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.
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.
To calculate the change in enthalpy for the given reaction, we need to manipulate the given reactions and combine them in a way that cancels out the unwanted compounds and replicates the target reaction.
First, let's write down the given reactions and their corresponding enthalpy changes:
Reaction 1: 2Al (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H2 (g); ∆H = -1049 kJ
Reaction 2: HCl (g) → HCl (aq); ∆H = -74.8 kJ/mol
Reaction 3: H2 (g) + Cl2 (g) → 2HCl (g); ∆H = -1845 kJ/mol
Reaction 4: AlCl3 (s) → AlCl3 (aq); ∆H = -323 kJ/mol
Now, let's analyze the target reaction:
Target Reaction: 2Al (s) + 3Cl2 (g) → 2AlCl3 (s)
We need to change the states of the compounds in the given reactions and reverse certain reactions to match the target reaction. Here's how we can manipulate the reactions:
1. Reaction 1 is already quite close to the target reaction. We need to convert AlCl3 (aq) to AlCl3 (s). To do this, we can reverse Reaction 4 (multiply it by -1). This gives us:
Revised Reaction 1: 2Al (s) + 6HCl (aq) → 2AlCl3 (s) + 3H2 (g); ∆H = -1049 kJ
2. We can use Reaction 2 to convert HCl (g) to HCl (aq) by reverse multiplying it:
Revised Reaction 2: HCl (aq) → HCl (g); ∆H = +74.8 kJ/mol
3. Now, we need to balance the number of moles of Cl2 in the target reaction, which is 3. For this, we can use Reaction 3 and multiply it by 3/2 to get the desired number of moles of Cl2:
Revised Reaction 3: 3/2 H2 (g) + 3/2 Cl2 (g) → 3HCl (g); ∆H = -3 * (-1845 kJ/mol) = +5535 kJ
Now, let's add up the revised reactions:
Revised Reaction 1: 2Al (s) + 6HCl (aq) → 2AlCl3 (s) + 3H2 (g); ∆H = -1049 kJ
Revised Reaction 2: HCl (aq) → HCl (g); ∆H = +74.8 kJ/mol
Revised Reaction 3: 3/2 H2 (g) + 3/2 Cl2 (g) → 3HCl (g); ∆H = +5535 kJ
By summing up these revised reactions, we obtain the target reaction:
2Al (s) + 3Cl2 (g) → 2AlCl3 (s); ∆H = -1049 kJ + 5535 kJ + 74.8 kJ/mol = 4636.8 kJ
Therefore, the change of enthalpy for the reaction 2Al (s) + 3Cl2 (g) → 2AlCl3 (s) is 4636.8 kJ.
To summarize the manipulations:
- We reversed Reaction 4 to convert AlCl3 (aq) to AlCl3 (s).
- We reverse multiplied Reaction 2 to convert HCl (g) to HCl (aq).
- We multiplied Reaction 3 by 3/2 to balance the desired number of moles of Cl2.
I hope this explanation helps you understand the process of calculating the change in enthalpy for a given reaction.