Using the Hess Law and the thermochemical equations below, MgO (s) + + 2HCl (aq)

Question

Using the Hess Law and the thermochemical equations below,

     MgO(s)   +   + 2HCl(aq)   ------> MgCl2(g)   + H2O(l)         Hrxn = -111.7 kJ/mol      

     Mg(s)    + 2 HCl(aq)     ------>      MgCl2(g)   + H2(g)         Hrxn = -548.3 kJ/mol

     H2(g)    + 1/2 O2(g)     ------>      H2O(l)          Hrxn = -142.9 kJ/mol


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

         Mg(s)    + 1/2 O2(g)     ------>      MgO(g)      Hrxn =   ?

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

Explanation / Answer

the data given is
     MgO(s)   +   + 2HCl(aq)   ------> MgCl2(g)   + H2O(l)         Hrxn = -111.7 kJ/mol       (1)

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

     Mg(s)    + 2 HCl(aq)     ------>      MgCl2(g)   + H2(g)         Hrxn = -548.3 kJ/mol (2)

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

     H2(g)    + 1/2 O2(g)     ------>      H2O(l)          Hrxn = -142.9 kJ/mol   (3)

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

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

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

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