Predict whether the entropy value will increase or decrease during the following processes....\n1. H2O (l) -> H20 (s)\n2. 02 (g) -> 2 O2 (g)\n3. NaCl (s) -> NaCl (aq)\n4. NH3 (g) -> NH3 (aq) \n\nIs there any work to show for these and how do you do them??
129839
Basically its just when something goes from a liquid to a solid the entropy decreases and so you show the decrease in entropy values which should be in your chemistry. And when something goes from a solid to a liquid, the entropy increases and liquid to a gas, it increases. You should use the entropy values and show the change between the states. That would be the work I would show
463459
In a closed vessel the change in partial pressure for each gas during reaction scales to the stiochiometric coefficients.\n\nSo for this reaction, if 23 torr/min of NO is being used, then there must be 23 torr/min of NOCl appearing since both coefficients are 2 and they are on opposite sides of the equation.\n\nLikewise, if 23 torr/min of NO is being used, then 11.5 torr/min of Cl2 is also being used since they are on the same side of the equation.  It is only 11.5 torr/min because the coeffient for Cl2 is only 1, so it is half the amount of NO.\n\nThe total pressure change will be the sum of the amounts:\n\npNO + pCl2 + pNOCl = -23 -11.5 + 23 = -11.5\n\nKeeping two sig figs, this would mean that the pressure is decreasing by 12 torr/min.
668339
1) exothermic reaction gives off heat\n2) a catalyst speeds up a rxn\n3) chemical reaction\n4) 1  + 4 --> 1 + 1 + 2\n\ncoefficiant of HCl is one
589019
Here's what I wrote you, hopefully some nice thermodynamics expert out there can comment for you.\nΔG = ΔH - T ΔS is the equation for gibb's free energy, it's a number chemists make up to describe the system as a combination of enthalpy (which is the energy a system has available to heat transfer) and expansion energy, temperature x entropy or ΔS.  The TΔS part indicates how much expansion can take place, because entropy is determined by how many molecules the reaction "sets free" kind of.\nThe pre-equilibrium state can be calculated by finding amounts of gas produced and the temperature at any given moment during the move toward equilibrium.  Then they take a bunch of points and find the slope/rate of change and find the area under the curve.  I suppose that it is possible for that function to be a log of the products/reactants(the usual way to write equilibriums) depending on the order of the reaction and the products.  I think it would have to be gases producing a solid, 1st order.  Then the slope/rate would be a function 1/x and the integral would be logx.  But that wouldn't be true for all reactions.
208233
Entropy is a measurement of probability.  Say that there's two chambers of pink gas connected by some tubing.  The pink gas is free to move around between the two chambers.  \nIs it possible that every particle of gas will rest neatly in the first chamber?  Yes.  Is it at all probable?  No way.  The chances of that happening are nearly nothing.  That's because of entropy.  Entropy is disorder, and everything likes to go to higher disorder.
551990
ok--i'm not completely sure about this--but i can at least try-- *cough cough*unlike some people*cough*cough*catweazle*\n\nwe did this type of stuff last year.  but i think that you might want to use the ideal gas law?? we always called it pervnert... PV=nRT... does this sound familiar?    \n\nif not, i suggest asking your teacher, or some other knowledgable chemistry person.\n\ni hope this helps!
532272
For the reaction:\n\nMgO + 2 HCl  -->  MgCl2 + H2O\n\nThe temperature should go up, not down.  Only the initial and final temperature are needed.\n\nTypical values for students are between 110 an 150 kJ/mol MgO.  This is typical, not the best that can be done.
342239
First of all, transliterate this into coherent English.\n\nSecond of all, when you find the theory of entropy in those holy books, you let me know.
215628
see dear ,\nwhile calculating the energy changes we need to know the hidden heat { latent heat } that is absorbed during the change in state . for eg when we are calculating the heat absorbed or given out during a change in state , like ice to water , just first calculate the latent heat ie \nq=mL{ heat=mass *latent heat } now add this value to \nq=mct{heat=mass*specific heat*change in temp} keep in mind that specific heat is a constant value \nwater's specific heat is 4180j/kg/degree Celsius.
481315
Im a little rusty on the thermodynamics, but lets see if this helps.\nV/T (initial)=V/T (final) since no gas is lost or added  Raising the temperature on a container that is free to expand does not change the pressure, so you can find the change in volume with temperature.  Since V=A*h, you can find h.\n\nNow, the problem is that the vessel is not entirely free to expand, and acts on a spring, with force equal to F=k*h.  This force is related to the pressure created by the expanding vessel as simply F/A.\n\nAll values are given, so I think you are set.
650014
Problem \ntake D as Delta(change)\nEstimate the change in enthalpy, DH, for the following reaction:\n\nH2(g) + Cl2(g) --> 2 HCl(g) \n\nSolution \n\nTo work this problem, think of the reaction in terms of simple steps:\n\n\nFirst, the reactant molecules, H2 and Cl2, break down into their atoms\nH2(g) --> 2 H(g)\nCl2(g) --> 2 Cl(g)\n\n\nNext, these atoms combine to form HCl molecules\n2 H(g) + 2 Cl(g) --> 2 HCl(g) \n\nIn the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:\n\nDH1 = +(436 kJ + 243 kJ) = +679 kJ\n\nBond breaking requires energy, so we expect the value for DH to be positive for this step.\n\nIn the second step of the reaction, two moles are H-Cl bonds are formed. Bond breaking liberates energy, so we expect the DH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:\n\nDH2 = -2(431 kJ) = -862 kJ\n\nBy applying Hess's Law, DH = DH1 + DH2\n\nDH = +679 kJ - 862 kJ\n\nDH = -183 kJ\n\n\nAnswer\n\nDH = -183 kJ
96123
Kc values can be manipulated in the same way that any equilibrium constants are manipulated, in a way similar to the way that thermodynamics values are manipulated.  \n\nNow, the relationship between the first reaction and the second reaction is that the first reaction was reversed and all the coefficients were cut in half.  That means that we can inverse the Kc value of the first reaction and take the square root of it to get the Kc value of the second reaction.  \n\n1/79000=1.266e-5\nsqrt(1.266e-5)=.00356\n\n.00356 is the Kc value for the second reaction.
