7. Equilibria

14/4/15. There’s a post here: Electrode potentials under non-standard conditions. which is a kind of “cross topic” issue. It has an “applied equilibrium” dimension to it, hence I’m posting it here.

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Feel free to leave any other comments / questions on topic 7. Equilibria.

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9 thoughts on “7. Equilibria

    • A homogeneous reaction is one in which the states of all species is the same.
      *think of homo meaning ‘the same’)
      e.g. H2(g) + I2(g) ⇌ 2HI(g) – all species in this case are gaseous
      You could of course have all liquid reaction, etc….

      heterogeneous refers to different states being present. (think of Hetero = different)
      An example of such an equilibrium is CaCO3(s) ⇌ CaO(s) + CO2(g)
      note we have solids and gases in the same reaction
      (also note the eqn of this reaction is only noticeable in a closed (and usually small) system)

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  1. CH3CH2OH(l) + CH3CH2COOH(l) ——–(equilibrium arrows) CH3CH2COOCH2CH3(l) + H2O(l)
    Kc=7.5 at 50degrees celsius. If 50.0g of ethanol is mixed with 50.0g of propanoic acid, what mass of ethyl propanoate will be formed at equilibrium?

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    • This is a “special case” It’s a 1+1 –> 1+1 reaction. There is a balance in the stoichiometry on LHS and RHS so moles can be used in the Kc equation as the volumes in numerator and denominator will cancel. Can always do this when Kc is unitless

      Then something like the ICE method….

      This kind of question would be (perhaps) one of the hardest calc based questions that ypu’ll get on this topic. I’m still not too familiar with CIE exam patterns but I’d be pretty sure this kind of question will appear on almost every set of exams (possibly alternating with a Kp question)

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    • Just to remind people of the Haber process N2(g) + 3H2(g) ⇌ 2NH3(g) H= -92 kJ mol-1

      Reaction rate is proportional to temperature, so high temperatures increase the rate of production of ammonia, but, the as the reaction is exothermic, increased temperatures will lower the equilibrium constant so yield will decrease. A compromise temperature of about 400oC is used which gives the best rate against yield tradeoff

      Application of increased pressures cause the instantaneous pressure equilibrium quotient, Q(p), to be further away from the pressure equilibrium constant, Kp, when there is less moles of gaseous reactant (denominator) than gaseous product (numerator). Hence use of elevated pressures caused the position of the equilibrium will shift to the right and yield of ammonia will increase. But high pressure equipment and its maintenance is costly, reducing profits. A compromise pressure of about 200 atm is used to get greater yield at not too costly a price.

      Comments: This is probably one of the most (there are a couple more) “wordy” aspects of chemistry you’ll come across. It’s kind of hard to be very very brief when answering. A ‘method’ of answering is to state the effects on rate and yield for the variables like temp and pressure (note I left off the inc. rate dur pressure as the effect in this case is regarded as one of the more minor effects) and then stating the compromise (a ‘key word’) is employed and give the actual compromise conditions used.

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  2. The solubility product, Ksp, of magnesium hydroxide has a numerical value of 2.0 × 10–11. Use the value of Ksp given to calculate the concentration of Mg(OH)2 in a saturated solution.
    Explain whether magnesium hydroxide would be more or less soluble in 0.1 mol dm–3 MgSO4(aq) than in water.

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    • You should be able to state a Ksp equation for any salt.

      It is the product of the concentration of the aqueous ions, each ion raised to the power of it’s respective stoichiometric amount in the salts formula.

      Ksp for Mg(OH)2 is therefore given by…

      Ksp = [Mg2+(aq)]^1 x [OH-(aq)]^2

      the symbol ‘^’ is computer symbology for “raised to the power of”

      The Ksp equation relates to the concentration of those ions at saturation

      These questions very often require the student to expressing the concentration of one of the ions in terms of the other (see below). This allows is to reduce the number of mathematical ‘unknowns’, enabling us to solve for just one ion.

      e.g. In Mg(OH)2 for every one mole of salt, we et one mole of Mg2+ and two moles of OH-
      The concentration of OH- is twice [Mg2+]

      // I can’t be bothered to write in the (aq) symbols, but they should be there //

      so if z = [Mg2+]
      then [OH-] = 2z

      putting these into the Ksp equation we get…
      Ksp = (z) x (2z)^2

      so Ksp = 4z^3 (typo corrected)

      Rearrange to solve for z
      and then use the 1 salt = 1 Mg2+ : 2OH-
      ratio’s to work out the concentration of the Mg(OH)2 at saturation.

      The second part of the question refers to “the common ion effect” You should be able to get the answer after reading about it.

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