Weekly Review Week of 9/24-9/28 Geoffrey Geberth 10/1/18

Chemical Equilibria

Activity (ai) Measure of how the free energy of a compound changes as a reaction proceeds Unitless Gas 𝑎 𝑖 = 𝑃 𝑖 𝑃 ° Solutions 𝑎 𝑖 = 𝑖 𝐶 ° Pure solids and liquids 𝑎 𝑖 =1

Mass Action Expression xX+𝑏𝐵⇄𝑐𝐶+𝑑𝐷 𝑎 𝐶 𝑐 𝑎 𝐷 𝑑 𝑎 𝑋 𝑥 𝑎 𝐵 𝑏 Remember: Products/Reactants! Can be done with concentrations or pressures 𝐶 𝑐 𝐷 𝑑 𝑋 𝑥 𝐵 𝑏 𝑃 𝐶 𝑐 𝑃 𝐷 𝑑 𝑃 𝑋 𝑥 𝑃 𝐵 𝑏 Remember: The standard state of a pure solid or liquid is 1 These expressions give us the Reaction Quotient (Q)

Practice 𝑁 2 𝑔 +3 𝐻 2 𝑔 ⇌2 𝑁𝐻 3 ⇒ 𝑁𝐻 3 2 𝑁 2 𝐻 2 3 𝑁 2 𝑔 +3 𝐻 2 𝑔 ⇌2 𝑁𝐻 3 ⇒ 𝑁𝐻 3 2 𝑁 2 𝐻 2 3 𝐶𝐻 3 𝑂𝐻(𝑠𝑙𝑛)+ 𝐶 4 𝐻 9 𝑂𝐻(𝑠𝑙𝑛) 𝑐𝑎𝑡𝑎𝑙𝑦𝑡𝑖𝑐 𝐻 + 𝐶 4 𝐻 9 𝑂𝐶𝐻 3 (𝑠𝑙𝑛)+ 𝐻 2 𝑂(𝑠𝑙𝑛) ⟹ 𝐶 4 𝐻 9 𝑂𝐶𝐻 3 𝐻 2 𝑂 𝐶𝐻 3 𝑂𝐻 𝐶 4 𝐻 9 𝑂𝐻 𝐶𝑂 2 (𝑔)+𝐶(𝑠)⇌2𝐶𝑂(𝑔) ⟹ 𝐶𝑂 2 𝐶𝑂 2

I know a thing or two about K’s Equilibrium Constant Value representing the reaction quotient when the system is at equilibrium K > 1 -> favors products K < 1 -> favors reactants Can be huge or tiny, but is never negative Different values for concentrations: 𝐾 𝑐 = 𝐶 𝑐 𝐷 𝑑 𝑋 𝑥 𝐵 𝑏 𝐾 𝑝 = 𝑃 𝐶 𝑐 𝑃 𝐷 𝑑 𝑃 𝑋 𝑥 𝑃 𝐵 𝑏 𝐾 𝑝 = 𝐾 𝑐 (𝑅𝑇 ) ∆𝑛 Arrived here from the ideal gas law (see gchem) I know a thing or two about K’s

Manipulating K There are 3 main ways we manipulate reactions and wish to know the new K Reverse the reaction 𝐴+𝐵⇌𝐶 𝐾 1 ⟹𝐶⇌𝐴+𝐵 𝐾 2 = 1 𝐾 1 Multiply stoichiometric coefficients 𝐴+𝐵⇌𝐶 𝐾 1 ⟹ x𝐴+𝑥𝐵⇌𝑥𝐶 𝐾 2 = 𝐾 1 𝑥 Add subsequent reactions 𝐴+𝐵⇌𝐶 𝐾 1 ;𝐶⇌𝐷 𝐾 2 ⟹𝐴+𝐵⇌𝐷 𝐾 3 = 𝐾 1 𝐾 2

Q and K Remember: Q is at any point in the reaction and moves, but K is only at equilibrium and is fixed (it depends on ΔG) Comparing Q and K let’s us figure out in what direction a reaction will proceed, if at all Q = K -> at equilibrium (forward rate = reverse rate) Q < K -> Too many reactants (reaction produces more products) Q > K -> Too many products (reaction produces more products) K Q

