Newest questions tagged enthalpy - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-08-19T00:04:25Z https://chemistry.stackexchange.com/feeds/tag?tagnames=enthalpy&sort=newest http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://chemistry.stackexchange.com/q/119263 0 Effect of the reaction stoichiometric coefficients on Δn user208973 https://chemistry.stackexchange.com/users/74517 2019-08-16T12:17:48Z 2019-08-16T12:17:48Z <p>We have ΔH=ΔE+RTΔn for any chemical reaction (suppose ideal gases) where Δn is the difference between the stoichiometric coefficients of the gaseous products and gaseous reactants.</p> <p>Note that ΔH in J/mol must be a constant value for a reaction for nonvariant given conditions, and for sure independent of multiplying or dividing the reaction coefficients by a certain natural number.(or this is what i think)</p> <p>Consider this reaction from my chemistry book: (sorry for the bad format)</p> <p>2C3H5N3O9(l) = 5H2O(l) + 6CO2(g) + 3NO2(g) + 0.5O2(g)</p> <p>The question is to calculate ΔH at 298K, given ΔE in J/mol.</p> <p>The teacher considered Δn to be (6+3+1/2) - (0 ) = 9.5</p> <p>What i was confused about is that the stoichiometric coefficient of nitroglycerin (the reactant) is not 1. </p> <p>Shouldn't we divide the reaction coefficients by 2 to calculate Δn? And if not, why? And what are the criteria to determine the correct "order" of <strong>any</strong> reaction stoichiometric coefficients? </p> <p>Thanks a lot. Any hint or comment is appreciated.</p> https://chemistry.stackexchange.com/q/119161 -1 Two methods for calculating ΔH yield different results [on hold] user208973 https://chemistry.stackexchange.com/users/74517 2019-08-13T16:23:27Z 2019-08-13T17:13:06Z <p>A problem in my book says:</p> <blockquote> <p><span class="math-container">$\pu{140 g}$</span> of <span class="math-container">$\ce{N2}$</span> (perfect gas) changes from <span class="math-container">$\pu{200 K}$</span> to <span class="math-container">$\pu{400 K}$</span> at constant volume. Calculate the change in internal energy <span class="math-container">$\Delta E$</span> and the change in enthalpy <span class="math-container">$\Delta H.$</span> Given: <span class="math-container">$C_V = \pu{20.77 J K-1 mol-1}$</span></p> </blockquote> <p>There is no problem in calculating <span class="math-container">$ΔE:$</span></p> <p><span class="math-container">$$\Delta E = nC_V\,\Delta T = \pu{20770 J}$$</span></p> <p>However, if I say</p> <p><span class="math-container">$$\Delta H = nC_p\,\Delta T = n(R + C_V)\,\Delta T$$</span></p> <p>it sure does not equal <span class="math-container">$\Delta E.$</span> But <span class="math-container">$\Delta H = \Delta E + p\,\Delta V$</span> must be applicable and we have <span class="math-container">$\Delta V = 0.$</span></p> <p>Can someone please help?</p> https://chemistry.stackexchange.com/q/119145 1 What is total enthalpy? David Lu https://chemistry.stackexchange.com/users/80503 2019-08-13T09:44:57Z 2019-08-13T09:44:57Z <p>I've seen that it's defined as <span class="math-container">$h_0 = h + \frac{v^2}{2}$</span>, where <span class="math-container">$h_0$</span> is the total enthalpy per unit mass. I can see that the <span class="math-container">$\frac{v^2}{2}$</span> term probably has something to do with kinetic energy, but I am still really confused. Doesn't <span class="math-container">$h$</span> already account for flow work? And also, how can both sides of the equation have the same units if <span class="math-container">$h_0$</span> is "per unit mass"?</p> https://chemistry.stackexchange.com/q/118982 0 How do you find the heat capacity ratio of combustion products? David Lu https://chemistry.stackexchange.com/users/80503 2019-08-08T10:14:37Z 2019-08-08T10:14:37Z <p>For example, if gaseous oxygen <span class="math-container">$(O_2)$</span> and paraffin wax <span class="math-container">$(C_{32}H_{66})$</span> were to stoichiometrically combust at STP, how would one go about finding the heat capacity ratio (<span class="math-container">$\gamma={c_p \over c_v}$</span>) of the resulting mixture of gases? I'm guessing that it involves finding the adiabatic flame temperature? (but idk how lol)</p> https://chemistry.stackexchange.com/q/118597 4 Infinite Increase in Entropy when Energy added to Absolute Zero Liam https://chemistry.stackexchange.com/users/73990 2019-07-28T12:39:10Z 2019-07-30T23:20:32Z <p>My textbook states the following:</p> <blockquote> <p>If a system were at absolute zero, an additional small amount of heat energy would lead to an infinite increase in entropy. Such a state is impossible. Absolute zero can never be achieved.