Calculate the standard enthalpy change for the following reaction at 25 °C.
MgCl2(s) + H2o(l) ---> MgO(s) + 2HCl(g)
4 answers
dHrxn = [n*dHf products] - [n*dHf reactants]
Could you elaborate? I don't quite understand what that means.
MgCl2(s) + H2o(l) ---> MgO(s) + 2HCl(g)
Look up delta H formation for MgO(solid), HC(gas), MgCl2(solid), and H2O(liquid).
Then n = mols and those are the coefficients. The products are MgCl2 and HCl. Reactants are MgCl2 and H2O.
So you have (1*dHf MgO + 2*dHf HCl)as the products and subtract from that (1*dHf MgCl2 + 2*dHf HCl) as the reactants. That will give you the dH reaction. It all goes together as this.
dHrxn = (n*dHf products) - (n*dHf reactants)
Look up delta H formation for MgO(solid), HC(gas), MgCl2(solid), and H2O(liquid).
Then n = mols and those are the coefficients. The products are MgCl2 and HCl. Reactants are MgCl2 and H2O.
So you have (1*dHf MgO + 2*dHf HCl)as the products and subtract from that (1*dHf MgCl2 + 2*dHf HCl) as the reactants. That will give you the dH reaction. It all goes together as this.
dHrxn = (n*dHf products) - (n*dHf reactants)
For anyone who stumbles upon this page, DrBob222 made a mistake. Take note that the second reactant is H2O (l), not another 2HCl (g). Make sure to look up the dHf of H2O instead.
1 answer