Question

Using the Hess Law and the thermochemical equations below,

     MgO_(s)   +   + 2HCl_(aq)   ------> MgCl_2(g)   + H₂O_(l)         ΔH_rxn = -111.7 kJ/mol      

     Mg_(s)    + 2 HCl_(aq)     ------>      MgCl_2(g)   + H_2(g)         ΔH_rxn = -548.3 kJ/mol

     H_2(g)    + 1/2 O_2(g)     ------>      H₂O_(l)          ΔH_rxn = -142.9 kJ/mol


find the heat of reaction of the following: (0.25 pt.)

         Mg_(s)    + 1/2 O_2(g)     ------>      MgO_(g)      ΔH_rxn =   ?

b) If the theoretical enthalpy of this reaction is -602 kJ/mol, calculate the percent error (0.05 pt.)

Answer 1

the data given is
     MgO_(s)   +   + 2HCl_(aq)   ------> MgCl_2(g)   + H₂O_(l)         ΔH_rxn = -111.7 kJ/mol       (1)

reversing the reaction-1, MgCl2(g)+ H2O(l)------>MgO(s)+ 2HCl(aq), ΔH_rxn= 111.7 KJ/mole     (1A)

     Mg_(s)    + 2 HCl_(aq)     ------>      MgCl_2(g)   + H_2(g)         ΔH_rxn = -548.3 kJ/mol (2)

Eq.1A+ Eq.2 gives H2O(l)+Mg(s)---->MgO(s)+ H2(g), ΔH_rxn =111.7-548.3 =-436.6 KJ/mole (2A)

     H_2(g)    + 1/2 O_2(g)     ------>      H₂O_(l)          ΔH_rxn = -142.9 kJ/mol   (3)

Addition of Eq.2A and 3 gives Mg(s)+0.5O2(g)------>MgO(s), ΔH_rxn =-142.9-436.6 =-579.5 Kj/mole

given Mg(s)+0.5O2(g)------>MgO(s), ΔH_rxn =   -602. 2 Kj/mole

% error= 100*{-602.5-(-579.5)/-602.5}=3.82%