# Calculating enthalpy
**URL:** https://boards.straightdope.com/t/calculating-enthalpy/679715
**Category:** Factual Questions
**Created:** [January 25, 2014, 11:59pm UTC](https://boards.straightdope.com/t/calculating-enthalpy/679715 "2014-01-25T23:59:10Z")
**Posts on this page:** 5
**Page:** 1
### Author:  [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)
#### Post date: [January 25, 2014, 11:59pm UTC](https://boards.straightdope.com/t/calculating-enthalpy/679715/1 "2014-01-25T23:59:10Z")
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.
> [@](#):
>
> When 25.0 mL of 1.0 M H[sub]2[/sub]SO[sub]4[/sub] is added to 50.0 mL of 1.0 M NaOH at 25.0º C in a calorimeter, the temperature of the aqueous solution increases to 33.9º C.
>
> Assuming that the specific heat of the solution is 4.18 J/(g\*ºC), that its density is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔH in kilojoules for the reaction:
>
> H[sub]2[/sub]SO[sub]4/sub+2NaOH→2H[sub]2[/sub]O(l)+Na[sub]2[/sub]SO[sub]4/sub
>
> Express your answer using two significant figures.
OK… q = mass \* specific heat \* ∆T (∆H = q for 1 mol.)
Given:
[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]
So:
∆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? 😕
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### Author:  [@Ronald\_Raygun](https://boards.straightdope.com/u/Ronald_Raygun)
#### Post date: [January 26, 2014, 12:45am UTC](https://boards.straightdope.com/t/calculating-enthalpy/679715/2 "2014-01-26T00:45:38Z")
> [@Johnny\_L.A](#):
>
> 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.)
>
> Given:
> [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]
> So:
>
> ∆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? 😕
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.
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### Author:  [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)
#### Post date: [January 26, 2014, 1:09am UTC](https://boards.straightdope.com/t/calculating-enthalpy/679715/3 "2014-01-26T01:09:08Z")
> [@Ronald\_Raygun](#):
>
> 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?
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### Author:  [@Ronald\_Raygun](https://boards.straightdope.com/u/Ronald_Raygun)
#### Post date: [January 26, 2014, 4:33am UTC](https://boards.straightdope.com/t/calculating-enthalpy/679715/4 "2014-01-26T04:33:52Z")
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).
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### Author:  [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)
#### Post date: [January 26, 2014, 5:24pm UTC](https://boards.straightdope.com/t/calculating-enthalpy/679715/5 "2014-01-26T17:24:37Z")
Thanks. I’ll consider that when I wake up a bit. (Long night last night.)