Topic 11

Topic 11

1. Calculating Kc

   • [H2O] is omitted from Kc expression as the value is constant
   • When working out the units of Kc, the units can cancel out hence you would write no units.
   • When the value of Kc is greater than 1 then at equilibrium the concentrations of products is higher than the concentrations go reactants. This means there is a higher yield of products. The graph shows equilibrium lying to the right.
   • When Kc is less than 1 then at equilibrium the concentration of reactants is greater than the concentration of products, hence there is a lower yield of products. The graph shows equilibrium lying to the left.

2. ICE Tables- Calculating Kc

   • Method:
     • 1) Draw ICE table and fill in what you can
       • 2) Deduce concentrations at equilibrium
         • 3) Write out Kc expression
           • 4) Substitute in concentrations in Kc expression
             • 5) Work out the units based off Kc expression
   • I- Initial moles, C- Change in moles, E- moles at equilibrium, [ ]- concentration. If there was no volume equilibrium=concentration

3. Calculating Kp

   • PARTIAL PRESSURE- Pressure a gas would exert if it were on its own
   • PARTIAL PRESSURE = MOLE FRACTION X TOTAL PRESSURE
     • MOLE FRACTION OF A= NUMBER MOLES OF SPECIES A / TOTAL NUMBER OF MOLES OF ALL SPECIES
   • 0.546 moles PCl3 and 0.373 moles Cl2 were heated. At equilibrium there were 0.178 moles PCl5. Ptot=225kPa. What is the Kp?
     • PCl3 +. Cl2. ⇌. PCl5
       • 0.546. 0.373. 0
         • -0.178. -0.178. +0.178
           • 0.368. 0.195. 0.178
             • 0.4966... 0.2631.... 0.2402...
               • 111.740... 59.21... 54.048...
                 • Ptot = 225kPa
                 • Kp= p(PCl5) / (p(Cl2) X p(PCl3))= 8.17X10-3 kPa-1
             • 0.741
     • I
       • C
         • E
           • Xf
             • PP

4. Manipulating K

   • Increasing the concentration of the reactant- The value of Kc is unaffected as only temperature affects is. If the concentration of A increases then the quotient would have a larger denominator hence the quotient would be less than K. The quotient would have to shift to the right to restore the value of K.
   • Increase concentration of product- The value of of Kc is unaffected since only temperature can affect it. If concentration of D increases then the quotient would have a larger numerator hence the quotient would be greater than K. The quotient would have to shift to the left to restore the value of K
   • Effect of pressure on Kp
     • 1) K remains the same as isn't affected by pressure
       • 2) Which has changed: Denominator or numerator
         • 3) Value of Q vs K (which is bigger?)
           • 4) How will equilibrium shift?
   • Increasing temperature (endothermic reaction)- An increase in temperature would affect Kc value. The value of K would increase so the old Kc value becomes the quotient. The quotient shifts to the right to restore K
   • Decreasing temperature (endothermic reaction)- A decree in temperature would affect Kc value. The value of K would decrease so the old Kc value would become the quotient. The quotient would then shift to the left to restore K.
   • Increasing temperature (exothermic reaction)- An increase in temperature would affect Kc value. Increasing temperature causes Kc to decrease so old Kc becomes quotient. Quotient shifts let to restore K.

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