Gibbs Free Energy: Equilibrium Constant?
816266
At constant temperature and pressure, a reaction will occur spontaneously if the Gibbs free energy change of the reaction is less than zero.\n\nThe Gibbs free energy change, delta-G, is given by:\n\ndelta-G = delta-H - T*delta-S\n\nwhere T is the absolute temperature, \n\ndelta-H is the enthalpy change for the reaction at the temperature and pressure of interest, and \n\ndelta-S is the entropy change for the reaction at the temperature and pressure of interest.\n\nWe are given the enthalpy and entropy changes for the reaction of interest in this question.  I will assume that the enthalpy change and entropy change for this reaction are independent of temperature. Note also that the values of both of these thermodynamic parameters are given in "per mole" units in this question.  I will assume these values are normalized to one mole of one of the reactants, and not one mole of the product.\n\nPlugging in the enthalpy and entropy values, and setting delta-G to be less than zero:\n\n0 > delta-G = 180500 J/mol - T*24.8 J/(mol*K)\n\nT  > (180500 J/mol)/(24.8 J/(mol*K))\n\nT > 7278 K\n\nThis is an extremely high temperature, and the assumption that we made above that the enthalpy and entropy changes were independent of temperature is probably not very good given the large temperature extrapolation being made here.
922031
The reaction for the Fritz-Haber-Bosch process is:\n\n3H2(g) + N2(g) <-> 2NH3(g)\n\nThe equilibrium constant for this reaction can be written as:\n\nKp = ((p_NH3)^2)/((p_N2)*(p_H2)^3) = 7.16*10^-5 atm^-2,\n\nwhere p_xxx stand for the partial pressure of the gas "xxx" at equilibrium.\n\nWe are told that initially (before any reaction) the partial pressures of H2 and N2 gases are 3.7 atm and 5.4 atm, respectively.  Because each mole of NH3 produced consumes 1/2 mole of N2 and 3/2 mole of H2, we know from mass balance that at equilibrium,\n\np_N2 = 5.4 atm - 0.5*p_NH3\n\nand\n\np_H2 = 3.7 atm - 1.5*p_NH3\n\nPlugging these expressions into the equilibrium constant, we have that:\n\n7.16*10^-5 atm^-2 = ((p_NH3)^2)/[(5.4 atm - 0.5*p_NH3)*(3.7 atm - 1.5*p_NH3)^3]\n\n\nlet p_NH3 = x, so the algebra is easier to write out.\n\n7.16*10^-5 atm^-2 = (x^2)/(273.53 atm^4 - x*357.99 atm^3 + x^2*165.67 atm^2 - x^3*30.712atm + x^4*1.688)\n\n\n(7.16*10^-5 atm^-2)/x^2 = 1/(273.53 atm^4 - x*357.99 atm^3 + x^2*165.67 atm^2 - x^3*30.712atm + x^4*1.688)\n\nx^2/(7.16*10^-5 atm^-2) = 273.53 atm^4 - x*357.99 atm^3 + x^2*165.67 atm^2 - x^3*30.712atm + x^4*1.688\n\nx^2*1.397^4 atm^2 = 273.53 atm^4 - x*357.99 atm^3 + x^2*165.67 atm^2 - x^3*30.712atm + x^4*1.688\n\n273.53 atm^4 - x*357.99 atm^3 - x^2*1.38*10^4 atm^2 - x^3*30.712atm + x^4*1.688 = 0\n\nThis is an ugly 4th order polynomial in x.  The most direcet way of solving for the roots is to simply plot it, and look for the zeros of the function.  I'll admit to using a numerical solver to find that the roots are at:\n\nx = -81.762 atm, -0.1548 atm 0.1284 atm, and 99.98 atm.\n\nOnly the positive roots have a physical meaning, and the root at 99.98 atm is not possible because the starting pressures of the reactants are not high enough to generate a partial pressure of the product that is this high.  We have, then, that x = p_NH3 = 0.1284 atm.\n\nWe can check that this is a correct answer by calculating the equilibrium partial pressures of N2 and H2 and plugging all the values back into the equilibrium constant:\n\nK = ((0.1284 atm)^2)/((5.4 atm - 0.5*0.1284 atm)*(3.7 atm - 1.5*0.1284 atm)^3)\n\nK = 7.16*10^-5 Check!\n\n--------------\n\nThe previous answerer had the right idea, but has the wrong expression for the equilbrium constant because (s)he wrote the chemical reaction so that one mole of NH3 is produced (i.e., the reaction I wrote divided by two).  The equilbrium constant for that reaction would have units of atm^-1, not atm^-2, as we are given in the question.  Had the previous answerer used the square root of the equilibrium constant given in the question, then his/her answer would have been correct.
