H2 Chem MCQ - Equilibria

• gohby

  6 Aug 15, 18:00

  Hi UltimaOnline,

   

  Continuing from the previous thread, I have further questions on equilibria to seek clarification:

   

  Q1:

   

  http://img.photobucket.com/albums/v700/gohby/Chemistry/SRJC_zpswcv8i3zn.jpg

   

  Answer: C

  Remarks: Where pH=pKa, [weak acid] = [conjugate base]. Number of moles of CH3COONa = 0.0001. So shouldn’t the answer be D, because at point D the number of moles of excess ethanoic acid when 20cm³ of it has been added be 0.0001 too?

  
  

  Q2: Bleaching solutions are manufactured by dissolving chlorine gas in sodium hydroxide solution to give the following reaction. Cl2 (g) +  2OH­â€� (aq) ↔ OCl­â€� (aq) + Cl­â€�  (aq) + H2O (l) Users are warned not to mix the bleach with other cleaning solutions to prevent evolution  of hazardous chlorine gas. Which of the following actions will lead to liberation of chlorine  gas?

   

  A Addition of water to bleach

  B Mixing of an alkali with bleach

  C Shaking bleach with table salt, NaCl

  D Subjecting bleach to high pressure

   

  Answer: C

   

  Remarks: Why doesn’t A result to the liberation of chlorine gas too since the equilibrium will shift to the left upon the addition of water? Wrt choice D, if I were to subject the equilibrium system to high pressure, am I right to say that the equilibrium will shift to the right so as to reduce pressure by reducing the number of gaseous particles instead?

   

  Q3: A car burning lead­â€�free fuel has a catalytic converter fitted to its exhaust. On analysis, its exhaust gases are shown to contain small  quantities of nitrogen oxides.  Which modifications would result in lower exhaust concentrations of nitrogen oxides?

  1 an increase in the surface area of the catalyst  in the converter.

  2 an increase in the rate of flow  of the exhaust gases through the converter.

  3 a much higher temperature of combustion in the engine

   

  Answer: 1 only

   

  Remarks: My understanding is this: 2 is wrong because an increase in the rate of flow of gases does not increase the number of active sites on the catalyst, which will not lead to an increase in the rate of reaction. 3 is also wrong because combustion is not a reversible reaction so the higher temperature will not affect the concentration of products. Am I right?

   

  Q4: Pure NOCl gas, was heated at 320°C in a 2 dm³ vessel. At equilibrium, 30% of the NOCl  gas had dissociated according to the equation below and the total pressure was p atm. 2NOCl (g) ↔ 2NO (g) + Cl2 (g)  What is value of Kp?

   

  Answer: 0.0120p

   

  Remarks:

   

  My workings:

   

                       2NOCl (g) ↔ 2NO (g) + Cl2 (g)

  Initial no of moles        a                  0        0       

  Change “        -0.3a        0.3a        0.15a

  Eqm “            0.7a        0.3a        0.15a

  Mole Ratio        0.7/1.15    0.3/1.15    0.15/1.15

  Kp = (0.15/1.15)p  x  (0.3/1.15)²p²

          _______________________

             (0.7/1.15)²p²                

   

  That works out to ~0.024p, and I can’t figure out where the mistake lies.

   

  Thank you for your time, UltimaOnline :)

• UltimaOnline

  6 Aug 15, 18:35

  Hi Gohby,
  
  As always, you've a very good style of asking questions, with your own mathematical working and thought processes indicated clearly.
  
  Q1. You're right, the correct answer is D.
  
  Q2. While you're right that adding water will shift the p.o.e to the LHS, but the effect is less significant than option C, because as indicated by (aq) state symbol, water is already present in very large excess. For D, notice the qn states increasing pressure on bleach, not the entire system. Since bleach is aqueous, increasing pressure on an aqueous species has no effect on p.o.e.
  
  Q3. Options 2 & 3 will *increase* the molarities of the oxides of nitrogen. Option 2 affords less time and opportunity available for the catalysts to do their work. Option 3 provides more energy above the activation energy barrier for oxides of nitrogen to be generated.
  
  Q4. Your working and answer is correct. Volume of container affects the answer only for Kc, not Kp.
• gohby

  19 Aug 15, 12:05

  Hi UltimaOnline,

  I have further questions on Ionic Equilibria -

  Q5: HA is a weak acid and can have different degree of acidity in aqueous solution and in liquid ammonia. The respective equations that represent their dissociations are as follows:

  
  

  HA + H2O ↔ A- + H3O+

  HA + NH3 ↔ A- + NH4+

  
  

  Which of the following statements is correct

  
  

  1: Degree of dissociation of HA is identical in aqueous solution and liquid ammonia.

  2: PKa of NH3 is larger than that of H2O, hence HA is a stronger acid in liquid ammonia.

  3: Kb of NH3 is larger than that of H2O, hence HA is a stronger acid in liquid ammonia.

  
  

  Remarks: The answer is 3.

  
  

  1 is wrong because even though the Ka is constant be it in aqueous solution or liquid ammonia, the degree of dissociation would differ if [H2O] or [NH3] differs, is that right?

  2 is wrong because we do not talk about PKa for bases?

  Why is 3 correct? If Kb of NH3 is larger than that of H2O, wouldn’t that only tell us that HA is a weaker base in liquid ammonia, rather than it being a stronger acid?

  
  

  Q6:

  
  

  http://img.photobucket.com/albums/v700/gohby/Chemistry/YJC_zpshsaymyxb.jpg

  
  Remarks: My workings are as follows - the salt, sodium propanoate, is alkaline. Hence, I obtained the Kb of the salt, used the ICE calculations to obtain the [OH-] and subsequently the pH, which works out to be 8.94. However, how does that tell us about the suitability of the indicators for the titration?

  
  Thank you for your help. :)

• UltimaOnline

  19 Aug 15, 22:45

  Q5. It's not the molarity that matters, but the identity of the solvent. By solvent, we assume present in very large excess, such that its molarity is irrelevant. Yes, Ka of NH3 is irrelevant here, as NH3 functions as a base here. Since NH3 is a stronger base than H2O, it forces the deprotonation of HA to a greater extent compared to H2O, hence the differing Ka values in different solvents.
  
  Q6. A suitable indicator would be one whose own pKa value is close to the pH of the solution at equivalence point, and would therefore change colour when equivalence point is near. Hence, only phenolphthalein can be used for this titration. Bear in mind that endpoint only approximates equivalence point.