Question

Using the enthalpy of formations from the lab introduction, determine the overall change in enthalpy for the reaction between HCl and magnesium. Give your answer in kilojoules (kJ).

Answer 1

Mg(s) + Cl2(g)  MgCl2(s) ∆Hƒ ̇= -640.7 kJ/mol
1/2 H2(g) + 1/2 Cl2(g)HCl(aq) ∆Hƒ ̇= -84.0 kJ/mol

Answer 2

Hi again.

Answer 3

First of all, take note that standard enthalpy change of formation is the energy absorbed by one mole of a substance when formed by its REACTANTS IN THEIR STANDARD STATES, and that the elements in their standard states have a 0 energy.
According to your first equation, all the elements have 0 energy except for MgCl2, meaning the energy obtained is totally for it.
For your second equation, the same thing happens. The -84Kj/mol is for HCl, because the rest are elemnts in THEIR STANDARD STATES.
Thirdly, reconsidering the equation Mg +2HCl - MgCl2 + H2, the energy for HCl is now 2(-84) because you have 2 moles of it.
The standard enthalpy vchange of formation formula is= -energy of reactants + energy of products. Reactants is always is the Left Hand Side and the product, the opposite side. [0 +2(-84)] + [0+(-640.7)]=work it out. The answer will be in Kj/mol.