</p> </blockquote> <p>It also provides the equation:</p> <p><span class="math-container">$$\Delta S_\text{surroundings} = \frac{-\Delta H_\text{system}}{T}$$</span> Where <span class="math-container">$T$</span> is given in kelvin.</p> <p>From the statement, will the entropy of the <strong>system or surroundings</strong> increase; from what I can deduce, I would say the entropy increase would be within the system as the surroundings is losing energy, but I am not sure.</p> <p>The second part of my question is, <strong>why would this lead to an infinite increase in entropy?</strong> Please provide a comprehensible mathematical explanation and also, preferably, an analogy.</p> <p>Textbook: Pearson Baccalaureate: Higher Level Chemistry, 2nd Edition. By Catrin Brown and Mike Ford Pages: 254-255</p> https://chemistry.stackexchange.com/q/118347 3 Facing a problem in enthalpy calculation bscripts https://chemistry.stackexchange.com/users/81159 2019-07-22T11:50:42Z 2019-07-22T17:37:29Z <blockquote> <p>If 1 mole of gaseous carbon atoms are converted into diamonds, calculate enthalpy change of process given that bond enthalpy of <span class="math-container">$\ce{C-C}$</span> bond is <span class="math-container">$\pu{400 kJ mol-1}.$</span></p> </blockquote> <p>According to me, the answer should be that since C(diamond) has 4 bonds and C(gas) doesn't have any bonds, 4 bonds are being formed so enthalpy change should be </p> <p><span class="math-container">$$ΔH = 4\cdot (\pu{-400 kJ mol-1})$$</span></p> <p>But in the correct solution it says each bond is counted twice, so answer will be </p> <p><span class="math-container">$$ΔH = \frac{4\cdot (\pu{-400 kJ mol-1})}{2}$$</span></p> <p>I don't get why are we doing this. In other questions we don't consider the double counting of bonds. For example, another question I did had 1 mole phosphorus gas converting into <span class="math-container">$\ce{P4}$</span> molecule, and even though <span class="math-container">$\ce{P4}$</span> has 6 bonds, we did not consider what we did with diamond, so what is the reason for that?</p> https://chemistry.stackexchange.com/q/118337 0 Calculating bond energy given the enthalpy value of burning reaction user5721565 https://chemistry.stackexchange.com/users/79878 2019-07-22T03:26:05Z 2019-07-23T13:03:41Z <blockquote> <p>The burning enthalpy (exothermic) of <span class="math-container">$\ce{NO +O2 -&gt;NO2}$</span> is <span class="math-container">$\pu{57 kJ mol^-1}$</span> and <span class="math-container">$\ce{O=O}$</span> bond energy is <span class="math-container">$\ce{494 kJ mol^{-1}}$</span>, then what is the bond energy of <span class="math-container">$\ce{N-O}$</span> ?</p> </blockquote> <p>So I balanced the equation</p> <p><span class="math-container">$$\ce{2NO + O2 -&gt;2NO2}$$</span></p> <p>wrote down Lewis structures and got</p> <p><span class="math-container">$$(2\Delta H(\ce{N=O}) + \Delta H(\ce{O=O}) - 2(\Delta H(\ce{N-O}) + \Delta H(\ce{N=O})) = \pu{-57 kJ mol^-1}$$</span></p> <p>(<span class="math-container">$\pu{-57 kJ mol^-1}$</span> since it is an exothermic reaction), so we are left only with</p> <p><span class="math-container">$$\Delta H(\ce{O=O}) - 2\Delta H(\ce{N-O}) = \pu{-57 kJ mol^-1}$$</span></p> <p>thus</p> <p><span class="math-container">$$\Delta H(\ce{N-O}) = \frac{494+57}{2}\pu{kJ mol^-1} = \pu{275.5 kJ mol^-1}$$</span></p> <p>However the right answer is</p> <p><span class="math-container">$$\frac{494-2\cdot 57}{2} \pu{kJ mol^-1}= \pu{190 kJ mol^-1}$$</span></p> <p>and I really don't get what am I doing wrong? </p> https://chemistry.stackexchange.com/q/118263 -1 Reaction of sulphur with calcium carbonate klj 1 https://chemistry.stackexchange.com/users/79396 2019-07-19T21:19:23Z 2019-07-21T13:45:24Z <p>Can sulphur react with calcium carbonate to form sulphur trioxide and calcium carbide, as the energy of 3(S=O) is greater than ((C=O) +2(C-O))?</p> <p>EDIT: i forgot to say that this is in an oxygen-free environment. </p> https://chemistry.stackexchange.com/q/118092 1 Why enthalpy changes in phase transition when there is no change of temperature? Yahel Abraham https://chemistry.stackexchange.com/users/79964 2019-07-16T09:38:25Z 2019-07-16T11:55:30Z <p>Why there is a change in enthalpy in phase transition?</p> <p>I'm asking because enthalpy depends only on temperature:</p> <p><span class="math-container">$$\mathrm dH = C_p\,\mathrm dT,$$</span></p> <p>and the change in temperature is <span class="math-container">$\mathrm dT = 0.