Constant-Volume Calorimetry

In a bomb calorimeter, a chemical reaction takes place at constant volume, and the resulting change in temperature can be used to evaluate the change in internal energy associated with the reaction.

[@rinsim]

Hi Steven, brilliant video as always! I think the 11 "mol/mol" works because we express U and H per mole of sucrose and because the stoichiometric coefficient of sucrose is 1 in the reaction. I suppose what we want to calculate is "delta(H) of the sucrose + oxygen system per mole of sucrose". Would it be possible accurate calculate the delta(pV) term just measuring the pressure difference in the internal container, as the volume is constant? Here we have ignored the solid and considered the gases ideal. How does a real bomb calorimeter work? Is it really used to calculate delta(H) or it is more useful for delta(U) only? If I'm not mistaken, at 11:16 "pv work" should be more correctly "pv delta" as the work is always 0 considering any system (water only, sucrose+oxygen, overall system). Lastly, we are ignoring the work of the electrodes. Can this be justified considering the initial conditions as the instant immediately after the combustion has been ignited?

[@PhysicalChemistry]

Such thorough questions! I love the level of detail you're bringing to the discussion. Yes, the stoichiometric coefficient 11 mol/mol can be thought of as either 11 moles of gaseous products per mole of 1 mole of forward reaction or (since the stoichiometric coefficient of sucrose is 1) 11 moles of gas per mole of sucrose consumed. We're measuring the molar enthalpy of combustion, so your description is correct: the ΔH of the entire system per mole of sucrose reacted. And you're right, you could in principle measure VΔP to get the value of Δ(PV) for this constant-volume process, if you had easy access to the pressure data. (In practice, that's probably a bit complicated because it changes very quickly, and drops as the heat is transferred to the surroundings and steam condenses to liquid water.) You're correct that I misspoke @11:16. I should have said "change in the PV product", rather than "PV work". There is no PV work being done in this constant-volume process. Thanks for pointing that out. Finally: the neglect of the electrodes is because we are considering the system as only the sucrose and oxygen (and water and CO₂). We ignore the electrodes, wires, metal container, etc. There certainly was some transfer of energy from the electrodes to the system, to start the combustion reaction. But it's not necessary to consider the source of that activation energy in understanding the thermodynamics of the chemical reaction. It's reasonable, as you say, to confine our attention to only what happened after the reaction was initiated,.