Enthalpy of reaction problem? Please help?

Question

Can you please help me?? I am so lost on how to solve this, and would love to see exactly how to do it. I know you have to break up the compounds into ions if they are aq, but not much else. In addition, I don't have a Hf value for NO3, so is that something you look up or have to memorize? Anyways, here is the question:

 

 

Using the values for ?Hf, determine the ?H(enthalpy) for the following reaction: Na2CO3 (aq) + Mg(NO3)2 (aq) → 2 NaNO3 (aq) + MgCO3 (s)

 

Substance (state)          Hf (kJ/mol)

Na+ (aq)                      -240.1

Cl- (aq)                         -167.2

 

 

MgCO3 (s)                     -1096

Mg+2 (aq)                     -466.9

CO3 2- (aq)                   -677.1

 

 

Thank you so much! I really want to learn what to do! 

Answer 1

Angelica,

These values are not something you should have to memorize. The only thing you have to memorize is where you can find these values.

 

Having said that, let's solve this problem.

 

You said that you know you have break up the compounds if they are aqueous, so let's do that.

 

2Na⁺(aq)  + CO₃^2-(aq)  +  Mg²⁺(aq)  +  2NO3^-(aq)  ->  2Na⁺  +  2NO₃^-(aq)  +  MgCO₃(s)

 

I see Na⁺ and NO₃^- on both sides of the reaction equation, so let's rewrite the reaction as a net ionic reaction:

 

CO₃^2-(aq) + Mg²⁺(aq)  ->   MgCO₃(s)

 

Now, the enthalpy of the reaction can be calculated by adding up all of the enthalpies of formation for the products,

adding up all of the enthalpies of formation of the reactants, and then subtracting the latter from the former.

 

 

ΔH_reaction    =   -1096 kJ/mol   -  ( -677.1 kJ/mol  +  -466.9 kJ/mol )   =

 

       -1096 kJ/mol   -   ( -1144 kJ/mol )   =    48 kJ/mol