$</span></p> https://chemistry.stackexchange.com/q/118017 0 Calculating bond dissociation enthalpy of F-F bond AfiJaabb https://chemistry.stackexchange.com/users/78533 2019-07-14T10:34:51Z 2019-07-14T11:43:08Z <blockquote> <p><span class="math-container">$$\ce{N2 + 3 F2 -&gt; 2 NF3} \quad ΔH = \pu{-263 kJ mol-1}$$</span></p> <p>The bond dissociation enthalpies of the nitrogen triple bond and nitrogen–flourine bond are <span class="math-container">$\pu{946 kJ mol-1}$</span> and <span class="math-container">$\pu{272 kJ mol-1}.$</span> Find the bond dissociation enthalpy of the <span class="math-container">$\ce{F-F}$</span> bond.</p> <p><strong>A</strong>. <span class="math-container">$\pu{-423 kJ mol-1}$</span><br> <strong>B</strong>. <span class="math-container">$\pu{-393 kJ mol-1}$</span><br> <strong>C</strong>. <span class="math-container">$\pu{-141 kJ mol-1}$</span><br> <strong>D</strong>. <span class="math-container">$\pu{+141 kJ mol-1}$</span><br> <strong>E</strong>. <span class="math-container">$\pu{+423 kJ mol-1}$</span> </p> </blockquote> <p>I know that since bond dissociation enthalpy is endothermic it has to be one of the last two.</p> <p>I tried substracting bond dissociation enthalpies of products from reactants and equating that to the enthalpy change but my answer is nowhere near the given ones. Is my method wrong?</p> https://chemistry.stackexchange.com/q/117923 2 I can't find reaction enthalpy ∆H for water formation reaction Bruno Machado https://chemistry.stackexchange.com/users/80901 2019-07-11T15:48:06Z 2019-07-11T20:38:52Z <p><span class="math-container">$$\ce{1/2 O2(g) + H2(g) → H2O (l)}$$</span></p> <p>The tabulated value is <span class="math-container">$\Delta H = \pu{−285.8 kJ/mol}$</span></p> <p>I'm using the bond energies values for each molecule and for some reason, I am getting a different result, not even close to <span class="math-container">$\pu{-285 kJ/mol}$</span>.</p> <p>Bond dissociation energies:</p> <p><span class="math-container">$\ce{H-H} = \pu{436 kJ/mol}$</span></p> <p><span class="math-container">$\ce{O=O} = \pu{499 kJ/mol}$</span></p> <p><span class="math-container">$\ce{O-H} = \pu{464 kJ/mol}$</span></p> <p>My attempt to calculate from bond dissociation energies: </p> <p>(500.1/2) + 436 → 464.2 </p> <p>250 + 436 → 928</p> <p>(<span class="math-container">$∆H$</span> found is positive)</p> https://chemistry.stackexchange.com/q/117521 1 Question about calculating enthalpy change king https://chemistry.stackexchange.com/users/58796 2019-07-01T01:47:05Z 2019-07-01T02:26:13Z <p><span class="math-container">$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$</span></p> <p><span class="math-container">$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$</span></p> <p>Both <span class="math-container">$\Delta H$</span>'s given by the above formulas should be equivalent. However, I found one special case that the above 2 formulas give 2 different results. <span class="math-container">$$\ce{2 H+ + 2 e- → H2}$$</span> <span class="math-container">$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)= 0-0=0$$</span> <span class="math-container">$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)&lt;0$$</span> because there is only bond formation, no bond break. How to explain this difference? Can I say the heat is released during this process, because <span class="math-container">$$\Delta H$$</span> is smaller than 0? </p> https://chemistry.stackexchange.com/q/117456 4 Exothermic: enthalpy or heat? Flufflette https://chemistry.stackexchange.com/users/80553 2019-06-28T22:56:44Z 2019-06-29T23:37:29Z <p>I am used to thinking of exothermic reactions as being defined as having negative <span class="math-container">$ΔH.$</span> However, I also am used to hearing/thinking of exothermic reactions as "releasing heat". </p> <p>Since change in enthalpy <span class="math-container">$∆H = Q + VΔP,$</span> it seems that enthalpy and heat flow <span class="math-container">$(Q)$</span> are not strictly the same thing. Is it not therefore possible to have a reaction which absorbs heat <span class="math-container">$(+Q),$</span> but has a decrease in pressure, such that the product <span class="math-container">$-VΔP &gt; Q,$</span> thus both absorbing heat and being exothermic (or if exo-/endothermicity is determined by <span class="math-container">$Q,$</span> then an exothermic reaction with positive <span class="math-container">$∆H)?