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Given [concentration] of all species |
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Write the Kc expression |
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Fill in the expression accordingly |
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Solve for Kc |
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Given (original concentration) of one species
and [EC] of another |
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Write the Kc expression |
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Set up chart filling in what you know |
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Figure out change: [EC] – (OC) |
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Determine change for all species |
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Calculate [EC] for each species |
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Fill in the expression accordingly |
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Solve for Kc |
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Given (original concentration) of one species
and how much of it was used |
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Write the Kc expression |
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Set up chart filling in what you know |
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Figure out change: (OC) x % of change |
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Determine change for all species |
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Calculate [EC] for each species |
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Fill in the expression accordingly |
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Solve for Kc |
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OCQ:
Original Concentration Quotent |
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Use OCQ to determine direction of the rxn when
(OC) for all species is given |
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If OCQ > Kc rxn |
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If OCQ < Kc rxn |
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Given Kc and [concentration] of all species
except one. |
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Write the Kc expression |
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Fill all known values into the expression and
solve for the unknown |
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Given Kc and only the (OC) for all species |
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Set up chart filling in what you know |
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Let “x” represent the change – determine the
direction of the rxn if necessary
(Use OCQ) |
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Let “x” be negative on the side reacting and
positive on products side. |
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Given Kc and only the (OC) for all species (continued) |
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Multiply “x” by coefficients in the equation for
that species |
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Substitute the [OC +/- change] as [EC] value and
solve for “x” |
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Substitute value of “x” to get actual [EC] |
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When a system at equilibrium is subjected to a
stress, the system will shift in a direction so as to relieve the stress. |
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What are some of these “stresses”? |
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Adding or subtracting a species |
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change in volume of the container |
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change in the pressure |
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change in the temperature |
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Adding a species. |
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Rxn will shift in the direction away from the
species added. |
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Removing a species. |
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Rxn will shift in the direction toward the
species removed. |
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Changing the volume of the container. |
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Increase in the volume. |
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Rxn will shift toward the side with the most
moles of gas. |
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Decreasing the volume. |
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Rxn will shift toward the side with the fewest
moles of gas. |
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Changing the volume will make no difference if
there are the same number of moles of gas on each side. |
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Changing the pressure on the container. |
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Increase in the pressure. |
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Rxn will shift toward the side with the fewest
moles of gas. |
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Decreasing the pressure. |
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Rxn will shift toward the side with the most
moles of gas. |
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Changing the pressure will make no difference if
there are the same number of moles of gas on each side. |
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Changing the temperature. |
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Affect determined by sign of DH |
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New value for Kc also a result |
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Increase in temperature shifts the reaction in
the endothermic direction |
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Decrease in temperature shifts the reaction in
the exothermic direction. |
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Kp is the equilibrium constant based on
equilibrium partial pressures in atm. |
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It is related to Kc by the equation: |
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Kp = Kc(RT)Dng where R is the Ideal Gas Constant
and T is the temperature. Dng is the change in number of moles of gas as
the reaction is read, left to right. In other words, # mol of gas products
- # mol of gas reactants. |
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Mathematical relationship is: DG = -0.0191 T
logK |
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K stands for every equilibrium constant used in
different types of equilibria. It
applies to Kp and NOT Kc for gaseous equilibria. |
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If DG < 0, then log K > 0 and K > 1 |
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when all species are at unit concentrations the
reaction is spontaneous to the right. |
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If DG > 0, then log K < 0 and K < 1 |
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the reaction is spontaneous to the left |
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If DG = 0, then log K = 0 and K = 1 |
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the reaction is at equilibrium |
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