9701 MCQ Solution 50

08:44 equilibria 0 Comments

Before the reaction has even started there were 0.20 mol of hydrogen and and 0.15 mol of iodine these moles are therefore initial , at equilibrium it said that there are 0.26 mol of hydrogen Iodide which is the only product. since its moles we can use the same expression as for concentration. 0.20 mol of hydrogen , 0.15 mol of iodine and 0 mol of hydrogen iodide were present as initial , 0 mol of hydrogen iodide because its initial and the reaction hasn’t even started therefore no product was formed. However at equilibrium it was found that there were 0.26 mol of hydrogen iodide. the ratio between hydrogen , iodine and hydrogen iodide is : 1:1:2. since it has been found out there is 0.26 mol of hydrogen iodide it means that number of moles of hydrogen iodide has increased from 0 to 0.26. and this can only happen when the number of mols of the reactants has decreased to form the product. number of moles of hydrogen iodide has increased by 0.26 but hydrogen iodide is twice both the reactants. so divide by 2 therefore 0.13. this means that each of hydrogen and iodine has lost 0.13 moles to form hydrogen iodide therefore at equilibrium number of moles of hydrogen is the initial minus the number of moles lost to form hydrogen iodide therefore 0.20-0.13=0.07. as for iodine its 0.15-0.13=0.02. so at equilibrium only 0.07 mol of hydrogen and 0.02 mol of iodine remained after equilibrium has been established. therefore to calculate Kc the product will be in the numerator and reactants at dominator. and if there is two moles we put square and so on therefore. Kc= (0.26) square / 0.07 x 0.02. Answer is C