$</span></p> <p>As a sort of corollary, is there an example of a thermodynamic process that changes only the pressure (thus <span class="math-container">$V∆P)$</span> with no heat flow or <span class="math-container">$PV$</span> work? <span class="math-container">$(Q = 0 = PΔV)$</span></p> https://chemistry.stackexchange.com/q/117350 1 What makes determining entropy changes theoretically so hard compared to enthalpy? Nathan Matthieu Tang https://chemistry.stackexchange.com/users/76548 2019-06-26T11:24:05Z 2019-06-26T11:24:05Z <p>It is really hard to determine the entropy change of a reaction accurately because of subtle solvent effects and a lack of standard entropies. On the other hand, enthalpy is relatively easy and through a thermodynamic cycle, it can usually be calculated. But what about our knowledge of entropy limits our ability to measure it? Without using straight empirical data, why is using force field and electronic interactions not very reliable? Many reactions face this problem, like the mixing of liquids.</p> https://chemistry.stackexchange.com/q/117119 1 Enthalpy of combustion of 1-butanol Alif Papp https://chemistry.stackexchange.com/users/80318 2019-06-20T10:46:26Z 2019-07-23T07:01:35Z <p>1-butanol <span class="math-container">$\ce{C4H10O}$</span> </p> <p>So I have absolutely no idea how to find the enthalpy and I did some looking on the internet with no use since I don't understand.</p> <p>I came here hoping someone generous could do the sample calculations so I can understand and do the rest my calculations and test for my experiment.</p> <p><a href="https://i.stack.imgur.com/qUEQa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qUEQa.png" alt="enter image description here"></a></p> <p>That's the data for one the mixture of alcohol.</p> <p>How would I figure out enthalpy for test 1? </p> <p>(maybe I could cut out test 2, 3 and average in the picture since its irrelevant in figuring out the enthalpy of test 1, and causes confusion)</p> <p>For temperature before and after. The solution was used to heat up 200 ml of water. By burning the mixture, igniting it.</p> https://chemistry.stackexchange.com/q/117111 1 Reaction enthalpy using bond energy in resonating molecules RandomAspirant https://chemistry.stackexchange.com/users/79319 2019-06-20T06:49:57Z 2019-06-20T09:13:12Z <p>To find out and enthalpy of a reaction using bond energy we use the formula</p> <p><span class="math-container">$$\Delta H = \sum\Delta_\mathrm{bond}H_\mathrm{reactants} - \sum\Delta_\mathrm{bond}H_\mathrm{products}$$</span></p> <p>since we use all the individual <span class="math-container">$\Delta H$</span> of the bonds to calculate the enthalpy, e.g. </p> <p><span class="math-container">$$\Delta H_\ce{C2H6} = 2\Delta_\mathrm{bond}H_\ce{C-C} + 6\Delta_\mathrm{bond}H_\ce{C-H}$$</span></p> <p>But if a molecule is in <strong>resonance</strong>, how is the bond enthalpy taken? For example,</p> <p><a href="https://i.stack.imgur.com/zNO4u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zNO4u.png" alt="putative resonance structures"></a></p> <p><a href="https://i.stack.imgur.com/yxBDE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yxBDE.png" alt="resonance structures"></a></p> <p>Is bond enthalpy the sum of the bond energies of that canonical structure which is the most stable or the average of all canonical structures or something else?</p> https://chemistry.stackexchange.com/q/117025 4 Why are Born-Haber enthalpy calculations inaccurate for transition metal compounds? Tony B https://chemistry.stackexchange.com/users/80272 2019-06-18T13:40:28Z 2019-06-18T13:40:28Z <p>A senior high/secondary school level exam question is about discrepancies between calculated and empirical enthalpies of formation. It asks why a discrepancy “usually” exists for d-block compounds such as copper (II) chloride. </p> <p>For something like AgCl or AlCl3 the answer is covalent character. But that’s not true for CuCl2, is it?</p> https://chemistry.stackexchange.com/q/117012 2 Iron(III) nitrate and hydrogen peroxide reaction user80262 https://chemistry.stackexchange.com/users/80262 2019-06-18T09:15:05Z 2019-06-18T21:40:12Z <p>When iron(III) nitrate is added to hydrogen peroxide as a catalyst, the temperature initially decreases and begins to increase after 30 seconds. Why is this the case? (is there any intermediate reactions of <span class="math-container">$\ce{H2O2}$</span> and <span class="math-container">$\ce{Fe(NO3)3}?)