Determining enthalpy and internal energy of formation for potassium superoxide - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-08-20T07:48:55Z https://chemistry.stackexchange.com/feeds/question/2438 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://chemistry.stackexchange.com/q/2438 1 Determining enthalpy and internal energy of formation for potassium superoxide Fiire https://chemistry.stackexchange.com/users/715 2012-10-24T02:36:47Z 2018-02-28T13:57:27Z <blockquote> <p>Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.</p> <p>We know:</p> <ul> <li>The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$</li> <li>The mass of water inside it: $1450\ \mathrm g$</li> <li>The change in temperature: $2.62\ \mathrm K$</li> <li>The end product is $\ce{K2O}$</li> </ul> </blockquote> <p>The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is <em>extravagant</em>; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.</p> <p>Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?</p> https://chemistry.stackexchange.com/questions/2438/-/2440#2440 2 Answer by ManishEarth for Determining enthalpy and internal energy of formation for potassium superoxide ManishEarth https://chemistry.stackexchange.com/users/22 2012-10-24T09:14:27Z 2018-02-28T13:57:27Z <p>Now, when we say that the temperature rose by $\pu{2.62K}$, this is the temperature of the calorimeter and the water together. We are assuming that they have been given time to reach thermal equilibrium (and generally water and copper calorimeters take relatively less time to reach equilibrium, it's pretty safe to neglect the time). Calorimeters can be filled with varied amounts of water, so the calorimeter's constant is for just the calorimeter, not the water in it.</p> <p>Note that if we were to assume that the specific heat of potassium/oxide were not negligible, we would have to know the specific heats of the two, as well as the temperature at the start of the reaction and the temperature at which they want $\Delta H^o$ to be calculated at. "Standard state" doesn't fix temperature, though when you say $\Delta H^o$, you generally mean $\Delta H^o_{\pu{298K}}$. Using these, you can form a reaction pathand calculate the answer (I can elaborate on this if you want, it's pretty interesting)</p> <p>With these assumptions, one gets the heat released to the order of 10 kJ, and the energy is to the order of 100 kJ/mol. Not too extravagant, remember that a joule is a rather tiny unit of energy. And formation enthalpies/energies are generally to the order of 100 or 1000 kJ.</p> <p>Finding the enthalpy change now is pretty easy. Since $\Delta H=\Delta U+ \Delta(PV)=\Delta U+\Delta(nRT)$. Using this and looking at the overall reaction, it shouldn't be too hard :)</p>