Calculating enthalpy

OK, I’m stuck. Yes, it’s a homework question. I’m not asking for the answer, but where I’m going wrong in my calculations.

OK… q = mass * specific heat * ∆T (∆H = q for 1 mol.)

[ul][li]75 mL solution (25 mL added to 50 mL)[/li][li]Since density is 1.00 g/mL, mass is 75 * 1.00 = 75.00 g.[/li][li]Specific heat is 4.18 J/(g*ºC)[/li][li]∆T = 33.9 - 25.0 = 8.9ºC[/ul][/li]

∆H = 75.00 g * 4.18 J/(g*ºC) * 1 kJ/1000 J * 8.9ºC = 2.79 kJ

2.8 kJ is incorrect. Negative 2.8 kJ is incorrect. Where am I screwing up? :confused:

You’ve found the enthalpy change for the specific reaction you’ve got going on in the calorimeter – that is, for a certain amount of sulfuric acid. However, the problem asks for the enthalpy change of the reaction in general, regardless of whether you start with 1 mol sulfuric acid or 1 mmol sulfuric acid. So you’ll need to give an answer in units kJ/mol, or the enthalpy change for each mole of sulfuric acid you have.

I’m not entirely sure I follow you. A given amount of acid makes a given amount of heat. Are you saying that I arrived at q, and now I have to scale q to find ∆H?

Now that I’m looking at the problem on a laptop rather than a smart phone and see that the answer needs to be in kJ, I’m not sure.

But what I was getting at was that when you look up enthalpy changes, typically they are given per mole of reaction. This makes sense – if my reaction involves twice as much stuff (e.g., I started with 50 mL of a 1 M solution instead of just 25 mL of a 1 M solution), it’d give off twice as much heat.

Since you start with 1/40 mole H2SO4 in the calorimeter, I would take my 2.8 kJ and multiply it by 40. Your answer will then be in kJ/mol, that is, the change in enthalpy starting with 1 mol H2SO4. I’m not sure if this is where you’re going wrong though, but give it a shot (if it’s something you can check easily).

Thanks. I’ll consider that when I wake up a bit. (Long night last night.)