Question

Part A. Calculate the enthalpy of the reaction

2NO(g)+O2(g)?2NO2(g)

given the following reactions and enthalpies of formation:

12N2(g)+O2(g)?NO2(g),   ?H?A=33.2 kJ

12N2(g)+12O2(g)?NO(g),  ?H?B=90.2 kJ

Part B. Calculate the enthalpy of the reaction

4B(s)+3O2(g)?2B2O3(s)

given the following pertinent information:

B2O3(s)+3H2O(g)?3O2(g)+B2H6(g),    ?H?A=+2035 kJ

2B(s)+3H2(g)?B2H6(g),                            ?H?B=+36 kJ

H2(g)+12O2(g)?H2O(l),                ?H?C=?285 kJ

H2O(l)?H2O(g),                                          ?H?D=+44 kJ

Answer 1

Part A : 2NO(g)+O₂(g) 2NO₂(g) : H=?   ------------ (1)

given that

1/2 N₂(g)+O₂(g) NO₂(g)    :H _A = 33.2 kJ       --------- (2)

1/2 N₂(g)+1/2O₂(g) NO(g),      :H _B = 90.2 kJ    ---------- (3)

Equation (1) is obtained from Eqn(2) & Eqn(3) are as follows:

Eqn (1) = [2x Eqn (2)] + [2 x reverse of Eqn(3)]

So H = [2xH _A ]+ [2x(-H _B )]

            = [2x(33.2 kJ ) ]+ [2x(-90.2 kJ)]

            = -114 KJ

Part B : 4B(s)+3O₂(g)    2B₂O₃(s)     : H=?   ------------ (1)

given that

B₂O₃(s)+3H₂O(g) 3O₂(g)+B₂H₆(g)    :H _A =+2035 kJ                  ------------------ (2)

2B(s)+3H₂(g) B₂H₆(g)                        :H _B =+36 kJ                     ------------------ (3)

H₂(g)+1/2 O₂(g) H₂O(l)                :H _C = - 285 kJ                           ------------------ (4)

H₂O(l) H₂O(g)                         :H _D =+44 kJ                             ------------------ (5)

Equ (1) can be obtained from remaining equations as follows:

Eqn (1) = [2xreverse of Eqn(2)]+[2xEqn (3)]+[6x reverse of Eqn (4) ] + [6x reverse of Eqn(5)]

So   H = [2 x (-H _A )]+[2x(H _B )]+[6x (-H _C ) ] + [6x (-H _D )]

            = [2 x (-2035 kJ)]+[2x(+36 kJ)]+[6x (-285 kJ) ] + [6x (-44 kJ )]

            = -5972 KJ