I'm self-studying and can't solve this problem in Ball's Physical Chemistry. Now, the errata shows that the answer in the book is not correct; still, I can not make sense of it.

$(2.55)$ Benzoic acid, $\ce{C_6H_5COOH}$, is a common standard used in bomb calorimeters, which maintain a constant volume. If $1.20\text{g}$ of benzoic acid gives off $31,723\text{J}$ of energy when burned in the presence of excess oxygen at a constant temperature of $24.6^\circ\text{C}$, calculate $q, w, \Delta H,$ and $\Delta U$ for the reaction.

$(2.56)$ $1.20\text{g}$ of benzoic acid, $\ce{C_6H_5COOH}$, is burned in a porcelain dish exposed to the air. If $31,723\text{J}$ of energy is given off and the system temperature is $24.6^\circ\text{C}$, calculate $q, w, \Delta H, $ and $\Delta U$.

The answers(corrected from errata) are,

$(2.55)$ $q = -31,723\text{J}, \Delta U = -31,723\text{J}, w = 0, \Delta H = -31,735\text{J}$

$(2.56)$ $q = -31,723\text{J}, \Delta H = -31,723\text{J}, w = 12, \Delta U = -31,711\text{J}$

The book never talks about heats of combustion prior. Computing using a value I found on the internet of $26.4 \,\text{kJ}/\text{g}$, I get a value for $q$, I get that $\Delta U = q + w$ then and we can solve for work. This seems to be an incorrect approach since the energy given off as heat seems to already be given.

Why is the enthalpy the same as the heat in the second case? I would assume it is because the pressure is constant, but only the external is constant, which is not what is used to determine that enthalpy is the same as heat.

Where does this work come from? The volume changes in the non-calorimeter experiment, so work on the acid is done. But how does on obtain this change?

How do I solve this problem correctly?

  • $\begingroup$ Welcome to Chemistry.SX! If you are interested in specific formatting tricks available here, check out this page, this page and this one on how to format your future posts better with MathJax and Markdown. Good luck! $\endgroup$ – andselisk Dec 21 '17 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.