Question

Please help me with this-- I don't understand it.

Calculate the change of enthalpy for the reaction 2Al (s) + 3Cl2 (g) --> 2AlCl3 (s) from the following reactions:


Reaction 1: 2Al (s) + 6HCl (aq) --> 2AlCl3 (aq) + 3H2 (g);Change in enthalpy: -1049 kJ


Reaction 2: HCl (g) --> HCl (aq);Change in enthalpy: -74.8 kJ/mol


Reaction 3: H2 (g) + Cl2 (g) --> 2HCl (g);Change in enthalpy: -1845. kJ/mol


Reaction 4: AlCl3 (s) --> AlCl3 (aq);Change in enthalpy: -323. kJ/mol


I have to include the following:

The numerical answer with correct units.

State which reactions, if any, you had to "Flip", or reverse.

State which reactions you had to multiply, if any, to get the correct amount of the compound. Also, include how much you multiplied the reaction by.

Answer 1

Follow these instructions:
Add eqn 1 as is to 6 times eqn 2 to 3 times eqn 3 to 2 times the reverse of eqn 4. Then add the delta values. If you multiply an equation multiply the dH value by the same multiplier. If you reverse and equation then change the sign of dH. When you get through with the equation manipulation, cancel those items that appear on BOTH sides of the equation and make sure you have the eqn you want; i.e., 2Al (s) + 3Cl2 (g) --> 2AlCl3 (s)
Post your work if you get stuck.