Equilibrium Constant Equations: Calculations of Kc, Kp, and Equilibrium Concentrations, Study Guides, Projects, Research of Chemistry

Download Equilibrium Constant Equations: Calculations of Kc, Kp, and Equilibrium Concentrations and more Study Guides, Projects, Research Chemistry in PDF only on Docsity! Calculations Involving Equilibrium Constant Equation Basically, there are three types of calculations involved in equilibrium constant equation: (a) Calculation of equilibrium constant Kc or Kp (b) Calculation of Kc or Kp given Kp or Kc (c) Calculation of equilibrium concentrations (molar concentrations or partial pressures) Let us illustrate these three types of calculations with examples. Calculation of Equilibrium Constant Example Consider the following equilibrium reaction 3 2 4( ) ( ) ( ) ( )NH aq H O l NH aq OH aq+ −+ + If [NH3]= 0.02M, [NH4 +] =0.05 M, and [OH-] = 0.6 M, what is the equilibrium constant Kc for this reaction? Answer The equilibrium constant equation excluding the water (remember pure liquids do not appear in equilibrium constant expression) for the above reaction is 4 3 [ ][ [ ]c NH OHK NH + − = ] Substitute the given concentrations to evaluate Kc. Hence 4 3 [ ][ ] (0.05)(0.6) 1.5 [ ] 0.02c NH OHK NH + − = = = Example Consider the following heterogerneous equailibrium reaction: 3 2( ) ( ) ( )CaCO s CaO s CO g+ At 8000C, the pressure of CO2 gas is 0.236 atm. Calculate (a) Kp and (b) Kc for the reaction at this temperature. 1 Answer Since CaCO3 and CaO are solids, their concentrations do not enter into the equilibrium constant expression. Also, the pressure of CO2 is given, and therefore, first we solve for Kp and then solve for Kc. (a) Kp = PCO2 = 0.236 (b) The relation between KP and Kc is Kp = Kc (0.0821 x T)Δn Here T = 273 +800 = 1073 K, and Δn = 1. So, we substitute these values into the above equation to yield, 0.236 = Kc (0.0821 x 1073) Kc = 2.68 x 10-3 Calculation of Kc or Kp given Kp or Kc Example The equilibrium constant (Kc) for the reaction 2 4 2( ) 2 ( )N O g NO g is 4.63x10-3 at 250C. What is the value of Kp for this reaction at this temperature? Answer Relationship between Kp and Kc is Kp = Kc (0.0821 x T) Δn Here T = 25 + 273 = 298 K, and Δn = 2 – 1 = 1. Thus Kp = 4.63 x 10-3 ( 0.0821 x 298) = 0.113 Example For the reaction 2 2 3( ) 3 ( ) 2 ( )N g H g NH g+ 2