Entropies affect
Explain how the change affects the entropy of a system (increases or decreases).
Temperature increases
Entropy increases with increase in temperature
Energy considerations favor ___ (A-B or A-A?) bonds
A-B
For which reaction do you expect the change in entropy to be negative?
a. 4C(s) + 2 O2 (g) -> 4CO(g)
b. Br2 (s) -> Br2 (l)
c. H2O (l, 25degreesC) -> H2O (l, 30 degrees C)
d. Cl2 (g) + 2HI (g) -> I2 (s) + 2HCl (g)
d. Cl2 (g) + 2HI (g) -> I2 (s) + 2HCl (g)
Calculate the enthalpy change for the reaction of ethene and hydrogen, given the following bond energy values in kJ/mol:
C2H4 + H2 ---> C2H6
H−H 436; C−H 412; C=C 612; C−C 348
reactant bonds broken: four C−H bonds, one C=C bond, one H−H bond
product bonds made: one C−C bond, six C−H bonds
ΔH = −124 kJ
What is Gibbs Free Energy Equation?
dG = dH - TdS
Explain how the change affects the entropy of a system (increases or decreases).
The disorder of the phase increases
Entropy increases as the disorder of the phase increases (solid is most ordered, gas is least ordered)
In general, energetic stability ________ with decreasing (more negative) potential energy of a substance.
increases
For which of these processes would delta S be expected to be the most positive?
a. O2 (g) + 2 H2 (g) -> 2 H2O (g)
b. H2O (l) -> H2O (s)
c. N2O4 (g) -> 2 NO2 (g)
d. NH4NO2 (s) -> N2 (g) + 2 H2O (g)
d. NH4NO2 (s) -> N2 (g) + 2 H2O (g)
Problem #12: Calculate the C=C bond energy in ethene:
H2C=CH2(g) + H2(g) --> H3C−CH3(g) ΔH = −138 kJ/mol
Bond enthalpies (kJ/mol): C−C = 348; H−H = 436; C−H = 412
1) Hess' Law for bond enthalpies is:
ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken
2) Let's insert symbols, not numbers:
ΔH = Σ ([(C=C) + (4) (C−H) + (H−H)] − Σ [(6) (C−H) + (C−C)]
3) Cancel 4 C−H bonds:
ΔH = Σ ([(C=C) + (H−H)] − Σ [(2)(C−H) + (C−C)]
4) Put numbers in place and solve:
−138 = Σ ([(x) + (436)] − Σ [(2)(412) + (348)]
x = 598 kJ
For thermodynamically favored processes, dG ___ 0
<
Explain how the change affects the entropy of a system (increases or decreases).
Dissolving a solid or liquid
Dissolving a solid or liquid in a liquid increases entropy
_________ processes tend to be more energetically favored
Exothermic
a. Fe (s) -> Fe (l)
b. Fe (s) + S (s) -> FeS (s)
c. 2 Fe (s) + 3/2 O2 -> Fe2O3 (s)
d. HF (l) -> HF (g)
e. 2 H2O2 (l) -> 2 H2O (l) + O2 (g)
c. 2 Fe (s) + 3/2 O2 -> Fe2O3 (s)
Problem #13: The following two equations produce methane and ethane:
C + 4H ---> CH4 ΔH = −1652 kJ/mol
2C + 6H ---> C2H6 ΔH = −2825 kJ/mol
(a) Calculate the bond enthalpy of a C−H bond.
(b) Calculate the bond enthalpy of a C−C bond.
In the first equation, 4 C−H bonds are formed.
−1652 kJ/mol divided by 4 = 413 kJ <--- that's the bond enthalpy of a C−H bond
Note that bond enthalpies are expressed as a positive value (energy put into the bond to break it), so I ignored the minus sign on the 1652 value.
For the second reaction, note that six C−H bonds are formed and one C−C bond is formed.
413 times 6 = 2478
2825 − 2478 = 347 kJ <--- that's the bond enthalpy of a C−C bond
for dG to be more negative, dH is ______ and dS is _______
Also, demonstrate on a PE and S graph.
dH is negative and dS is positive.
demonstrating on the board...
Explain how the change affects the entropy of a system (increases or decreases).
Dissolving a gas
Dissolving a gas into a liquid or solid decreases entropy
_________ (Oxygenated or deoxygenated) fuels will produce less energy per mole
Oxygenated
Which has the greatest absolute entropy?
a. five moles of C(s) at 50 degrees C
b. four moles of CH3Cl (l) at 100 degrees C
c. one mole of C2H6 (g) at 25 degrees C
d. three moles of CH3Cl (l) at 75 degrees C
c. one mole of C2H6 (g) at 25 degrees C
Calculate the bond energy of the Cl−F bond using the following data:
Cl2 + F2 ---> 2ClF ΔH = −108 kJ
Bond enthalpies (in kJ/mol): Cl−Cl (239); F−F (159)
Hess' Law for bond enthalpies is:
ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken
−108 = [239 + 159] − 2x
−2x = −506
x = 253 kJ/mol
Note the use of 2x because there are two ClF molecules.
Hydrogen reacts with nitrogen to form ammonia (NH3) according to the reaction
3 H2(g) + N2(g) ⇋ 2NH3(g)
The value of ΔH° is –92.38 kJ/mol, and that of ΔS° is –198.2 J/mol · K. Determine ΔG° at 25°C.
a. +5.897 × 104 kJ/mol
b. +297.8 kJ/mol
c. –33.32 kJ/mol
d. –16.66 kJ/mol
e. +49.5 kJ/mol
c. –33.32 kJ/mol
Explain how the change affects the entropy of a system (increases or decreases).
Atomic size and molecular complexity
Entropy increases with mass and molecular complexity
The decomposition of stable compounds tends to be ___________
endothermic
Which of the following represents an increase in entropy?
a. freezing of water
b. boiling of water
c. crystallization of salt from a supersaturated solution
d. the reaction 2 NO (g) -> N2O2 (g)
e. the reaction 2 H2 (g) + O2 (g) -> 2H2O (g)
b. boiling of water
The decomposition reaction of tetrahedral P4 is as follows:
P4(g) ---> 2P2(g); ΔH = +217 kJ
If the bond energy of a single P−P bond is 200 kJ mol¯1, what is the energy of the PP triple bond in P2?
(note there are 6 P-P bonds in P4)
Say you break all 6 P−P bonds in P4, that is 6 x 200 = +1200.
1200 − 217 = 983 which is released when two P≡P bonds form.
So 983 / 2 = bond energy of a P≡P bond. To three sig figs, the answer is 492 kJ/mol.
for the reaction: CO (g) + H2O (g) -> CO2 (g) + H2 (g)
dH = -41.2 kJ and dS = -135 J/K
At what temperature does dG = 0?
is this reaction favored entropically or energetically?
which reaction is favored above this temperature?
which reaction is favored below this temperature?
305 K
energetically due to the exothermic nature and the negative dS
the reverse reaction is favored above this temperature
the forward reaction is favored below this temeprature