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Lesson 2 Equilibrium Law

Lesson 2 Equilibrium Law. The Equilibrium Expression. Mass action expression. Equilibrium constant. Kc =. Introduction to the Equilibrium law. [H 2 O]. [H 2 O] 2. [H 2 ] 2 [O 2 ]. [H 2 ] 2 [O 2 ]. 2H 2 (g) + O 2 (g)  2 H 2 O (g). 2 H 2 (g) + O 2 (g)  2H 2 O (g).

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Lesson 2 Equilibrium Law

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  1. Lesson 2 Equilibrium Law

  2. The Equilibrium Expression

  3. Mass action expression Equilibrium constant Kc = Introduction to the Equilibrium law [H2O] [H2O]2 [H2]2[O2] [H2]2[O2] 2H2(g) + O2(g)  2H2O (g) 2H2(g) + O2(g)  2H2O (g) 2H2(g) + O2(g)  2H2O (g) 2H2(g) + O2(g)  2H2O (g) 2H2(g) + O2(g)  2H2O (g) Step 1: Set up the “equilibrium law” equation Step 2: Product concentrations go in numerator Step 3: Concentration in mass action expression is raised to the coefficient of the product Step 4: Reactant concentrations go in denominator Step 5: Concentrations in mass action expression are raised to the coefficients of reactants

  4. Equilibrium law: important points • State (g, l, s, aq) may or may not be added at this point since we will only be dealing with gases for this section. Later it will matter. • Homogeneous equilibrium: A reaction with reactants and products in the same phase. • Hetrogeneous equilibrium: A reaction with reactants and products in the different phase. • The equilibrium law includes concentrations of products and reactants in mol/L (M) • By substituting equilibrium concentrations into equilibrium law, we can calculate Kc …

  5. Equillibrium constant and Temperature • The value of Kc will depend on temperature, thus this is listed along with the Kc value. • Changing temperature changes rate of forward and reverse reactions by different amounts because Ea for forward and reverse reactions are different. • Numerical value of Kcdoes not depend on whether starting point involves reactants or products. • Kcdoes not depend on starting concentrations.

  6. Example 1 Given: CO + 2H2⇌ CH3OH Write an appropriate equilibrium expression.

  7. Equilibrium law: CO(g) + 2H2(g)  CH3OH(g) • Calculate K for the reaction above if the equilibrium concentration of the reactants and products respectively are [CH3OH]= 0.00261M, [CO]= 0.105 M and [H2] = 0.250 M

  8. Kc = [0.00261] ,Kc= [0.105][0.250]2 Equilibrium law: [CH3OH] CO(g) + 2H2(g)  CH3OH(g) =0.398 [CO][H2]2

  9. Equilibrium law 2. N2(g) + 3H2(g)  2NH3(g) Calculate K for the reaction above if the equilibrium moles of the reactants and products in a 2 L container respectively are [NH3] = 0.150 moles, [H2] and [N2] have equal moles of 0.300 moles.

  10. Kc = [x] ,Kc= [0.210][0.100]2 Equilibrium law- Finding [ ]eq given K 1) What is the concentration of Methanol if K = 0.500 and the the concentrations at equilibrium of CO = 0.210 M and H2 = 0.100 M respectively? CO(g) + 2H2(g)  CH3OH(g) [CH3OH] = 0.500 [CO][H2]2 0.00105 M x = [0.210][0.100]2 (0.500) =

  11. Finding [ ]eq K for the reaction below at 2800 C is 64. At equilibrium the [NH3] = 0.280 M and the [N2] = 0.00840 M. What is the concentration of [H2] at equilibrium? N2(g) + 3H2(g)  2NH3(g)

  12. R I C E Using ICE tables or RICE table to find Kc. For the following reaction, H2 + I2 2HI calculate K, if 0.100 M each of H2 and I2 are placed in a container and at equilibrium only 0.020 M of I2 is present. Ratio, Initial, Change, Equilibrium H2 I2 HI 1 1 2 0.100 0.100 0 -x -x +2x 0.100 –x 0.100 –x 2x

  13. The Magnitude of K The value of K helps to predict the relative concentrations of reactants and products in an equilibrium system.

  14. Equilibrium constant for reverse reaction • K’ = 1/K K = Equilibrium for forward reaction K’= Equilibrium for reverse reaction

  15. Homework • Read 7.2 • P. 437 #3,4,7,8 • P. 442 #1 • P. 444 #3-6

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