ΔG and Equilibria At equilibrium, ΔG = 0 ∆𝐺= ∆𝐺 𝑟 ° +𝑅𝑇𝑙𝑛 𝑄 The sign on ΔG tells you which direction the reaction will move ΔG < 0: Reaction moves to products ΔG > 0: Reaction moves to reactants ∆𝐺= ∆𝐺 𝑟 ° +𝑅𝑇𝑙𝑛 𝑄 At any instant in the reaction ∆𝐺 𝑟 ° =−𝑅𝑇𝑙𝑛 𝐾 At Eq 𝐾= 𝑒 −∆ 𝐺 𝑟 ° 𝑅𝑇

Let’s look at this equation more 𝐾= 𝑒 −∆ 𝐺 𝑟 ° 𝑅𝑇 If ∆ 𝐺 𝑟 ° =0, K=1 Temperature can change the equilibrium point! Direction depends on ∆ 𝐻 𝑟𝑥𝑛 Endothermic (ΔH > 0), increased T increases K Exothermic (ΔH < 0), increased T decreases K

RICE tables How we solve for changes in a system towards equilibrium R: Reaction I: Initial Conditions C: Change E: Equilibrium Example: 𝐴+𝐵⇌2𝐶 initially at .1 M in A and B (no C present)

This reaction is important in steel production! Practice 𝐹𝑒𝑂 𝑠 +𝐶𝑂 𝑔 ⇌𝐹𝑒 𝑠 + 𝐶𝑂 2 𝑔 ; 𝐾 𝑝 =0.259 @ 1000 𝐾 𝑃 𝐶𝑂,𝑖 =1.000 𝑎𝑡𝑚; 𝑃 𝐶 𝑂 2 ,𝑖 =.500 𝑎𝑡𝑚 Find the equilibrium partial pressures 𝐾 𝑝 =.259= 𝑃 𝐶𝑂 2 𝑃 𝐶𝑂 = .500+𝑥 1.000−𝑥 .0259−.0259𝑥=.500+𝑥 ⟹𝑥= −.241 1.259 =−𝟎.𝟏𝟗𝟏 → 𝑃 𝐶𝑂 =1.000− −.191 =𝟏.𝟏𝟗𝟏 𝒂𝒕𝒎; 𝑃 𝐶𝑂 2 =.500+ −.191 =.𝟑𝟎𝟗 𝒂𝒕𝒎 Rxn FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g) Intl P (atm) - 1.000 .500 Change (atm) -x +x Eq P (atm) 1.000 - x .500 + x Screenshot taken by me in Fallout 4 This reaction is important in steel production!

More practice! NO is a pollutant produced in car engines from the reaction: 𝑁 2 (𝑔)+ 𝑂 2 (𝑔)⇌2𝑁𝑂(𝑔) 𝐾 𝑐 =1.7∗ 10 −3 @ 2300 𝐾 If the initial concentrations of N2 and O2 are both 1.40 M, what are the equilibrium concentrations of all 3 compounds? 𝐾 𝑐 =1.7∗ 10 −3 = 𝑁𝑂 2 𝑁 2 𝑂 2 = (2 𝑥) 2 1.40−𝑥 1.40−𝑥 = (2 𝑥) 2 1.40−𝑥 2 → 1.7∗ 10 −3 = 2𝑥 1.40−𝑥 Rxn N2 (g) + O2 (g) ⇌ 2NO (g) Intl conc (M) 1.40 Change (M) -x +2x Eq conc (M) 1.40 - x 0 + 2x

More practice! NO is a pollutant produced in car engines from the reaction: 𝑁 2 (𝑔)+ 𝑂 2 (𝑔)⇌2𝑁𝑂(𝑔) → 1.7∗ 10 −3 = 2𝑥 1.40−𝑥 .0412 1.40−𝑥 =2𝑥 .058−.0412𝑥=2𝑥→2.041𝑥=.058 𝑥=.𝟎𝟐𝟖 𝑴 𝑁 2 = 𝑂 2 =1.40−.028=𝟏.𝟑𝟕𝑴; 𝑁𝑂 =2 .040 =.𝟎𝟓𝟔𝑴 Rxn N2 (g) + O2 (g) ⇌ 2NO (g) Intl conc (M) 1.40 Change (M) -x +2x Eq conc (M) 1.40 - x 0 + 2x

Takeaways Products/Reactants Keep track of your units Q is at any time, K is at equilibrium Q seeks K The ideal gas law is used to convert between Kp and Kc ΔGrxn = 0 at equilibrium