Midterm 1! Flashcards
a) Explain the term elementary step.
An elementary step is one that cannot be subdivided into simpler steps.
c) Give the rate law for rate determining step. Why does this step determine the rate?
d) Express the overall rate law in terms of reactants of the net reaction.
c) Give the rate law for rate determining step. Why does this step determine the rate?
d) Express the overall rate law in terms of reactants of the net reaction.
d) What is the overall order of the reaction? Justify your answer
e) What would happen to the reaction rate if [BrO3-]i were halved? Explain.
vnet = k3k2k1[BrO3-][H+]2[Br-] /(k-1k-2)
d) What is the overall order of the reaction? Justify your answer
1+2+1 = 4th order overall this is the sum of exponents on concentration in the rate law.
e) What would happen to the reaction rate if [BrO3-]i were halved? Explain. The rate would decrease by 1/2 since the above rate law is first order in [BrO3-].
- (3) Write the following general reactions: Autodissociation of water:
Base Hydrolysis (Let the base be the same as conjugate base of the weak acid in the next reaction.)
Acid Disscoaition:
Write the equilibrium expressions for the above reactions (Kw, Ka, and Kb) and use them to show the mathematical relationship that relates all three equilibrium constants to each other.
- (3) List the strong acids (all 7, and be specific about what is strong for the one diprotic acid).
- (3) List the strong acids (all 7, and be specific about what is strong for the one diprotic acid).
HCl, HBr, HI, HClO4, HClO3, H2SO4 (1st H+ only), HNO3
- (3) Give an example of a super acid or a super base and explain what makes it “super.”
- (3) Give an example of a super acid or a super base and explain what makes it “super.”
Superbase: H:- (hydride). or RO- (deprotonated alcohols) are examples we have discussed.
Super bases are stronger bases than OH- in H2O. When exposed to water, they are completely destroyed and result in the production of OH-.
Superacids: Protonated alcohols. There are many others we haven’t discussed.
Super acids are stronger acids than H3O+ in H2O. When exposed to water, they are completely destroyed and result in the production of H3O+.
- (5) Give. The pH of a 1.0 M solution of HClO (Ka = 3.0 × 10-8).
- (8) For the equilibrium: A(g) + 2 B(g) ⇌ C(g) + 2 D(g) K=3.5x10-20
If you combine C (0.400 M) with D (0.500M) with no A or B initially present, calculate the equilibrium concentrations of all substances. Clearly state any assumptions you make to solve this problem (there are two important assumptions that vastly simplify this problem)!
- (8) For the equilibrium: A(g) + 2 B(g) ⇌ C(g) + 2 D(g) K=3.5x10-20
If you combine C (0.400 M) with D (0.500M) with no A or B initially present, calculate the equilibrium concentrations of all substances. Clearly state any assumptions you make to solve this problem (there are two important assumptions that vastly simplify this problem)!
Since the reactants are vastly favored by K<<1, we will comes a solution without having to solve a high order (in this case, 4th order) polynomial if we reframe the question such that our starting conditions are mostly reactants instead of the starting conditions consisting mostly of products that are given.
To do this, we need to consider limiting reactants, and convert products to reactants while considering stoichiometry.
By inspection, D is limiting.
- (6) To answer the following questions, either refer to the Maxwell-Bolztzman distribution, or a molecular scale point of view. Pictures and graphs will aid in your explanations.
a) Lowering temperature decreases rates of all reactions in the gas, solution state, or liquid.
b) Reaction rates slow as reactions near completion.
- (6) To answer the following questions, either refer to the Maxwell-Bolztzman distribution, or a molecular scale point of view. Pictures and graphs will aid in your explanations.
a) Lowering temperature decreases rates of all reactions in the gas, solution state, or liquid.
b) Reaction rates slow as reactions near completion.
- (8) Given the above tabulated experimental information for the below [ClO 2] i (M) reaction, answer the following questions.
2ClO2(aq) + 2OH-(aq) → ClO3-(aq) + ClO2-(aq) + H2O(l)
a. Determine the rate law for the reaction. Show your work and/or explain your reasoning.
b. Calculate the value of the rate constant for this reaction. Note that it is sufficient for you to do so for only one of the above experiments.
- (6) The α-amylase enzyme in the bacterium, Bacillus halmapalus, will denature when it is in a Ca2+ ion deficient aqueous environment, indicating that Ca2+ is involved in maintaining the structure of the enzyme.1 When an enzyme or other protein is denatured, it changes shape, or otherwise degrades from the fully functional form that the enzyme has in its natural environment. In a 2003, paper, A. D. Nielsen and coworkers conducted kinetics studies on the denaturation of the enzyme, and published the plot below.
a) Using the above plot, estimate the activation energy only (give units) for the denaturation of the α-amylase enzyme. In doing so, clearly indicate the coordinates you are using as data points.
b) The above reaction is third order. Give units for the frequency factor and justify your answer.
- (2) For reasonably dilute aqueous reaction mixtures, why does [H2O(l)] not appear in equilibrium expressions? Materials of what other phase are also left out of the equilibrium expression?
- (2) For reasonably dilute aqueous reaction mixtures, why does [H2O(l)] not appear in equilibrium expressions? Materials of what other phase are also left out of the equilibrium expression?
Because all concentration terms in the equilibrium expression are divided by the standard state concentration for that substance. For liquids, this is the concentration of the pure liquid, 55.4 M for H2O at 298 K. For reasonably dilute solutions, [H2Osolvent] ≈ 55.4 M, so the term in the equilibrium expression for H2O approaches unity (one) and is left out. Solids are also left out.
- (5) Consider the reaction below. If 16.0 g of CO2 is dissolved in 1.00 L of pure water, the equilibrium concentration of H2CO3(aq), is 6.17 x 10-4. Give the equilibrium constant
- For the reaction: HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq), Keq = 6.3 x 10-4
b) (2) If Ag+ is added to the above reaction system, resulting in the precipitation of F- in the form AgF, how will the equilibrium system adjust to reestablish equilibrium.
- For the reaction: HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq), Keq = 6.3 x 10-4
b) (2) If Ag+ is added to the above reaction system, resulting in the precipitation of F- in the form AgF, how will the equilibrium system adjust to reestablish equilibrium.
The addition of Ag+(aq) will result in the removal of F-(aq) from solution, and thus, by L’Chatellier’s Principle will result in the above reaction shifting towards products to reestablish equilibrium.
- For the reaction: HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq), Keq = 6.3 x 10-4
c) (2) How will the addition of a small amount of water affect the above equilibrium reaction? Explain.
- For the reaction: HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq), Keq = 6.3 x 10-4
c) (2) How will the addition of a small amount of water affect the above equilibrium reaction? Explain.
K = [H3O+][F-]/[[HF]. The addition of a small amount of water will result in an increase in volume, and a decrease in concentration of all species in solution. Mathematically, this has a larger effect for products (numerator in the equilibrium expression). Thus, the system will shift towards products to reestablish equilibrium.
