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Equilibrium Constant. Example H 2 + F 2  2HF The equilibrium constant for the reaction above is found to be 2.1 x 10 3 at a specific temperature. At.

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Presentation on theme: "Equilibrium Constant. Example H 2 + F 2  2HF The equilibrium constant for the reaction above is found to be 2.1 x 10 3 at a specific temperature. At."— Presentation transcript:

1 Equilibrium Constant

2 Example H 2 + F 2  2HF The equilibrium constant for the reaction above is found to be 2.1 x 10 3 at a specific temperature. At equilibrium, the concentrations of both hydrogen and fluorine are found to be 0.0045 M. What is the equilibrium concentration of hydrogen fluoride?

3 Example A mixture of 5.00 x 10 -3 mole of H 2 and 1.00 x 10 -2 mole of I 2 is placed in a 5.00L container and allowed to come to equilibrium. Analysis of the equilibrium mixture shows the concentration of HI is 1.87 x 10 -3 M. Calculate the K c at 448°C for this reaction. H 2 + I 2  2HI

4 Example Sulfur trioxide decomposes at high temperatures in a sealed container as shown: 2SO 3  2SO 2 + O 2 Initially the vessel is charged at 1000K with 6.09 x 10 -3 M SO 3. At equilibrium the SO 3 concentration is 2.44 x 10 -3 M. Calculate the K c at 1000K.

5 K c and K p When concentration is expressed in M use K c as the constant. When concentration is expressed in atm use K p aA  cC + dD K p = P c c P D d / P A a

6 Example The reaction for the formation of nitrosyl chloride 2NO(g) + Cl 2 (g)  2NOCl(g) was studied at 25°C. The pressures at equilibrium were found to be P NOCl = 1.2atm, P NO = 5.0x10 -2 atm, and P Cl =.3atm. Calculate the K p for this reaction.

7 K p from K c K p = K c (RT)  n  n – total coefficients of gaseous products minus total coefficients of gaseous reactants R =.0821 L atm/mol K

8 Example The K c for the reaction N 2 + 3H 2  2NH 3 is.105. Calculate the K p at 472°C.

9 Solids and Liquids The concentrations of pure solids and liquids in a reaction does not really change They are not included in the equilibrium expression

10 Example Write the equilibrium expression for: CO 2 (g) + H 2 (g)  CO(g) + H 2 O(l) SnO 2 (s) + 2CO(g)  Sn(s) + 2CO 2 (g)


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