Presentation on theme: "1 Applications of Equilibrium Constants Predicting the Direction of Reaction We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium."— Presentation transcript:
1 Applications of Equilibrium Constants Predicting the Direction of Reaction We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium.
Applications of Equilibrium Constants Predicting the Direction of Reaction IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4
3 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (go left) If Q < K then the forward reaction must occur to reach equilibrium (go right)
Calculating Equilibrium Concentrations 1.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 2.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 3.Having solved for x, calculate the equilibrium concentrations of all species. 14.4
5 Calculating Equilibrium Constants Steps to Solving Problems: 1.Write an equilibrium expression for the balanced reaction. 2.Write an ICE table. Express the equilibrium concentrations of all species in terms of the initial concentrations. 3.Use stoichiometry (mole ratios) to express change in concentration with respect to the unknown x on the change in concentration line.
4.Solve for x and calculate the equilibrium concentrations of all species. Usually, the initial concentration of products is zero. (This is not always the case.)
At 1280 0 C the equilibrium constant (K c ) for the reaction Is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x [Br] 2 [Br 2 ] K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 Solve for x 14.4
K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac 2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = =.00015 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M 14.4
9 Example Problem: Calculate Keq This type of problem is typically tackled using the three line approach: 2 NO + O 2 2 NO 2 Initial: Change: Equilibrium:
10 Example Problem: Calculate Concentration Note the moles into a 10.32 L vessel stuff... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M 2 HI H 2 + I 2 Initial: Change: Equil: 0.242 M 00 -2x+x+x 0.242-2x xx What we are asked for here is the equilibrium concentration of H 2...... otherwise known as x. So, we need to solve this beast for x.
11 Example Problem: Calculate Concentration And yes, its a quadratic equation. Doing a bit of rearranging: x = 0.00802 or –0.00925 Since we are using this to model a real, physical system, we reject the negative root. The [H 2 ] at equil. is 0.00802 M.
12 Approximating If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants. 0.20 – x is just about 0.20 is x is really dinky. If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.
13 Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x With an equilibrium constant that small, whatever x is, its near dink, and 0.20 minus dink is 0.20 (like a million dollars minus a nickel is still a million dollars). 0.20 – x is the same as 0.20 x = 3.83 x 10 -6 M More than 3 orders of mag. between these numbers. The simplification will work here.
14 Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x These are too close to each other... 0.20-x will not be trivially close to 0.20 here. Looks like this one has to proceed through the quadratic...