Question

5. K, for the isomerization of butane (C,Hro, molar mass 58 g/mol) to isobutane (aliso C,Htp ) is 7.94 at 25° C C4H10 (butane) C4H10 (isobutane) # If 11.6 g of butane are placed in a 1 L flask, how many grams of isobutane will be present at equilibrium? What is the % conversion to isobutane. (10 points) (Ans. 10.3g, 89.0%) 6 In a study of interhalogen compounds, 0.20 mol of ICI (iodine chloride) were placed in a 2 L container where it decomposed to chlorine and iodine. 2 ici (g) Cl2 (g) + 2 (g) Kc= 0.110 Calculate the equilibrium concentrations of ICI, 2 and Cl2. (10 points) Ans. [21 (CI2] - 0.02 mol/L: ICI 0.06 molL)

Answer 1

C_9 H_10 (Butane) rightwardsharpoonoverleftwardsharpoon C_9 H_10 (isobutane) The mass of butane = 11.6 g The molar mass of butane = 58 gm/mol therefore The no.of moles of butane = mass of butane/molar mass of butance = 11.6/58 = 0.2 mol since, butane is placed in a perpendicular L flask the molarity of butane = 0.2 mol/1 lit = 0.2 mol/lit = 0.2 (m) Now, K_e = 7.94 at 25 degree C therefore K_e = [Iso butane]_eq/[Butane]_eq = x/0.2 - x or, 7.94 = x/0.2 - x or, (7.94 times 0.2) - 7.94 x = x or, 8.94 x = 7.94 times 0.2 or, x = 0.1776 (m) Therefore concentration of isobutene = 0.1776 (M) = 0.1776 l mol/lit therefore Grams of Iso butane = No. Of moles times molar mass = 0.1776 times 58 = 10.3 g Calculation of % conversion to isobutene: If we assume there are 1 mol butane and converted to x mol isobutene then are can write.