Exam 3

Wednesday 3/29
7:00-8:30pm
WEL 2.224


avg = 79.4



🔑 Here are all the KEYS to Exam 3


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gChem Chapters:
Chapter 10: Solubility Equilibria
Chapter 11: Kinetics

Learning Outcomes for Solubility Equilibria

Students will be able to...

  1. Understand the concept of saturation and how it relates to the solubility product, Ksp.
  2. Write total and net ionic equations to identify spectator ions.
  3. Quantitatively determine molar solubility from Ksp.
  4. Quantitatively determine Ksp from molar solubility.
  5. Convert general (common) solubility terms to molar solubility.
  6. Understand and apply the "common ion effect" on solubility.
  7. Given concentrations of specific ions, predict if a precipitate will form (amount or concentration) using Qsp vs Ksp.

Learning Outcomes for Chemical Kinetics

Students will be able to...

  1. Understand the concept of rate of change associated with a given chemical reaction and how it can be measured.
  2. Determine rate law of chemical change based on experimental data.
  3. Be able to identify the reaction order for a chemical change.
  4. Understand the concept of pseudo-first order kinetics and when they apply.
  5. Apply integrated rate equations to solve for the concentration of chemical species during a reaction of different orders.
  6. Understand the concept of mechanism and using rate law data predict whether or not a proposed mechanism is viable or not.
  7. Recall and explain why certain factors such as concentration, temperature, medium and the presence of a catalyst will affect the speed of a chemical change.
  8. Interpret a reaction coordinate diagram and determine if such a diagram supports a given single or multistep mechanism, including the concept and depiction of any transition states and reaction intermediates.
  9. Understand the concept of an activation energy in the context of the transition state and be able to calculate the activation energy given some experimental data.
  10. Recall, manipulate and properly employ the Arrhenius Law.
  11. Explain the function and purpose of a catalyst.

Solubility Equilibria

Pretty much all cations and anions in aqueous solutions are ±1, ±2, or ±3. There are really almost no charges above +3 or below -3. For that reason, the only salt ratios that we will encounter for MxXy type salts are    1 : 1    1 : 2    1 : 3   or   2 : 3 . And, since the cation / anion absolute charges could be reversed the flipped ratios are possible as well    2 : 1    3 : 1   or   3 : 2 .

Solubility Product (\(K_{\rm sp}\)) and Molar Solubility (\(x\))

salt
ratio		salt
formula		 aqueous
cation		 aqueous
anion		 equilibrium
formula		 Ksp(x) molar
solubility
1:1 MX(s) ⇌ M+(aq) + X–(aq) Ksp = [M+][X–] \(K_{\rm sp} = x^2\) \(x = \sqrt{K_{\rm sp}}\)
1:2 MX2(s) ⇌ M2+(aq) + 2 X–(aq) Ksp = [M2+][X–]2 \(K_{\rm sp} = 4x^3\) \(x = \root 3 \of {K_{\rm sp}\over 4}\)
1:3 MX3(s) ⇌ M3+(aq) + 3 X–(aq) Ksp = [M3+][X–]3 \(K_{\rm sp} = 27x^4\) \(x = \root 4 \of {K_{\rm sp}\over 27}\)
2:3 M2X3(s) ⇌ 2 M3+(aq) + 3 X2–(aq) Ksp = [M3+]2[X2–]3 \(K_{\rm sp} = 108x^5\) \(x = \root 5 \of {K_{\rm sp}\over 108}\)

and the "flipped" ratios...

salt
ratio		salt
formula		 aqueous
cation		 aqueous
anion		 equilibrium
formula		 Ksp(x) molar
solubility
2:1 M2X(s) ⇌ 2 M+(aq) + X2–(aq) Ksp = [M+]2[X2–] \(K_{\rm sp} = 4x^3\) \(x = \root 3 \of {K_{\rm sp}\over 4}\)
3:1 M3X(s) ⇌ 3 M+(aq) + X3–(aq) Ksp = [M+]3[X–] \(K_{\rm sp} = 27x^4\) \(x = \root 4 \of {K_{\rm sp}\over 27}\)
3:2 M3X2(s) ⇌ 3 M2+(aq) + 2 X3–(aq) Ksp = [M2+]3[X3–]2 \(K_{\rm sp} = 108x^5\) \(x = \root 5 \of {K_{\rm sp}\over 108}\)

Reaction Quotient vs Equilibrium Constant

forward\(\rightarrow\)
\(Q_{\rm sp} < K_{\rm sp}\) which means more solid will dissolve if there is any or nothing happens because the solution is undersaturated.
equil\(\rightleftharpoons\)ibrium
\(Q_{\rm sp} = K_{\rm sp}\) which means the solution concentrations will hold constant and there is no net change.
\(\leftarrow\)reverse
\(Q_{\rm sp} > K_{\rm sp}\) means the solution is oversaturatured and a precipitation will occur until the concentrations drop to saturation level.