$</span> </p> https://chemistry.stackexchange.com/q/117006 2 Energy absorbed by a system in isothermic process in which the phase changes from liquid to gas user5721565 https://chemistry.stackexchange.com/users/79878 2019-06-18T04:55:23Z 2019-06-18T20:51:22Z <p>A system contains <span class="math-container">$x$</span> mole of material, with surface area <span class="math-container">$S$</span> is pressed by a mass <span class="math-container">$M_1$</span> (no fraction and no any other outer forces involves). Let <span class="math-container">$\Delta H_\mathrm{l\rightarrow g}$</span> be the transition enthalpy from liquid to gas. The system is position in <span class="math-container">$(T,P_1)$</span> such that the material phase is liquid for those temperature and pressure values (<span class="math-container">$T$</span> and <span class="math-container">$P_1$</span>). Now instantaneously the mass <span class="math-container">$M_1$</span> is decreased to to an <strong>unknown</strong> value which we denote <span class="math-container">$M_2$</span> so the pressure decreases. In case that the temperature stays constant until the inner pressure is equal to the outer pressure <span class="math-container">$(T,P_2:=\frac{M_2g}{S})$</span> system gets to back to equilibrium, and the material phase in <span class="math-container">$(T,P_2)$</span> is gas, I wish to calculate how much energy was absorbed by the system till it gets to pressure equilibrium <span class="math-container">$(T,P_2)$</span>.</p> <p>In the question is says that the volume of the liquid can be neglected when compared to gas' volume. And that the volume of the water is independent of the pressure.</p> <p>What I did so far:</p> <p>The chemical equilibrium pressure (i.e the pressure in the phase transition point) can be calculated because the exercise gives the triple point <span class="math-container">$(T_\triangle , P_\triangle)$</span> so using the transition enthalpy constant and the constant <span class="math-container">$T$</span> of the process, one can calculate <span class="math-container">$(P_\mathrm{l\rightarrow g, eq},T)$</span> (using Clausius–Clapeyron relation). And by <span class="math-container">$PV=nRT$</span> one may calculate <span class="math-container">$V_\mathrm{eq} = \frac{nRT}{P}$</span> the volume in the chemical equilibrium point. By the unstruction <span class="math-container">$V_\mathrm{eq}$</span> is the total volume of the material (we neglect the liquid volume)</p> <p>We know the starting inner pressure which is <span class="math-container">$\frac{M_1g}{S} \text{[Pa]}$</span> and by the given instruction the pressure doesn't change the liquid volume we learn that until the phase transition point, the volume of the system stays constant <span class="math-container">$V_\mathrm{eq}$</span> which we calculated above.</p> <p>When the system gets to pressure equilibrium point <span class="math-container">$(T,P_2=\frac{M_2g}{S})$</span> we don't know the volume or <span class="math-container">$P_2$</span> because <span class="math-container">$P_2$</span> is unknown how every we not that its larger then <span class="math-container">$V_\mathrm{eq}$</span>.</p> <p>I'm pretty much stuck here, I don't see how to translate this information to evaluate the amount of energy absorbed. It seems to me that maybe one more piece of information is needed, but the exercise doesn't give one.</p> https://chemistry.stackexchange.com/q/115842 3 Relation between equilibrium constant and entropy change user226375 https://chemistry.stackexchange.com/users/76375 2019-05-24T02:34:07Z 2019-05-24T09:31:04Z <p>Like the Van't Hoff equation, which relates change in enthalpy to equilibrium constant, is there a similar equation for the relation between change in entropy and equilibrium constant?</p> <p>Consider the following specific case,</p> <blockquote> <p>For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by</p> <p>[A] With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive</p> <p>[B] With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases</p> <p>[C] With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative</p> <p>[D] With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases</p> <p><strong>Answer</strong>: (B) and (D)</p> </blockquote> <p><strong>My attempt:</strong></p> <p>Since equilibrium with surrounding is given, the reaction must be reversible.</p> <p>This implies, <span class="math-container">$∆S=0, ∆S_s=∆S_{surr}$</span>.</p> <p>Also, at equilibrium <span class="math-container">$∆G=0$</span>.</p> <p>This implies, ∆<span class="math-container">$H=T∆S_s=-T∆S_{surr}$</span></p> <p>(where the s subscript denotes system, and surr denotes surrounding)</p> <p>Writing the van't Hoff equation in differential form, I got</p> <p><span class="math-container">$= d(lnK)=∆H(dT/RT^2)$</span></p> <p><span class="math-container">$=d(lnK)=-∆S_{surr}(dT/RT)$</span></p> <p>Now, aren't options A and C true too? </p> <p>As on integrating, we can see directly the variation of <span class="math-container">$K$</span> with <span class="math-container">$∆S$</span>.