Question

hydrogen(a moles) and iodine (b moles) react to give 2x moles of HI at equilibrium .the total number of moles at equilibrium is

Answer 1

Calculation of total no. of moles of HI formed by the reaction of a moles of hydrogen and b moles of iodine

Explanation:

Initial moles of reactants and the products are as follow

Hydrogen = a moles

Iodine = b moles

HI = 0 moles

Finally,  the number of moles of reactants and the products are as follow

Hydrogen = x moles

Iodine = x moles

HI = 2x moles

At equilibrium, the number of moles of reactants and the products are as follow

Hydrogen =a- x moles

Iodine = b-x moles

HI = 2x -0 moles

Thus,

Total number of moles at equilibrium = (a-x) + (b-x) +2x

                                                                 = a-x+b-x+2x

                                                                 = a+b+2x-2x

                                                                 = a+b