1138290
Look up stuff like Arrhenius equations. I don't have my phys chem text book to hand.\n\nFor part two, I think it's D, as I'm fairly certain it's not any of the others.
1063841
W = PV = RTn, where P = pressure, V = volume, R = gas constant = 8.31 J/mol.K, T = temperature, n = moles of gas involved.\n\n1) Change in P = 4 x 10^5 Pa, then:\n\nChange in W = 4 x 10^5 Pa x 0.250 m^3 = 10^5 Pa.m^3 = 1 x 10^5 Joules.\n\n2) Change in V = 17.00 x 10^(-2), then:\n\nChange in W = 6 x 10^5 Pa x 17.00 x 10^(-2) m^3 = 102.0 x 10^3 Pa.m^3 = 1 x 10^5 Joules (note: there is only one significant number in result)
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
398060
As the previous answerer stated, there is, in general, no relationship between the standard Gibbs free energy change of a reaction, ∆Gº, and the reaction rate.  The reason is that most reactions involve an energy "barrier" between the reactants and products that must be overcome in order for the reaction to proceed.  The higher this barrier, the less likely it is that statistical fluctuations in the energy of the reactant molecules will enable them to surmount the barrier and the slower the rate.  Conversely, a low energy barrier leads to a fast reaction rate.  The free energy change for the overall reaction simply tells you what the difference in energy is between the products and reactants, and not how high an energy barrier separates them.\n\nThese concepts are embodied in a theory of kinetics called "transition state theory", in which a reaction is described by a continuous transition from the unreacted reactants to the fully reacted products.  Every state in this transition process has a free energy associated with it, and a reaction follows a lowest energy path from reactants to products.  The state of the system that has the highest free energy along this path is called the "transition state" (or "activated complex").  This state corresponds to the "energy barrier" described in the previous paragraph.\n\nIf one assumes that there can be a thermodynamic equilibrium between the reactants and the transition state , and that the rate of a chemical reaction is proportional to the concentration of species in the transition state, then one can show that the log of the reaction rate is proportional to Gibbs free energy difference between the transition state and the reactants, ∆G*.  This relationship is given by the Eyring equation:\n\nrate ~ (kT/h) exp (–∆G*/RT)\n\nwhere\n\nk is Boltzmann's constant\nT is the absolute temperature\nh is Plank's constant\nR is the gas constant\n∆G* is the Gibbs free energy difference between the activated complex and the reactants\n\nThus, there *is* a relationship between Gibbs free energy and reaction rates, but the free energy involved is not the free energy change of the overall reaction, but instead, the free energy change associated with going from reactants to the species present at the highest energy point on the reaction path.\n\nSee sources for more details.
