Calculate the standard enthalpy change for the reaction 2A+B?2C+2D Use the following data: Substance ?H?f (kJ/mol) A -249 B -389 C 217 D -483 ?H?rxn = 355 kJ For the reaction given in Part A, how much heat is absorbed when 3

2A + B <-------> 2C + 2D ;

Delta H⁰_rkn = 2*[delta H⁰_f D] + 2*[delta H⁰_f C] - [delta H⁰_f B] - 2*[delta H⁰_f A] = 355 kJ

Now, Assuming that while reacting with 3.1 moles of reactant A, all other reactants are in excess

Thus, for reacting with 3.1 moles of A, moles of B required = (1/2)*moles of A = 1.55

moles of C formed = moles of D formed = moles of a A reacting = 3.1

Thus, heat absorbed in the reaction

= 3.1*[delta H⁰_f D] + 3.1*[delta H⁰_f C] - 1.55*[delta H⁰_f B] - 3.1*[delta H⁰_f A]

(3.1*-483) + (3.1*217) - (1.55*-389) - (3.1*-249)= 550.25 kJ