How can you figure out the sign of delta G by analyzing the qualitative characteristics of an experiment?
403536
The sign of delta G can be figured out qulitatively.\n\nIf a certain reaction is spontaneous, then delta G is negative but if a certain reaction is non-spontaneous, then delta G will have a positive sign.
235402
See comment above.\n\nYou do need to do several trials at various concentrations though. You can then determine it through a linear regression.\n\nI also suggest that you consult a book on Analytical chemistry, because Beer's law has limitations. For example, it loses linearity at higher concentrations.
715037
Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
657704
gamma, delta, epsilon, etc.
730844
OK...You can start with the equation of DG (I'm using D instead of writing "delta").\n\nDG = DH -TDS  You've seen that equation, right?  OK..Since the boiling point of water is 100 degreed C at 1 atm pressure, we are really talking about an equilibrium situation.  So, DG= 0.\n\nConsequently, DH = TDS.  \n\nSince you know DH and you know T (in Kelvin), you can calculate DS.  Make sure that you convert DH into Joules (instead of kilojoules) before you finish the problem.
364473
Already been done.\n\nBased on current observable evidence, there is increasing perenctages of certainty regarding the timeline and history of the development of life on our planet.\n\nIt doesn't exclude G-d. It just outlines the observable and measurable properties of the phyiscal evidence.
1042274
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.
152826
Delta G refers to the change in free energy.\n\nWhen a reaction takes place there is a change in the free energy, like dehydrogenation of alcohols. The delta G value depends on reactants and products, i.e. delta G value of dehydrogenation of different alcohols will be different.\n\nEnzymes are biological catalysts. They act by lowering the energy of activation(quite a different parameter than delta G) of a reaction. \n\nEnzyme itself can't have delta G value.
799142
Hi Kitty,\nHere is what, I think, you need.\n\nExothermic, dH < 0, Entropy increases (dS > 0), Reaction is spontaneous and dG < 0\n\nExothermic, dH < 0, Entropy decreases (dS < 0), Reaction is spontaneous only at low temps, if |T dS| < |dH|\n\nEndothermic, dH > 0, Entropy increases (dS > 0), Reaction is spontaneous only  at high temps, if T dS > dH\n\nEndothermic, dH > 0, Entropy decreases (dS < 0), Reaction is not spontaneous, dG > 0
723847
I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
410633
2)\n\nAs deltaG=deltaH - TdeltaS\n\ndeltaH needs to be multiplied by 1000 to make it in J\n\nPut into the equation you get: T=279.26K\n\nie 6.1 degrees Celsius\n\nthe importance is that this is the temperature at which the reaction becomes thermodynamically spontaneous.  Basically the benzene will solidify at 6.1degrees.
1012436
Yes, knowing these values for the combusion of the fuel used on the internal combustion engine (ICE) will help you determine the maximum amount of energy which may be extracted from the engine and then from that, the engine's efficiency.\n\nThe maximum amount of energy which may be extracted out of a chemical reaction (the max work the reaction can do) can be calculated by finding the reactions change in the Gibbs free energy under the conditions which the reaction is under.\n\nThe Change in the Gibbs Free Energy (delta G) can be calculated as,\n(delta G) = (delta H) - T*(delta S)\nWhere (delta H) is the change in the enthalpy of the reaction, T is the temperature of the reaction in Kelvin, and (delta S) is the change in entropy of the reaction.\n\nIf the value of (delta G) comes out to be <0 (i.e negative), then the reaction is spontaneous under those conditions and work can be extracted up to the magnitude of the change in Gibbs free enery.\nIf the value of (delta G) is greater than zero (positive) then the reaction is not spontaneous and requires that much energy to be put INTO the system in order for the reaction to proceed under those conditions.  In this case, no work can be extracted, work actually needs to be spent in order to get the reaction to occur.
744648
Use Gibbs free energy equation\n\nDeltaG=DeltaH - TDeltaS\n\nYou want Delta G to be negative so if Delta H positive then you need a greater TDeltaS \nDelta S is -25 so in order for the reaction to be spontaneous the T has to be less than zero
344369
If it were not negative it would not take place.  That is the point of delta G - it is a measure of free energy based on a combination of the heat released (delta H) and the entropy (randomness) - delta S.  If delta G is not negative for a given process under a given set of conditions, then that process will not occur.  If the process does occur, then delta G is negative.  Now - be careful - just because delta G is negative does not mean a process will neccesarily occur - or at least occur at an appreciable rate - in amny cases you need activation energy to "get the process over the hump" to make things happen.  Example - matches - they ignite - that is a negative delta G process - but it doesn't happen spontaneously - at least not at room temperature - you need to activate the process by heat or friction.
804468
You need to look up the entropy for all compounds, multiply by the entropy by the stochiometric coefficients (e.g. 4 for NH3, 3 for O2). Then subtract the left side from the right side.\n\nVery simple actually, once you find the data.
735334
a = (v_f - v_i)/t\n-0.25g = (0-v)/t\n0.25g = v/t\nt = 4v/g\n\nd = v*t + 0.5*a*t^2\nd = v*4v/g + 0.5*(-0.25g)*(4v/g)^2\nd = (4v^2)/g - 0.125 *g* (16 v^2 )/g^2\nd = (4v^2)/g - (2v^2)/g\n\nd = (2*v^2)/g
1142668
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