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2. - ICE Tables - 2

3.  Previous chemistry courses, stoichiometric calculations were straightforward when reactions proceed to completion (called quantitative reactions – where virtually all of the limiting reagent is consumed)  When reversible reactions achieve equilibrium before all of the reactants become products, the stoichiometry needs more thought and organization (using ICE Tables) I – Initial C – Change E – Equilibrium 3

4. For systems composed of aqueous solutions or gases, I, means initial concentrations of reactants and products (before reaction) C, stands for the change in the concentrations of reactants and products between the start and the point at which equilibrium is achieved E, stands for the concentrations of reactants and products at equilibrium 4

5. Consider the following equation for the formation of hydrogen fluoride from its elements at SATP: H 2 (g) + F 2 (g)  2 HF (g) 5

6. If the reaction begins with 1.00 mol/L concentrations of H 2(g) and F 2(g) and no HF (g), calculate the concentrations of H 2(g) and HF (g) at equilibrium if the equilibrium concentrations of F 2(g) is measured to be 0.24 mol/L. List all given information: [H 2(g) ] initial = 1.00 mol/L [F 2(g) ] initial = 1.00 mol/L [HF (g) ] initial = 0.00 mol/L [F 2(g) ] equilibrium = 0.24 mol/L 6

7. H 2(g) + F 2(g) 2 HF (g) Initial concentration (mol/L) Change in concentration (mol/L) Equilibrium concentration (mol/L) 7

8.  The balanced chemical equation indicates that the change occurs in a 1:1:2 molar ratio, but we do NOT know what amount of the reactants is converted to product.  We choose the variable “x” to represent changes in the concentrations of reactants and products, with the coefficients of “x” corresponding to the coefficients in the balanced equation. 8

9. H 2(g) + F 2(g) 2 HF (g) Initial concentration (mol/L) Change in concentration (mol/L) Equilibrium concentration (mol/L) 9

10. Knowing that [F 2(g) ] equilibrium = 0.24 mol/L, we can determine the value of x. 1.00 mol/L – x = 0.24 mol/L - x = 0.24 mol/L – 1.00 mol/L - x = - 0.76 mol/L x = 0.76 mol/L 10

11. Now use the value of x to calculate the equilibrium concentrations of the other two entities. [H 2(g) ] = 1.00 mol/L – x = 1.00 mol/L – 0.76 mol/L = 0.24 mol/L [HF (g) ] = 2x = 2(0.76 mol/L) = 1.52 mol/L 11

12. The equilibrium concentrations of H 2(g) and HF (g) are 0.24 mol/L and 1.52 mol/L. Learning Tip Some problems do not provide you with M or mol/L. Instead they specify size of container and moles of reactant and product and you need to calculate molar concentration. C = molar concentration (mol/L) n = moles (mol) v = volume (L) 12

13. p. 437 Practice UC # 6, 7 p. 437 Section 7.1 UC # 1, 3, 4, 7, 8, 9(a) 13