Question

                        In each case, find the missing ∆Hº_rxn(kJ/mole)

                        a. If     2 Al(s) + Fe₂O₃(s) ---- > Al₂O₃(s) + 2 Fe(s) ΔH_rxn^o = -851.5kJ/mol

                                    then

                        1/8 Al₂O₃(s) + 1/4 Fe(s) --- >   1/4 Al(s) + 1/8 Fe₂O₃(s)     ΔH_rxn^o = ?

                        b. If    CaO(s) + 3 C(s) --- > CaC₂(s) + CO(g) ΔH_rxn^o = 464.8 kJ/mol

                                    then

                        6 CaO(s) + 18 C(s) --- > 6 CaC₂(s) + 6 CO(g)   ΔH_rxn^o = ?

                        c. If N₂(g) + 2 O₂(g) ---- > N₂O₄(g)                          ΔH_rxn^o = 9.2 kJ/mol

                        and 2 NO₂(g) ----- > N₂(g) + 2 O₂(g)                           ΔH_rxn^o = -33.2 kJ/mol

                                    then

                        3N₂O₄(g) ---- > 6 NO₂(g)                      ΔH_rxn^o = ?

Answer 1

a)

This reaction is -(1/8)*given reaction

So,

ΔHo rxn = -(1/8)*given ΔHo rxn

= -(1/8)*(-851.5)

= 106.4 KJ/mol

Answer: 106.4 KJ/mol

b)

This reaction is 6*given reaction

So,

ΔHo rxn = 6*given ΔHo rxn

= 6*(464.8)

= 2788.8 KJ/mol

Answer: 2788.8 KJ/mol

C)

Lets number the reaction as 1, 2, 3 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 3 = -3 * (reaction 1) -3 * (reaction 2)

So, ΔHo rxn for required reaction will be:

ΔHo rxn = -3 * ΔHo rxn(reaction 1) -3 * ΔHo rxn(reaction 2)

= -3 * (9.2) -3 * (-33.2)

= 72.0 KJ

Answer: 72.0 KJ