I am having real trouble understanding the first law of thermodynamics and I think the question, below, highlights most of my issues:

1) (a) A bomb calorimeter was calibrated by igniting a $0.825\ \mathrm{g}$ sample of benzoic acid in the presence of excess oxygen. The heat given out made the temperature of the calorimeter rise by $1.940\ \mathrm{K}$ from $298\ \mathrm{K}$. The internal energy of combustion of benzoic acid is $-3251\ \mathrm{kJ\ mol}^{-1}$. Calculate the heat capacity of the calorimeter.

(b) In two separate experiments in the same apparatus, $0.498\ \mathrm{g}$ of fumaric and $0.509\ \mathrm{g}$ of maleic acid (both of RMM 116) were ignited and gave temperature rises of $0.507\ \mathrm{K}$ and $0.528\ \mathrm{K}$ respectively. Use the heat capacity of the calorimeter, calculated in (a), to calculate (i) the molar internal energy of combustion, (ii) the molar enthalpy of combustion, and (iii) the molar enthalpy of formation of fumaric and maleic acids. Comment on the difference. The standard enthalpy of formation of water is $285.8\ \mathrm{kJ\ mol}^{-1}$ and of $\ce{CO2}$: $-393.5\ \mathrm{kJ\ mol}^{-1}$.

Part (a) was fine - the heat capacity at a constant pressure = change in internal energy/temperature change. Then divide by number of moles.

Part (b) is where I struggle conceptually. I got the answer correct but only by following lecture notes. Sticking with just fumaric acid because there's no point talking about both. It's easy to find the molar change in internal energy using the constant volume heat capacity ($-1338\ \mathrm{kJ\ mol}^{-1}$).

However, to find the molar enthalpy change, the change in $H = U + pV$. but since $\ce{pV=nRT; H=U+nRT}$ where n depends on the change in the number of gaseous moles from products to reactants - in this case +1. So $H = U + RT$ giving $-1336\ \mathrm{kJ\ mol}^{-1}$ (less exothermic).

Firstly: Why are the values negative. If the system is the fumaric acid/products of reaction then surely there is an increase in internal energy because the particles will be moving faster after the bonds have broken. (Am I right in thinking that the system is the chemical)?

Secondly: Why is the enthalpy value less exothermic? (I have a very poor grasp for what enthalpy is especially in a situation like this).

Any help with answering my questions or helping me with this concept would be greatly appreciated. Also, do you know any good books/videos I could watch to help me undertand?


1 Answer 1


Combustion releases energy from the system usually as heat and/or work to the surroundings, that's why the standard energy of combustion has a negative sign. The bomb is the system in this case (alternatively you can see the chemical as the system, but you will need to include the gaseous products in your system too, changing the volume of the system, which I find a less preferable viewpoint).

The value for the enthalpy will be lower since energy is trapped in the form of increased pressure introduced by the reactions products (mainly by $\ce{CO_2}$) in the bomb. This pressure can be used to do work on the surroundings (in an open system this will be usually on the atmosphere). If you determine the final pressure, you can calculate your $\Delta U$ back ($\Delta U = \Delta H - (p_{end}-p_{atm})V_{bomb}$).

What book do you use for the lectures?

  • $\begingroup$ Physical Chemistry by Peter Atkins. He used to be a lecturer here so they strongly advise using it. $\endgroup$
    – RobChem
    Nov 15, 2014 at 15:24
  • $\begingroup$ @user3764899 That book is fine, although I find the newer versions sometimes so focused on "pushing the material" that it becomes confusing (all the different colors and frames). Did my answer help you? $\endgroup$
    – Jori
    Nov 15, 2014 at 16:08
  • $\begingroup$ Yes, the answer was great. However, what do you mean by "you will have your delta U back"? Also, surely the bomb is the surroundings because as the chemical gives off energy, surely the internal energy of the bomb would increase not decrease? $\endgroup$
    – RobChem
    Nov 15, 2014 at 16:28
  • $\begingroup$ Also, do you know a better book? Perhaps even a good alternative? We have really good libraries here so I will definitely be able to find one. $\endgroup$
    – RobChem
    Nov 15, 2014 at 16:32
  • $\begingroup$ @user3764899 See edit. I believe the bomb is only the internal chamber of the bomb calorimeter, which has the diathermic wall (see e.g. precisionnutrition.com/wordpress/wp-content/uploads/2010/02/…). The bomb is usually made of a material that can withstand high pressure. Atkins is perfectly fine! It is regarded as an authoritative book within its field. $\endgroup$
    – Jori
    Nov 15, 2014 at 17:04

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