Chemical Kinetics

Normalized Rate of Reaction

We can measure the rate of a reaction via any of the species in the over reaction:

\(a {\rm A} + b {\rm B} \rightarrow c {\rm C} + d {\rm D}\)

The rate of the "overall reaction as balanced" means you are measuring the rate of entire reactions as written. So each reactant and product is running at a scaled up version of the overall reaction. The "scaled up" amount is the coefficient for that species. For example, reactant A is running "a" times faster than the overall reaction. So to get ALL species to match up, we normalize the rates to the rate of one mole of reaction (the whole thing).

rate of reaction = \(-{\Delta[{\rm A}]\over a\Delta t} = -{\Delta[{\rm B}]\over b\Delta t} = {\Delta[{\rm C}]\over c\Delta t} = {\Delta[{\rm D}]\over d\Delta t}\)

Make sure you can convert the rate of any one species into the rate of another species. This is really just saying to know how stoichiometry works when dealing with rates.

Rate Laws and "order"

Rates of reactions are dependent on the concentrations of the reactant (and sometimes product) species. This is generalized by showing the rate law for a reaction of A, B, and C going to products.

\({\rm rate\;of\;rxn} = k [{\rm A}]^x[{\rm B}]^y[{\rm C}]^z\cdots \)

\(x\) is the order in A, \(y\) is the order in B, and \(z\) is the order in C. Sum those orders all together and you'll have what is called the overall order. The overall order is important to know and the units for the rate constant, \(k\), depends on the overall order.

Kinetic Rate Formulas for:    A → products

namezero orderfirst ordersecond order
rate law \({\rm rate} = k\) \({\rm rate} = k[{\rm A}]\) \({\rm rate} = k[{\rm A}]^2\)
integrated
rate law		 \([{\rm A}]_0 - [{\rm A}] = kt\) \(\ln[{\rm A}]_0 - \ln[{\rm A}] = kt\)

   \(\ln\left({[{\rm A}]_0 \over [{\rm A}]}\right) = kt\)		 \({1\over[{\rm A}]} - {1\over[{\rm A}]_0} = kt\)
straight line form
   (\(y = mx + b\))		 \([{\rm A}] = -kt + [{\rm A}]_0 \) \(\ln[{\rm A}] = -kt + \ln[{\rm A}]_0 \) \({1\over[{\rm A}]} = kt + {1\over[{\rm A}]_0}\)
half life \(t_{1/2} = {[{\rm A}]_0\over 2k}\) \(t_{1/2} = {\ln 2\over k}\) \(t_{1/2} = {1\over [{\rm A}]_0 k}\)
to actually plot on x,y graph
\(y\) = conc
\(x\) = time		 \(y = -kx + {\rm A}_0 \) \(y = {\rm A}_0 e^{-kx} \) \(y = \left(kt + {1\over{\rm A}_0}\right)^{-1}\)

Rate Constant, k, and Temperature, T

Arrhenius Equation:    \(k = A\,e^{-E_{\rm a}/RT} \)

\(E_{\rm a}\) is the activation energy for the reaction and it is always positive. The \(A\) is known as the Arrhenius "pre-exponential factor" and is unique for every reaction - it has the same units as the rate constant, \(k\). The factor can be eliminated by setting a ratio for two values of k at two different temperatures. Dr. McCord calls this the "user friendly" version of the Arrhenius equation.

\[\ln\left({k_2\over k_1}\right) = {E_{\rm a}\over R}\left({1\over T_1} - {1\over T_2}\right)\]

Notice that this is the third time you've seen this basic formula in CH302. The first time, it was the Clausius-Clapeyron equation for vapor pressures at two temperatures. The second time, it was the van't Hoff equation for the equilibrium constant, K, at two temperatures. Now the same set up works again but for rate constants. You should now be a "pro" at using this type of formula.

One more thing... the Arrhenius equation can be rewritten to be in a "straight line plot" form, \(y=mx+b\). That form is the following

\(\ln k = {-E_{\rm a}\over R}\left({1\over T}\right) + \ln A \)

So your plot is ln(k) vs 1/T and the slope will be equal to –Ea/R, and the y-intercept is equal to ln(A).