Equilibrium Calculations Quiz

PDF page 753: reactant and product concentrations initially present, for how they change as the reaction proceeds, and for what they are when the system reaches equilibrium. The acronym ICE is commonly used to refer to this mathematical approach, and the concentrations terms are usually gathered in a tabular format called an ICE table.

PDF page 752: 3 Δ[NH ] = −Δ[N ] = −2x 3 2 ⎝ ⎠ 1 mol N 2 where the negative sign indicates a decrease in concentration.

PDF page 752: either NH3 only, or a mixture of any two of the three chemical species involved in the equilibrium. Regardless of its initial composition, a reaction mixture will show the same relationships between changes in the concentrations of the three species involved, as dictated by the reaction stoichiometry (see also the related content on expressing reaction rates in the chapter on kinetics). For example, if the nitrogen concentration increases by an amount x: Δ[N ] = + x 2 the corresponding changes in the other species concentrations are ⎛3 mol H ⎞

PDF page 752: 2 Δ[H ] = Δ[N ] = +3x 2 2 ⎝ ⎠ 1 mol N 2

PDF page 752: −x −x x C H (g) + 5O (g) ⇌ 3CO (g) + 4H O(g) 3 8 2 2 2 (c) x 5x −3x −4x Check Your Learning Complete the changes in concentrations for each of the following reactions: 2SO (g) + O (g) ⇌ 2SO (g) 2 2 3 (a)

PDF page 752: The equilibrium constant for a reaction is calculated from the equilibrium concentrations (or pressures) of its reactants and products. If these concentrations are known, the calculation simply involves their substitution into the K expression, as was illustrated by Example 13.2. A slightly more challenging example is provided next, in which the reaction stoichiometry is used to derive equilibrium concentrations from the information provided. The basic strategy of this computation is helpful for many types of equilibrium computations and relies on the use of terms for the

PDF page 756: Under certain conditions, the equilibrium constant Kc for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is 0.0211. What are the equilibrium concentrations of PCl5, PCl3, and Cl2 in a mixture that initially contained only PCl5 at a concentration of 1.00 M? Solution Use the stepwise process described earlier. Step 1. Determine the direction the reaction proceeds. The balanced equation for the decomposition of PCl5 is PCl (g) ⇌ PCl (g) + Cl (g) 3 2 5 Because only the reactant is present initially Qc = 0 and the reaction will proceed to the right. Step 2. Develop an ICE table.

PDF page 757: = 2 −0.0211 ± 0.291 = 2 The two roots of the quadratic are, therefore, −0.0211 + 0.291 x = = 0.135 2 and −0.0211 − 0.291 x = = −0.156 2 For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so x = 0.135 M. The equilibrium concentrations are

PDF page 758: Note in this case that the change in concentration is significantly less than the initial concentration (a consequence of the small K), and so the initial concentration experiences a negligible change: if x ≪ 0.15 M, then (0.15 – x) ≈ 0.15 This approximation allows for a more expedient mathematical approach to the calculation that avoids the need to solve for the roots of a quadratic equation: 2 (x)(x) x ≈ Kc = 0.15 − x 0.15 2 −10 x

PDF page 754: Finally, substitute the equilibrium concentrations into the K expression and solve: − [I ] 3 − Kc = [I ][I ] 2 −4 3.39 × 10 M = = 776 −4 −4

PDF page 754: (6.61 × 10 M)(6.61 × 10 M)

PDF page 757: = 2 −0.0211 ± 0.291 = 2 The two roots of the quadratic are, therefore, −0.0211 + 0.291 x = = 0.135 2 and −0.0211 − 0.291 x = = −0.156 2 For this scenario, only the positive root is physically meaningful (concentrations are either zero or positive), and so x = 0.135 M. The equilibrium concentrations are

PDF page 757: ⎣PCl ⎦ = x = 0.135 M

PDF page 757: ⎣PCl ⎦ = 1.00 − 0.135 = 0.87 M

PDF page 758: The value of x calculated is, indeed, much less than the initial concentration −6 8.6 × 10 ≪ 0.15 and so the approximation was justified. If this simplified approach were to yield a value for x that did not