a A +b B in Visual C#.NET

Reader Data Matrix ECC200 in Visual C#.NET a A +b B

a A +b B
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c C + dD
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It is important to remember that at equilibrium the concentrations of the chemical species are constant, not necessarily equal There may be a lot of C and D and a little A and B, or vice versa The concentrations are constant, unchanging, but not necessarily equal At any point during the preceding reaction, a relationship may be defined called the reaction quotient, Q It has the following form:
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[C]c [D]d [A]a [B]b
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The reaction quotient is a fraction In the numerator is the product of the chemical species on the right-hand side of the equilibrium arrow, each raised to the power of that species coefficient in the balanced chemical equation It is called the Qc in this case, because molar concentrations are being used If this was a gas-phase reaction, gas pressures could be used and it would become a Qp Remember: products over reactants
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Keywords and Equations
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KEY IDEA
Q = reaction quotient
[C]c [D]d , where aA + bB cC + dD [A]a [B]b
Equilibrium Constants: K = equilibrium constant Ka (weak acid) Kb (weak base) Kc (molar concentrations)
Kw (water)
Kp (gas pressure)
Ka =
[H+ ] [A ] [OH ] [HB+ ] Kb = [HA] [B]
Kw = [OH ] [H+] = 10 10 14 = Ka Kb at 25 C + pH = log [H ], pOH = log [OH ] 14 = pH + pOH
pH = pK a + log
[A ] [HA] [HB+ ] [B]
pOH = pK b +log
pKa = log Ka, pKb = log Kb
Kp = Kc(RT ) n, where n = moles product gas moles reactant gas
Gas constant, R = 00821 L atm mol 1 K 1
Equilibrium 213
Equilibrium Expressions
The reactant quotient can be written at any point during the reaction, but the most useful point is when the reaction has reached equilibrium At equilibrium, the reaction quotient becomes the equilibrium constant, Kc (or Kp if gas pressures are being used) Usually this equilibrium constant is expressed simply as a number without units, since it is a ratio of concentrations or pressures In addition, the concentrations of solids or pure liquids (not in solution) that appear in the equilibrium expression are assumed to be 1, since their concentrations do not change Consider the Haber process for the production of ammonia: N 2 (g ) + 3H 2 ( g) 2NH3 ( g)
The equilibrium constant expression would be written as: Kc = [ NH3 ]2 [ N 2 ][H 2 ]3
If the partial pressures of the gases were used, then Kp would be written in the following form: Kp=
2 PNH
3 PN PH
KEY IDEA
There is a relationship between Kc and Kp: Kp = Kc(RT ) , where R is the ideal gas constant (00821 L atm/mol K) and n is the change in the number of moles of gas in the reaction Remember: Be sure that your value of R is consistent with the units chosen for the partial pressures of the gases For the following equilibrium Kp = 190: C(s) = CO2(g) this equilibrium at 25 C C(s) + CO2 ( g ) K p = K c (RT ) n 1 90 = K c [(0 0826 L atm)(298K)](2 1) [( mol K)] 2CO(g) K p = 1 90 2 CO(g) Calculate Kc for
K c = 0 0777 The numerical value of the equilibrium constant can give an indication of the extent of the reaction after equilibrium has been reached If the value of Kc is large, that means the numerator is much larger than the denominator and the reaction has produced a relatively large amount of products (reaction lies far to the right) If Kc is small, then the numerator is much smaller than the denominator and not much product has been formed (reaction lies far to the left)
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^ Le Chatelier s Principle
At a given temperature, a reaction will reach equilibrium with the production of a certain amount of product If the equilibrium constant is small, that means that not much product will be formed But is there anything that can be done to produce more Yes, there is ^ ^ through the application of Le Cha telier s principle Le Cha telier, a French scientist, discovered that if a chemical system at equilibrium is stressed (disturbed) it will reestablish equilibrium by shifting the reactions involved This means that the amounts of the reactants and products will change, but the final ratio will remain the same The equilibrium may be stressed in a number of ways: changes in concentration, pressure, and temperature Many times the use of a catalyst is mentioned However, a catalyst will have no effect on the equilibrium amounts, because it affects both the forward and reverse reactions equally It will, however, cause the reaction to reach equilibrium faster
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