Equilibria: Part III. Reviewing the rules. Things to remember. When reacting species (both the reactants and the products) are in their most condensed form (solids), the units are mol/L (molarity) and the constant is expressed as Kc. Things to remember.
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Equilibria: Part III
Reviewing the rules
and the constant is expressed as
state, the concentrations can be
EXAMPLE: "CO(g) + 2H2(g)↔CH3OH(g) has a equilibrium constant of 10.5 at 220°C"
5. If you know the equilibrium constant going in one direction (say from left to right), the equilibrium constant in the other direction (say right to left) is the inverse of that value.
ClNO2(g) + NO(g) → NO2(g) + ClNO(g)
Kc = 1.3 x 10^4 at 25° C
NO2(g) + ClNO(g) → ClNO2(g) + NO(g)
Kc = 1/(1.3 x 10^4 at 25° C)
7.7 x 10^-5 at 25° C
6. This is how you convert an Equilibrium constant from Kc to Kp:
Kp = Kc (0.0821 T)^Δn
T is the temperature during the reaction in Kelvin. Remember that Kelvin has exactly the same degree of difference between integers as celsius, but it does have a different starting point. Kelvin's zero is 273° less than celsius. For example, 2° C is equal to 275° K. 27° C is equal to 300° K.
Δn is the sum of all stoichiometric coefficients belonging to products minus the sum of all stoichiometric coefficients belonging to reactants
8. If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium of the individual reactions.
Kc = (K'c) (K"c)
1st Step Reaction
2nd Step Reaction
1st step reaction
N2(g) + O2(g) ↔ 2 NO(g) Kc1 = 2.3 x 10^-19
2nd step reaction
2 NO(g) + O2(g) ↔ 2 NO2(g) Kc2 = 3 x 10^6
Write the equilibrium equation for this multi-step reaction:
Kc = Kc1 x Kc2 = (2.3 x 10^-19)(3 x 10^6)
Kc= 7 x 10^-13
N2(g) + 3 H2 (g)→ 2 NH3 (g)
Is 2.2 at this temperature, decide whether the system is at equilibrium. If not, predict which way the net reaction will proceed.
Nothing will happen. There will be no reaction.
LET'S LOOK AT HOW TO SOLVE THIS...
A = 1 mol/L
B = 2 mol/L
C = 0.02 mol/L
Therefore, in the future, we can just raise the Kc to a power equal to that of the stoichiometric coefficients belonging to the species.
Multiply the Kc' and Kc" to find the net Kc
Kc'Kc" = Kc = 0.25 x 10000 = 2500 (answer B)