and F6P are 120 μM and 28 μM respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
(a) Nonspontaneous as Δ𝐺∘>0
Δ𝐺=1.7×103+(8.314×310×ln28/120)=−2.1 kJ. The forward reaction to produce F6P is spontaneous under these conditions.