488061
Not sure the level with which you are learning, so feel free to ignore the thermodynamics explanation.  If you just want to answer the question, select D.\n\nLike most test questions, there is not enough conditions specified and truly, we could choose several anwers, but only after making assumptions that less likely to be in the testers mind.  I will assume a high school or intro level college course because above that this should be challenged as too vague.\n\nUse logic to realize that answer choice C is not true. Evaluating the others we realize that A means only heat is given off when bonds form and B means heat is absorbed or added during the formation of bonds.  D means conditions exist in which either situation is possible and E means that final product is in a state that has a higher value for entropy.  This only holds true when both Temp and pressure of a system is held constant, but we will eliminate later as well.\n\nBasically a bond can occur when the system goes from a high energy state to a lower energy state.  Formation of bonds is essentially a chemical reaction.  The Gibbs Free Energy equation:\n\n (delta)G=(delta)H-T(delta)S\n\nT=temp of system in Kelvin\n(delta)H is the heat energy change\n(delta)S is the change in entropy (disorder)\n\ndefines whether a reaction will occur spontaneously or not.  (deltaG) must be negative for this to be true. So we will assume \nusing this equation and recognizing that (delta)H tells us exo or endothermic if it is positive or negative , examination of the possibilities of (delta)H and (delta)S,\n\nIf (delta)H is negative and (delta)S is positive the equation will result in a negative value for (delta)G which means a favorible condition exists for bonding to occur.  Therefore, the condition of (delta)H negative (energy leaves as heat - exothermic) is true.  We can thus eliminate answer choice B.\n\nIf the opposite is true and (delta)H is positive and (delta)S is negative, the result is (delta)G that is positive and therefore not spontaneous.  no reaction or bonding occurs.\n\nIf both (delta)H is positive and (delta)S is positive, only a large enough (delta)S will overcome the positive (delta)H in the equation.  So if both are positive and T(delta)S is greater than (delta)H the bonding will occur.  Therefore, when (delta)H is positive (endothermic) the reaction may occur as well.  We can thus eliminate choice A.\n\nWe have shown that C is a true statement because conditions can be met in which the reaction is both exo or endothermic.  However, please note that in both conditions (delta)S is also positive, indicating an increase in entropy.  For the most part any system will always trend toward an insrease in entropy.  \n\nThe question then arises, what happens when (delta)S is negative (decreased entropy) and (delta)H is negative.  For most situations T(delta)S will be large enough to create a positive (delta)G from our equation.  In very rare cases will (delta)H - T(delta)S result in a negative answer.  However, the equation shows this as a possibility when (delta)G= large negative (delta)H - T x small negative (delta)S => large negative # minus a smaller negative => larger negative + smaller # = negative.  (Hope you follow my train of thought).  Therefore in the right situation - (delta) can be negative and bonding/reaction can occur when entropy decreases.  Eliminate E and C is the correct answer.\n\nCareful what you ask for.  Wikipedia site looks pretty informative but very detailed and you have to use a few different pages.
341245
1- find heat quantity\n Q=C*t\nQ = heat quantity of the reaction\nC= heat capacipy of calorimeter\nt=temperature increasing\n\n2- find heat of reaction \n H=(Q*1000)/(N*V)\nN = normality of acid and base\nV= volume of acid and base
369603
The answer is (e).\n\nTo calculate the enthalpy change, find the product of the specific heat of water, the mass of the water in grams and the temperature change of the water in kelvins. Since the change in degrees celcius equals the change in kelvins, use (-2).\n\nE (enthalpy change) = 4.18*200*(-2) = -1678 J \n\nI hope this helps!
344293
The condition for a reaction to proceed spontaneously at constant pressure and temperature is that the change in Gibbs free energy (delta-G) for the reaction is less than zero.\n\nThe Gibbs free energy change is given by:\n\ndelta-G = delta-H - T*delta-S\n\nwhere T is the absolute temperature (i.e., in kelvins).\n\nWe are given delta-H and delta-S for this reaction (the units should probably be in kJ/mole and J/(mol*K) respectively, not J and J/K), and need to find the temperature range (if it exists) for which delta-G is negative.  That is,\n\n0 > delta-G = -5500 J/mol - (-25 J/(mol*K)) * T\n\n5500 K > 25 *T\n\n220 K > T\n\nAnswer (b) is the correct one.
472799
OK...To answer this, you have to recognize that the free energy change at the boiling point is zero.  So, beginning with DG (D=Delta), you have:\n\nDG=DH-TDS=0\n\nSo, DS=DH/T \n\nwhere T is the boiling point is Kelvin.  If you substitute the values for DH and T, you can calculate the entropy change for the vaporization of trimethylamine.