Hess\'s Law-Part B indirect In B experiment, we will be finding the enthalpy off
ID: 501487 • Letter: H
Question
Hess's Law-Part B indirect In B experiment, we will be finding the enthalpy offormation for Mgos) using an the net method. According to Hess's Law, two or more reactions can be added to give a net reaction, AH for if reaction is simply the sum of the AH 's for the reactions which are added. Consider the following three reactions: AH1 H (1) va (2) Mg caal AH, (3) (4) Notice that when the three reactions are added, they form the formation reaction for Meo. Analogously, when we add the enthalpies for the three reactions, they give the enthalpy change associated with the formation reaction.Explanation / Answer
In the given documents, you already have calculations for Delta Ho1 and Delta Ho2.
Delta Ho1 = 6.075 kJ/mol
Delta Ho2 = -10.01 kJ/mol
Delta Ho3 = -285.9 kJ/mol
Delta Hof(MgO) nothing but Delta Ho4
According to Hess's Law,
Delta Hof(MgO) = Delta Ho4 = Delta Ho1 + Delta Ho2 + Delta Ho3
=> Delta Ho4 = 6.075 kJ/mol + (-10.01 kJ/mol) + (-285.9 kJ/mol)
=> Delta Ho4 = -289.8 kJ/mol
Thus, heat of formation of MgO is -289.8kJ/mol