136538
Bolztmann constant(k) = 1.380 6505(24) × 10−23 joules/kelvin\nRead more about it on the following link.\nhttp://en.wikipedia.org/wiki/Boltzmann_constant
708589
A system is in mechanical equilibrium when the sum of the forces, and torque, on each particle of the system is zero. \n\nA system is in dynamic equilibrium  when two reversible processes occur at the same rate. Many processes (such as chemical reactions) are reversible.\n\nIn thermodynamics, a thermodynamic system is said to be in thermodynamic equilibrium when it is in thermal equilibrium, mechanical equilibrium, and chemical equilibrium. The local state of a system at thermodynamic equilibrium is determined by the values of its intensive parameters, as pressure, temperature, etc. Specifically, thermodynamic equilibrium is characterized by the minimum of a thermodynamic potential, such as the Helmholtz free energy, i.e. systems at constant temperature and volume:\n\nA = U – TS \nOr as the Gibbs free energy, i.e. systems at constant pressure and temperature:\n\nG = H – TS \nThe process that leads to a thermodynamic equilibrium is called thermalization. An example of this is a system of interacting particles that is left undisturbed by outside influences. By interacting, they will share energy/momentum among themselves and reach a state where the global statistics are unchanging in time.\n\n\n\nChemical equilibrium is the state in which the concentrations of the reactants and products have no net change over time. Usually, this state results when the forward chemical reactions proceed at the same rate as their reverse reactions. The rates of the forward and reverse reactions are generally not zero but, being equal, there are no net changes in any of the reactant or product concentrations. This process is known as dynamic equilibrium (Atkins & Jones, 2001). Fundamentally, reactions do not go fully to completion, but reach an equilibrium, because of the entropy of mixing associated with reaction.
1006740
The equilibrium constant for this reaction is given by:\n\n[NO]^2/([N2]*[O2]) = 6.91*10^-4\n\nYou also know, from mass balance, that:\n\n[NO] + 2*[N2] = 2*0.75 mol/L \nand\n[NO] + 2*[O2] = 2*2.4 mol/L\n\n (i.e., the total concentration of nitrogen and oxygen atoms doesn't change in this system -- the number of nitrogen atoms at equilibrium after the gasses react must be the same as the number of atoms you started with)\n\nYou now have three equations in three unknowns ([NO], [O2], and [N2]).  These three unknowns are the concentrations of the gases at equilibrium.  \n\nTo solve these equations, we can rewrite the mass-balance equations as:\n\n[N2] = 0.75 - 0.5*[NO]\n[O2] = 2.4 - 0.5*[NO]\n\nsubstitute these into the equilbrium constant equation to eliminate [N2] and [O2]:\n\n[NO]^2/((0.75 - 0.5*[NO])*(2.4 - 0.5*[NO])) = 6.91*10^-4\n\n[NO]^2/{1.8-1.575*[NO]+0.25*[NO]^2} = 6.91*10^-4\n\n[NO]^2 = 0.0012438 -0.001088325*[NO] +0.00017275*[NO]^2\n\n0 = 0.0012438 - 0.001088325*[NO] - 0.99982725*[NO]^2\n\nUse the quadratic formula to solve for [NO].  In general, there will be two roots, but only one will be physically meaningful.\n\n[NO] = -0.03582 or 0.03473.  Obviously, only the positive root has a physical meaning, so at equilibrium [NO] = 0.03473 mol/L\n\nSubstitute this into the mass-balance equations to get the concentrations of [O2] and [N2]:\n\n[N2] = 0.75 - 0.5*0.03474 = 0.73263 mol/L\n[O2] = 2.4 -  0.5*0.03474 = 2.38263 mol/L\n\nYou can check these results by plugging them back into the equilibrium constant equation to make sure everything agrees:\n\n0.03473^2/[0.73263*2.38263] ?= 6.91*10^4\n\n6.91*10^-4 = 6.91*10^-4  Check!
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.
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
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
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
701698
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.
723769
In thermodynamics, the Gibbs free energy is a thermodynamic potential which measures the "useful" work obtainable from a closed thermodynamic system at a constant temperature and pressure. Under these conditions, it is minimized at equilibrium.\n\nThe Gibbs free energy was developed in the 1870s by the American mathematical physicist Willard Gibbs. The Gibbs free energy is defined as:\n\n \nwhere (in SI units)\n\nG is the Gibbs energy (joules) \nH is the enthalpy (joules) \nT is the temperature (kelvins) \nS is the entropy (joules per kelvin)
857221
No. The equilibrium constant is thermodynamic. Reaction rate is kinetic. The conversion of diamond to graphite at 1 atmosphere has a high equilibrium constant. Do you see diamonds turning rapidly to graphite?