March 28, 2024, 12:08:47 PM
Forum Rules: Read This Before Posting


Topic: endothermic reaction & effects of solid/le chatelier's  (Read 1797 times)

0 Members and 1 Guest are viewing this topic.

Offline scientific

  • Regular Member
  • ***
  • Posts: 19
  • Mole Snacks: +0/-0
endothermic reaction & effects of solid/le chatelier's
« on: September 20, 2017, 07:06:21 PM »


C(s) + CO2 (g)  ::equil:: 2CO(g)

ΔH for the reaction above is greater than zero. Assuming ΔH is independent of temperature, which of the following statemetns about the percent yield of CO(g) is true? 
1. It increases as the amount of C(s) increases
2.It doubles when the initial partial pressure of CO2 is doubled


The two here are definitely incorrect answers, but I can't figure out why:

1. I don't understand why this is incorrect; wouldn't increased amount of C increase product, since there's more reactants to make product?
2. Is the answer supposed to be that doubling pressure of CO2 would QUADRUPLE yield of CO? Or is it that we don't know the actual relationship?

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: endothermic reaction & effects of solid/le chatelier's
« Reply #1 on: September 21, 2017, 10:34:02 PM »
Just some quick thoughts. It's an interesting question because they want to know about the % yield, not total yield. That may be an important consideration.

1. - The activity of a solid is by definition = 1, so isn't part of the equilibrium expression. Increasing the amount of solid does not shift the equilibrium.

2. - I think it is important to remember that the question wants to know what happens to the percent yield, not the total yield. The theoretical yield of the reaction is given by the special case where the equilibrium constant approaches infinity (at equilibrium, all substances are products). The % yield will be related to the ratio of the actual equilibrium constant to the theoretically infinite equilibrium constant, and the value will be dependent on what the actual equilibrium constant is. But, we just need to show that this option is wrong, so we don't need to worry about precise values here. Regardless of the value of the equilibrium constant, doubling the carbon dioxide concentration does increase the amount of product at equilibrium as a consequency of La Chatelier's principle, but doubling this reactant concentration (pressure) does not quite double the amount of product formed at equilibrium except in the limit that the equilibrium constant gets very large - in which case every mole of reactant creates two moles of product. This is because there is a back reaction that will be increasingly competitive as the pressure of product molecules increases (because it is a biomolecular reaction). 

You could simulate this with some simple ICE tables using idealized equilibrium constants and starting concentrations. I.e., try an equilibrium constant of 1 and initial CO2 and CO concentrations/pressures of 1 and 0 respectively (units don't matter here). For the theoretical yield, you just use stoichiometry, although you can also just use a really high equilibrium constant, say 10^6, and this gives you about the same product yield as stoichiometry. Anyway, using these values I get a % yield of about 39%. If I double the starting CO2 concentration, the percent yield goes down to about 29%*, reflecting the fact that increasing the concentration results in a more competitive bimolecular back reaction.

*Note that the total yield does go up, from about 0.78 to 1.18. But the percent yield goes down, because the theoretical yield goes from 2 to 4 for starting CO2 concentrations of 1 and 2. All this assumes no volumes are changed, and that the amount of solid carbon is not a limiting reagent.

Here are the values I got by using a simple ICE table for the calculation - equilibrium CO concentrations for input K and initial CO2 concentrations. Feel free to point out any errors. I did it all very quickly.


What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Vidya

  • Full Member
  • ****
  • Posts: 839
  • Mole Snacks: +46/-62
  • Gender: Female
  • Online Chemistry and Organic Chemistry tutor
    • Online Chemistry Tutor
Re: endothermic reaction & effects of solid/le chatelier's
« Reply #2 on: September 23, 2017, 06:16:51 AM »
Quote
C(s) + CO2 (g)  ::equil:: 2CO(g)

ΔH for the reaction above is greater than zero. Assuming ΔH is independent of temperature, which of the following statemetns about the percent yield of CO(g) is true? 
1. It increases as the amount of C(s) increases
2.It doubles when the initial partial pressure of CO2 is doubled
Let me help you in simple way
1.For amount of carbon which is in solid form ...it is not the part of equilibrium expression and hence can not shift equilibrium in any direction.
2.Its true that theoretical yield quadruples on doubling pressure of carbon dioxide gas.However in the question it is the percent yield.

Sponsored Links