</p> <p>What is the correct way to solve this?</p> <p><strong>Edit:</strong> Where ever there is a big space, it implies ∆. Example, ' ' H=T ' ' , </p> <p>Means, ∆H=T∆S</p> https://chemistry.stackexchange.com/q/115528 0 Need to normalize lattice enthalpy by no. of nearest-neighbor (NN) interactions to equate to NN interaction strength? theorist https://chemistry.stackexchange.com/users/68027 2019-05-19T00:50:22Z 2019-05-19T05:33:02Z <p>I've seen electrostatic interaction energy (from <span class="math-container">$E\sim \frac{q_1 q_2}{r}$</span>) used to explain differences in lattice enthalpies between different crystalline solids. For example, I've seen it explained that <span class="math-container">$\ce{NaF}$</span>'s <span class="math-container">$\pu{910 kJ/mol}$</span> lattice enthalpy is less than <span class="math-container">$\ce{MgCl2}$</span>'s <span class="math-container">$\pu{2326 kJ/mol}$</span> because of <span class="math-container">$\ce{Mg}$</span>'s greater cationic charge (+2 vs. +1). </p> <p>Question: Isn't the lattice enthalpy defined per stochiometric unit, rather than per nearest-neighbor interaction? And if so, if one is going to use electrostatic models of nearest-neighbor interaction to explain differences in lattice enthalpy, shouldn't the respective lattice enthalpies first be divided by the number of nearest-neighbor interactions per stochiometric unit? </p> https://chemistry.stackexchange.com/q/115355 0 Thermodynamic data related to diazotization reaction Mohan N. Bhawe https://chemistry.stackexchange.com/users/79182 2019-05-15T12:44:13Z 2019-05-15T15:16:18Z <p>Is there any thermodynamic data available for diazotization of aniline and also for the reduction of the diazonium salt using sodium sulfite? I also want to know the heat of reaction for both the reactions.</p> https://chemistry.stackexchange.com/q/114937 1 Relation between ΔH and ΔE for the reaction in gaseous phase at constant temperature and pressure Aryan Stark https://chemistry.stackexchange.com/users/62265 2019-05-06T20:44:22Z 2019-05-09T01:55:09Z <blockquote> <p>Which of the following statements is correct for the reaction at constant temperature and pressure:</p> <p><span class="math-container">$$\ce{CO(g) + \frac{1}{2}O_2(g) -&gt; CO2(g)}$$</span></p> <ol> <li><span class="math-container">$\Delta H = \Delta E$</span></li> <li><span class="math-container">$\Delta H &gt; \Delta E$</span></li> <li><span class="math-container">$\Delta H &lt; \Delta E$</span></li> <li>None of these.</li> </ol> </blockquote> <p>Since this occurs at constant temperature, <span class="math-container">$\Delta E = 0$</span> and <span class="math-container">$W &lt; 0$</span>. So, <span class="math-container">$\Delta H$</span> turns out to be less than <span class="math-container">$\Delta E$</span>, but this is not the answer. Why?</p> https://chemistry.stackexchange.com/q/114897 0 Comparing strength of the C-H, N-H, and O-H bonds Evelyn Ip https://chemistry.stackexchange.com/users/78645 2019-05-06T06:40:58Z 2019-05-06T08:26:34Z <blockquote> <p><span class="math-container">\begin{align} \ce{CH4(g) + Cl(g) &amp;-&gt; CH3(g) + HCl(g)} &amp;\quad ΔH^\circ &amp;= \pu{-14 kJ mol-1}_\mathrm{rxn}\\ \ce{NH4(g) + Cl(g) &amp;-&gt; NH2(g) + HCl(g)} &amp;\quad ΔH^\circ &amp;= \pu{-36 kJ mol-1}_\mathrm{rxn}\\ \ce{H2O(g) + Cl(g) &amp;-&gt; OH(g) + HCl(g)} &amp;\quad ΔH^\circ &amp;= \pu{+40 kJ mol-1}_\mathrm{rxn} \end{align}</span></p> <p><strong>41</strong>. Based on the data above, what can be concluded regarding the strength of the <span class="math-container">$\ce{C-H}$</span>, <span class="math-container">$\ce{N-H}$</span>, and <span class="math-container">$\ce{O-H}$</span> bonds in the molecules shown?</p> <p>(<strong>A</strong>) The <span class="math-container">$\ce{C-H}$</span> bond is the strongest.<br> (<strong>B</strong>) The <span class="math-container">$\ce{N-H}$</span> bond is the strongest.<br> (<strong>C</strong>) The <span class="math-container">$\ce{O-H}$</span> bond is the strongest.<br> (<strong>D</strong>) Nothing can be concluded without knowing the strength of the <span class="math-container">$\ce{H-Cl}$</span> bond. </p> </blockquote> <p>I am not sure how to approach the problem <strong>41</strong>. In exothermic reactions, more energy is released when the bonds are formed in the products than is used to break the bonds in the reactants. So would the answer be <strong>B</strong> because the <span class="math-container">$ΔH$</span> is exothermic and is the greater than <span class="math-container">$ΔH$</span> of -14 for <strong>A</strong>?