Download Chemical Equilibrium Solved Problems / Numercials and more Exercises Chemistry in PDF only on Docsity! 1 Chapter 7 Chemical Equilibrium SOLVED PROBLEMS Example 7.1 Write down the expressions of equilibrium constant (Kc) for the following reversible reactions. (i) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) (ii) 2N2O5(g) 4NO(g) + 3O2 Solution: To write an equilibrium expression, we should have the balanced chemical equation. All products given in the equation should be placed on numerator each separately in square bracket while reactants on denominator. Then finally raise the concentration of each substance to the power of its coefficient in the balance chemical equation. Kc= [๐ต๐ถ]๐[๐ฏ๐๐ถ]๐ [๐ต๐ฏ๐]๐[๐ถ๐]๐ Kc= [๐ต๐ถ]๐[๐ถ๐]๐ [๐ต๐ ๐ถ๐]๐ Example 7.2 An essential step in contact process is the oxidation of SO2 to SO3 2S02(g) + O2(g) 2S03 If in an experiment, there are 5 moles of S02, 4 moles O2 and 2.8 moles of SO3 present at equilibrium state in a 2dm3 flask. Calculate Kc. Solution: Since equilibrium moles of all components in the reacting mixture are given, we first convert them into molar concentration and then put into equilibrium expression to find out Kc. [SO2]eq = 5 2 = 2.5 mol/dm3 [O2]eq = 4 2 = 2 mol/dm3 [SO3]eq = 2.8 2 = 1.4 mol/dm3 Kc expression for the given reaction may be written as Kc= [๐๐3]2 [๐๐2]2[๐2] By substituting the equilibrium concentrations, we get. Kc= [1.4]2 [2.5]2[2.0] = 0.157mol-1.dm3 2 Chapter 7 Chemical Equilibrium Example 7.3 Ethyl acetate is an ester of ethanol and acetic acid commonly use as an organic solvent. CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l) In an esterification process 180g of acetic acid and 138g ethanol were mixed at 298K and allowed to start reaction under necessary conditions. After equilibrium is established 60g of unused acetic acid were present in the reaction mixture. Calculate Kc. Solution: Moles of CH3COOH = 180 60 = 3 moles Moles of C2H5OH = 138 46 = 3 moles Conc. (mole/dm3) CH3COOH + C2H5OH CH3COOC2H5 + H2O Initial 3 3 0 0 Equilibrium 3-x 3-x x x But at equilibrium unused acetic acid is 60g which is equal to 1 mole. Therefore 3-x=1 and x=2 Now substituting values of equilibrium mixture in Kc expression. Now applying Law of mass action Kc= [๐ถ๐ป3๐ถ๐๐๐ถ2๐ป5] [๐ป2๐] [๐ถ๐ป3๐ถ๐๐๐ป] [๐ถ2๐ป2] Kc= [2] [2] [1] [1] Kc= 4 Example 7.4 Nitrosyl chloride is a yellow coloured gas prepared by the reaction of NO and Cl2 gases. 2NO(g) + Cl2(g) 2NOCl If at certain temperature, the partial pressure of equilibrium mixture is NO = 0.17 atm, Cl2 = 0.2 atm and NOCI = 1.4 atm, Calculate Kp Solution: The equilibrium expression of the given reaction is written as Kp = [๐๐๐๐ถ๐ฟ]2 [๐๐๐]2[๐๐ถ๐ฟ2] Substituting the partial pressure in equilibrium expression, we get Kp= (1.4)2 (0.17)2(0.2) Kp = 339.1
