Presentation on theme: "Applications of Equilibrium Constants"— Presentation transcript:
1 Applications of Equilibrium Constants Predicting the Direction of ReactionWe define Q, the reaction quotient, for a reaction at conditions NOT at equilibriumaswhere [A], [B], [P], and [Q] are molarities at any time.Q = K only at equilibrium.
2 Applications of Equilibrium Constants Predicting the Direction of ReactionIFQc > Kc system proceeds from right to left to reach equilibriumQc = Kc the system is at equilibriumQc < Kc system proceeds from left to right to reach equilibrium14.4
3 Applications of Equilibrium Constants Predicting the Direction of ReactionIf 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)
4 Calculating Equilibrium Concentrations Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.Having solved for x, calculate the equilibrium concentrations of all species.14.4
5 Calculating Equilibrium Constants Steps to Solving Problems:Write an equilibrium expression for the balanced reaction.Write an ICE table. Express the equilibrium concentrations of all species in terms of the initial concentrations.Use stoichiometry (mole ratios) to express change in concentration with respect to the unknown x on the change in concentration line.
6 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.)
7 ICE 14.4 At 12800C the equilibrium constant (Kc) for the reaction Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.Br2 (g) Br (g)Let x be the change in concentration of Br2Br2 (g) Br (g)Initial (M)0.0630.012ICEChange (M)-x+2xEquilibrium (M)xx[Br]2[Br2]Kc =Kc =( x)2x= 1.1 x 10-3Solve for x14.4
8 Kc =( x)2x= 1.1 x 10-34x x = – x4x x = 0-b ± b2 – 4ac2ax =ax2 + bx + c =0x = =.00015x =Br2 (g) Br (g)Initial (M)Change (M)Equilibrium (M)0.0630.012-x+2xxxAt equilibrium, [Br] = x = Mor MAt equilibrium, [Br2] = – x = M14.4
9 Example Problem: Calculate Keq This type of problem is typically tackled using the “three line” approach:2 NO + O2 2 NO2Initial:Change:Equilibrium:
10 Example Problem: Calculate Concentration Note the moles into a L vessel stuff ... calculate molarity.Starting concentration of HI: 2.5 mol/10.32 L = M2 HI H I2Initial:Change:Equil:0.242 M-2x +x +xx x xWhat we are asked for here is the equilibrium concentration of H2 ...... otherwise known as x. So, we need to solve this beast for x.
11 Example Problem: Calculate Concentration And yes, it’s a quadratic equation. Doing a bit of rearranging:x = or –Since we are using this tomodel a real, physical system,we reject the negative root.The [H2] at equil. is M.
12 ApproximatingIf 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 Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I More than 3 ExampleInitial Concentration of I2: 0.50 mol/2.5L = 0.20 MI IMore than 3orders of mag.between thesenumbers. Thesimplification willwork here.Initialchangeequil:-x x0.20-x xWith an equilibrium constant that small, whatever x is, it’s neardink, and 0.20 minus dink is 0.20 (like a million dollars minus anickel is still a million dollars).0.20 – x is the same as 0.20x = 3.83 x 10-6 M
14 Initial Concentration of I2: 0.50 mol/2.5L = 0.20 M I2 2 I ExampleInitial Concentration of I2: 0.50 mol/2.5L = 0.20 MI IThese are too close toeach other ...0.20-x will not betrivially close to 0.20here.Initialchangeequil:-x x0.20-x xLooks like this one has to proceed through the quadratic ...
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