</p> https://chemistry.stackexchange.com/q/114793 0 Trying to understand rationale behind Joule-Thomson experiment theorist https://chemistry.stackexchange.com/users/68027 2019-05-03T23:38:26Z 2019-05-04T16:34:47Z <p>The purpose of the original Joule (as opposed to JT) expansion experiment was to assess intermolecular interactions in real gases. In that experiment, Joule immersed twin glass globes into a water bath. The globes were connected by a closed valve. One globe held a pressurized gas, while the other was evacuated. Joule then opened the valve, allowing the gas in the filled globe to expand into the evacuated one. Since the gas is expanding into (approximately) a vacuum, it does no work (Pext = 0), and thus any temperature change is due to changes in intermolecular potential energy resulting from changes in intermolecular separation (and conservation of energy). This "free expansion" is thus called a "Joule expansion". Joule's experiment failed to show a temperature change, however, because he was measuring the temperature change of the water bath, and the latter's heat capacity far exceeded that of the gas. [I recall reading that modern calculations of Joule's experiment estimate the actual temperature change was ~0.001C, too small for Joule to measure.]</p> <p>The obvious improvement would be to insulate the glass globes and measure the temperature of the gas directly. Joule didn't do this, however, because he realized that even the heat capacity of the glass globes would be too much greater than that of the gas, precluding him from accurately measuring a temperature change (i.e., perfect adiabatic walls don't exist). [This is explained in Reif, <em>Fundamentals of Statistical and Thermal Physics</em>]</p> <p>According to Reif, the solution was the JT experiment, which was designed to allow a steady-state flow of gas. Such a flow allowed for the temperature of the insulated apparatus to equilibrate with that of the gas. Since the container is now at the same temp as the gas, the high heat capacity of the container walls is not an issue, i.e., heat is no longer flowing between the gas and the container. This equilibration is not possible with a single-event experiment in which gas is expanded into a vacuum, but it is possible if you have a high-P tank with a pressure regulator, venting gas continuously into the atmosphere. That's what the JT set-up gives you. It so happens that this set up results in an isoenthalphic, rather than isoenergetic, process.</p> <p>QUESTION: Did the fact that the JT experiment is isoenthalphic result purely from the need to have a continuous flow (i.e., was its isoenthalpic nature just an incidental consequence of the design), or was it specifically designed to be isoenthalpic because Joule and Thomson gained specific insights into intermolecular interactions because of its isoenthalpic nature -- insights not possible from a free expansion? [I don't think the latter is the case, since one can get inversions in both free and JT expansions and, when it comes to gaining molecular insights, the JT expansion muddies the waters vs. a free expansion.] Yes, this experiment gives us <span class="math-container">$(\partial H/\partial P)_T$</span>, but that can also be obtained from 𝛽: <span class="math-container">$(\partial H/\partial P)_T$</span> = V(1- 𝛽T). [Or is 𝛽 actually measured from <span class="math-container">$𝝻_{JT}$</span>?]. </p> <p>My suspicion is they designed it to be continuous for the reasons I described above (steady-state temp drop), and its isoenthalphic nature was a happy (but important) bonus.</p> https://chemistry.stackexchange.com/q/114671 1 Change in heat of photosynthesis UnhookedSchnook https://chemistry.stackexchange.com/users/78621 2019-05-01T11:28:16Z 2019-05-02T08:44:38Z <p>Consider a plant that uses the energy it receives from the sun to do work (to undergo photosynthesis) to produce glucose. The minimum amount of work this plant can do to produce 1 mole of glucose is equal to the change in Gibbs free energy that goes into producing one mole of glucose. I calculated the change in Gibbs free energy that goes into producing one mole of glucose to be <span class="math-container">$\Delta G =2878.94\ \mathrm{kJ}$</span>. Therefore the plant does <span class="math-container">$2878.94\ \mathrm{kJ}$</span> of work. Also if we calculate the change in enthalpy for the same process we get, <span class="math-container">$\Delta H = 2803.04\ \mathrm{kJ}$</span>. Therefore the heat from photosynthesis under these conditions is, <span class="math-container">$Q=\Delta H - \Delta W = -69.9\ \mathrm{kJ}$</span>.