What is the correlation between the magnitude of ∆Gº and the speed of the reaction?
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.
883310
At temperatures above 220 K the rxn would be spontaneous. To determine spontaneity, first determine the delta G value by using the formula deltaG=deltaH-Tdelta S. Now find out what temperature value would make the delta G negative which would indicate a spontaneous rxn, and in this case it is spontaneous above 220 K.
868428
ΔP.E. = Δmgh\nwith m and g constant,\nΔP.E. = mgΔh\n\n(a)\nE(A) = 50*10*9.8 = 4900 J\nE(B) = 0\nΔE = 4900 J\n\n(b) and (c) have similar solutions.
729117
You don't give the rate law, so it's impossible to tell. Suppose [H2O2] = concentration of H2O2 and [I-] = concentration of iodide. If rate = k[H2O2][I-], then the mechanism is one step. This means that both an H2O2 and an I- have to come together at the same time for the reaction to occur. If rate = k[H2O2], then it's a two-step process. This is because something has to happen to H2O2 in a slow, rate-limiting step, followed by the reaction of whatever with I-.
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
882396
ummmm, maybe increase by a factor of 5 or so. Its an exothermic reaction to begin with and you're not limited by the amount of water( i think, if not not all your drano will dissolve) so it should be about five times the energy released compared to just two grams.
818036
For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
955628
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.
444111
Rate of reaction depends upon the concentration of reactants. Since, 6M HCl has a greater conc. than 3M HCl, rate of reaction decreases when Mg reacts with 3M HCl
533621
(delta)H(rxn) = sum of heat of formation of products - sum of heat of formation of reactants.\n\na.  Hf Br2(g) = 30.91 kJ/mol, Hf Br2(l) = 0, so (delta)H = -30.91 kJ/mol\n\nb.  Hf CaCO3(s) = -1207 kJ/mol, Hf CaO(s) = -635.5 kJ/mol, Hf CO2(g) = -393.5 kJ/mol, so (delta)H = -635.5 - 393.5 + 1207 = +178 kJ/mol\n\nc.  Hf NO(g) = 90.25 kJ/mol, Hf O2(g) = 0, Hf NO2(g) = 33.2 kJ/mol, so (delta)H = 2 * 33.2 - 2 * 90.25 = -114.1 kJ/mol
947088
Pressure a higher pressure will push the molecules closer to each other so the reaction takes place faster cause the molecules colide more often.\n\nAmounts of the chemicals, will take longer as you increase each to fully react because there is more to react.\n\nConcentrations of the liquids a higher concentration will have more of the molecules to react.\n\nAnd the amount of activation energy if the amount is low the reaction will procede at a faster rate.
390546
I'll give you a hint.\n\nDelta G = Delta H - T Delta S\n\nUsing that formula, you can get 1\n\n(B) you can look up in any physical chemistry text:\n\nG(2) - G(1) = a formula I don't have available
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
896452
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.
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
746288
On these equations D is for Delta :\n\nDh = De +pDV (definition of enthalpy)\npDV = R DT (equation of state)\nDh = Cp DT  (definition of enthalpy)\nWith first law of thermodinamics :\nDe = Dq +Dw = Dq ) Cv DT\n\nSo Cp DT = Cv DT + RDT\n\nSo Cp = Cv +R  so Cp always greater than Cv \n\n :
827524
Easier compared to one with a higher heat capacity:\n\nH/(m*Cp) = (delta T), where m is the mass of the object, H is the heat that must be added (or removed) to raise (or lower) its temperature, delta T is its change in temperature, and Cp is its heat capacity. From the above formula, you can see that larger values of Cp result in small temp. changes, for the same heat amount.