University College - Chemistry Chem 112A Lecture Notes - Lecture 6: Acid Strength, Ion, Inert Gas

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harlequinimpala978

13 Dec 2016

School

WUSTL

Department

University College - Chemistry

Course

University College - Chemistry Chem 112A

Professor

Bleeke John

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Note this does not mean that an acid has to produce an anion. A strong acid will have a one-way arrow since the equilibrium lies virtually (aq) + a- (aq) all the way to the right. We have conjugate acid-base pairs (a pair of species that differ by one proton): acid 1 (ha) and base 1 (a-), acid 2 (h3o+) and base 2 (h2o) The proton donating power of ha is greater than that of h3o+ and therefore. Ha is a stronger acid than h3o: there are only a few strong acids: hi, hclo4, hclo3, hbr, hcl, h2so4, hno3, a weak acid (ha) is one that is only partially deprotonated in water. The equilibrium lies mostly to the left (small equilibrium constant) and does (aq) + a- (aq) (cid:3118)= [(cid:3119)+] (cid:3117) [ ](cid:3117) = [(cid:3119)(cid:3042)+][ : weak acids not go all the way to the right, generic weak acid: ha(aq) + h2o(l) h3o, weak acid equilibrium constant, =(cid:3119)+ .

Related Questions

2. Solve the following related problems involving the equilibrium between dinitrogen tetroxide and nitrogen dioxide.

N₂O₄ <==> 2 NO₂

The equilibrium constant for this reaction is K_p = 0.60 at 350 degrees Kelvin. A 1.0 L vessel contains the above two gases in equilibrium at 350 K. The total pressure of the two gases, P_(NO₂) + P_(N₂O₄), in the vessel is 0.50 atm. From this information, calculate the equilibrium partial pressure for each gas. This is the initial equilibrium position for the reaction, which you will use to consider the effects of the following perturbations.

This equilibrium mixture is subjected to the following four perturbations:

a. addition of N₂O₄ gas to the vessel (which had been at the above equilibrium) to give a total concentration (before readjustment of the equilibrium position) of 0.375 atm in the reactant gas;

b. decreasing the total volume of the vessel to 0.50 L (in answering b-d, go back to the initial equilibrium mixture before then applying the perturbation);

c. increasing the total pressure inside the vessel to 1.5 atm by adding 1.0 atm of N₂ gas;

d. increasing the temperature of the gas mixture to 700 degrees K, a temperature at which the equilibrium constant is 2.40.

3. Hints: Note that the equilibrium constant is a K_p value and that reactant and product concentrations are given in terms of partial pressures.

For each of these perturbations, you can solve for the equilibrium partial pressures of the two gases. To do this you may have to set up an ICE table. Note that in ï¿½bï¿½, the inverse relationship between P and V in the ideal gas law allows you to easily determine the partial pressures of the reactant and product gases immediately after the volume increase. Then you can calculate the final equilibrium concentrations following this perturbation.

Finally, the exact solution to part ï¿½dï¿½ is challenging. You should first consider the effect, on the partial pressures of the reactant and product gases, of an increase in temperature from 350 to 700 degrees K. (Remember that the ideal gas law indicates a direct proportionality of P and T.) Then you should consider a shift in the equilibrium concentrations, along with the new equilibrium constant for the higher temperature.

Even if you are not able to determine numerical values for the equilibrium positions following the perturbations, you still should understand the concepts and be able to prepare the following written response.

4. Now, prepare an explanation: Attempt to solve for the extent and direction of the shift in the equilibrium as a result of the above four perturbations. If you are unable to solve for numerical values of the final equilibrium partial pressures, you should still know the predicted directions of the shifts.

After studying and solving the above problem, write an explanation of the Le Chatelier Principle. This explanation should give a general statement of the principle and then should explain how the above four types of perturbations are rationalized in terms of this principle.

Le Chatelierâs principle describes the way that a reaction atequilibrium will respond if that equilibrium is somehow disturbed.Imagine that you have a reversible reaction you want to performthat produces a very valuable product but the equilibrium constantfor the process is very low. Using the 3 definitions of equilibriumpresented in lecture as your framework, explain in no more than 1paragraph how the yield of reaction can be greatly increased(meaning reactants are more extensively converted into products) byapplying Le Chatliers principle.

The 3 definitions we learned in lecture:

1. rate of forward reaction= rate of reverse reaction

2. no observable change in [reactants] or [products]

3. no change in energy

I know that Le Chatliers principle states that if a system atequilibrium is disturbed, then the system will adjust so thatequilibrium can be established. I also know that if reactants aresuddenly added to a reaction already at equilibrium, then potentialenergy has effectively been added to reactant side vs. product sideof reaction. Since the system has been pushed to a higher levelthan what would be experienced at equilibrium, the reaction willproceed forwards until the lowest energy point (equilibrium) hasbeen reached. I am not sure though if this is the answer myteacher is looking for.

Does this make sense?

For a reaction aA rightarrow Products, [A]_o = 13 M, and the first two half-lives are 35 and 70 minutes, respectively, a) Write the rate law for this reaction. b) Calculate k. c) Calculate [A] at t = 335 minutes The Model of Chemical Kinetics represents the reaction rate constant as: k = (2p) times exp(-Ea/RT). What does each variable in the equation mean- that is, in regard to their contribution towards the reaction rate constant? In other words, what does each variable actually mean/refer to? Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 63 degree C to 73 degree C? ln(k_2/k_1) = E_a/R(1/T_1 - 1/T_2) (i) What is a catalyst? (ii) What are the two types of catalysts? Explain specifically why they are different? (iii) What is one example of each type of catalyst? (i) What does it mean when a reaction is in chemical equilibrium? Specifically, within the context of reactants? products, and the forward and reverse reaction rates, what does chemical equilibrium mean? (ii) What is Le Chateleir's Principle? Within the context of reactant/product concentration, temperature, and pressure, how does Le Chatelier's Principle predict a reaction will respond to variations in reactant, product concentration, temperature, and pressure. (i) For the given reactions: N_2(g) + 3H_2(g) harr 2NH_3(g), H_2O(g) + CO(g) harr H_2(g) + CO_2(g), HCN(aq) harr H^+(aq) + CN^-(aq), 2KClO3(s) harr 2KCl(s) + 3O_2(g) and 2H_2 O(l) harr 2H_2(g) + O_2(g). Write the equation that governs each of their equilibrium rate constant K (ii) If N_2(g) + 3H_2(g) harr 2NH_3(g)'s (at 25 degree C) has K = 25, what would K be (at 25 degree C) for 2NH_3(g) harr N_2(g) + 3H_2(g)? (iii) What would K be if [5 times (N_2(g)+3H_2(g) harr 2NH_3(g)]? (i) For N_2(g) + 3H_2(g) harr 2NH_3(g): P_NH3: 1.9 times 10^-1 atm, P_N2 = 8.9 times 10^-2 atm, P_H2 = 1.9 times 10^-2 atm, what is K_p? (ii) Using K_p, calculated in the previous step, what is K at 45 degree C, where K_p = K(RT)^Delta n? (R = 0.08206 L atm/mol K). What is the difference between homogeneous and heterogeneous? (ii) If K is larger than 1, does the equilibrium life farther to the right (favoring the production of products) or to the left (favoring reactant formation)? (iii) If K is less than 1, does the equilibrium life farther to the right (favoring the production of products) or to the left (favoring the production of reactants)?

University College - Chemistry Chem 112A Lecture Notes - Lecture 6: Acid Strength, Ion, Inert Gas

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