</p> <p>This confuses me because it implies that photosynthesis under these conditions gives off heat. How can this be? </p> https://chemistry.stackexchange.com/q/114617 1 Heat and work of glucose combustion UnhookedSchnook https://chemistry.stackexchange.com/users/78621 2019-04-30T15:23:01Z 2019-05-01T04:06:39Z <p><span class="math-container">$$\ce{C6H12O6(s) + 6 O2(g) -&gt; 6 CO2(g) + 6 H2O(g)}$$</span> When looking at glucose combustion I found that the change in Gibbs free energy is greater than the change in enthalpy. If we assume the process is reversible we can set the change in Gibbs free energy to be the work available and if we assume there is no work we can set the enthalpy to be the heat. Why then would the work be less than the heat? I suspect that <span class="math-container">$\Delta G = \Delta H-\Delta (TS)$</span> is at work.</p> <p>Cool, so can you substitute work, W and heat, Q into the equation and write W=Q−TΔS to directly show that ΔS>0? Also would you please explain how an increase in entropy would imply the discrepancy between heat and work (not mathematically).</p> https://chemistry.stackexchange.com/q/114507 0 Is the enthalpy of a molecule determined by the energy of the electrons? [closed] Christopher U'Ren https://chemistry.stackexchange.com/users/29986 2019-04-28T19:31:14Z 2019-04-29T13:33:05Z <p>I understand that the strength of the bond of a molecule is determined by the potential and kinetic energy of the electrons. I also understand that a reaction where the bond strength of the products is stronger than the bond strength of the reactant releases energy (is exothermic).</p> <p>My textbook states that in a reaction where NaOH dissolves in water, the energy of the solid NaOH is greater than the energy of the dissolved ions in water. Is the "energy" of the reactants and products determined by the bond strength?</p> <p>Also, my textbook states that enthalpy is "the amount of energy, in the form of either kinetic or potential energy, a substance has." I'm a bit confused about enthalpy. My textbook enthalpy changes:</p> <p><a href="https://i.stack.imgur.com/8yAcq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8yAcq.jpg" alt="enter image description here"></a></p> <p>Are these enthalpy changes determined by the bond strength also? And is the enthalpy of a certain molecule also from the bond strength?</p> https://chemistry.stackexchange.com/q/114442 0 why is white phosphorous given to have ZERO standard heat of formation [duplicate] Snmohith Raju https://chemistry.stackexchange.com/users/77324 2019-04-27T16:59:36Z 2019-04-27T16:59:36Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/43141/phosphorus-standard-state" dir="ltr">Phosphorus standard state</a> <span class="question-originals-answer-count"> 1 answer </span> </li> </ul> </div> <p>why is white phosphorous given to have ZERO standard heat of formation??</p> <p>while it is well known fact that white phosphorous is highly reactive and is obviously less stable than red and black phosphorous</p> <p>also here=<a href="https://chemistry.stackexchange.com/questions/31664/which-allotrope-of-phosphorus-is-kinetically-the-most-stable">Which allotrope of phosphorus is kinetically the most stable?</a></p> <p>it says that black phosphorous is thermodynamically more stable</p> <p>also what is the relative stability of the allotropes [black red white purple]</p> https://chemistry.stackexchange.com/q/114347 3 What is meant by electrons moving to lower potential energy? Christopher U'Ren https://chemistry.stackexchange.com/users/29986 2019-04-26T02:15:01Z 2019-04-27T08:26:45Z <p>I am currently learning about chemical bonds in chemistry, and came across somebody who wrote this (<a href="https://socratic.org/questions/how-do-chemical-bonds-store-energy" rel="nofollow noreferrer">source</a>):</p> <p>"Chemical bonds certainly "contain" potential energy, and the atoms want to move to a lower potential energy (become more stable).</p> <p>When methane, CH 4 , forms, the valence electrons end up in more stable (lower energy) C-H bonds.</p> <p>These bonds are fairly strong, so methane is relatively inert.</p> <p>However, if you add energy to the methane in the form of a flame or a spark in the presence of oxygen, some of the molecules will have enough energy to overcome an activation energy barrier.</p> <p>Some of the C-H bonds will break.</p> <p>The electrons can then enter an even lower energy state by forming C=O and O-H bonds rather than staying as C-H and O=O bonds."</p> <p>What is meant by this? What does it mean when electrons enter a lower energy state, do they move to an even lower ground state?</p>