On my chem homework, a diagram called potention energy diagram is given. You need to be able to tell the potential energy of the products and reactants and the activation enrgy, the heat of reaction, and what would happen if a catalyst were added.
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Okay, cool!\nThe diagram is set up so that on the horizontal x-axis is the "reaction coordinate," which is basically just time as the reaction proceeds.  The vertical y-axis is energy.  \n\nThe potential energy of the reactants is where the graph curve starts, touching the y-axis.  The potential energy value is the y-value at this location.  \n\nWhere the graph curve levels off at the end is the products, so the potential energy of the products is the y-value where the curve levels off.  \n\nThe activation energy is how much energy had to be added into the system to get to the activated complex, located at the peak of the diagram's curve.  Take the y-value at the top of the curve and subtract from the the potential energy of the reactants.  \n\nThe heat of reaction is the difference in the potential energies of the products and reactants.  Take the potential energy of the products, and subtract from that the energy of the reactants.  \n\nIf a catalyst were added, the potential energy of the products and reactants remain unchanged, but the activation energy is less, so you draw a curve that starts in the same place and ends in the same place as the original diagram, but your curve just won't go up as high.  The hill for a catalyst is less steep.
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go to\nwww.chemistry.com
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Try these links:\n\nhttp://chemweb.calpoly.edu/chem/124/124Experiments/Thermochem/HessExp.html\n\nhttp://services.juniata.edu/ScienceInMotion/chem/labs/ap/hess.doc
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You will have to first find the standard enthalpy values of the reactants and products. They're given in the appendices of any standard GenChem or PhysChem books.\n\nThen, you subtract the sum of enthalpies of reactants from the sum of enthalpies of the product to get the enthalpy of the reaction.
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Kinetic, potential, nuclear, chemical and electrical.
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You may need to refer to Maxwell's distribution of molecular energies for this. Every reaction has its own specified activation energy which is the maximum amount of energy with which reacting molecules must collide to overcome the Ea barrier to form products. For some reactions, the Ea is very high and the molecules, in standard conditions, don't have enough internal energy needed to overcome that barrier - that's why they require a stimulus like the presence of light, to provide them with more internal energy.\n\nA catalyst acts in the same way, but unlike light which provides energy, it simply lowers the Ea for the reaction. The result is that the internal energy which the reactants possess is enough to get them through the barrier when they collide. Think of it as hurdles in a race being shortened down so that it's easier for the athletes to jump. With a lower Ea, more number of reacting molecules are able to cross the barrier per unit time, whereas if the Ea was not lowered, only a few molecules might gain the high Ea required and cross it, while the rest won't reach the transition state.\n\nBut the alternative transition state only has a lower Ea when you introduce a catalyst. In the presence of light, Ea will be the same, but the reacting molecules will get extra energy.
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You're forgetting something with your reasoning here.  It is true that in chemistry, creating bonds releases energy, but how do you get particles to create bonds between?  You have to break other bonds, which requires energy.  If catabolic pathways release energy when they break down complex molecules, the amount of energy it takes to break those bonds is less than the amount of energy released when new bonds are formed from the broken-up parts.
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For the forward reaction (from left to right), there is a decrease in enthalpy, which indicates it will tend to proceed as written, to products.\nThe entropy increases from left to right (products are gaseous), and this also favours the forward reaction to proceed to products.
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Calorimeter Info center\n\nhttp://www.scipeeps.com/Sci-Chemistry_Topics_C_-_Cl/Calorimeter.html
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The potential energy of a system is the energy which was spent to constitute the system from bringing its constituents closer\nto each other against the forces among them and this energy is stored in the syetem. The same way the interatomic energy is \nthe enrgy which was spent to bring the atoms closer against the vanderwaal forces.
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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
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Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
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The activation energy of a reaction is different for every reaction and is totaly independant of wether the reaction is exothermic or endothermic. For any reaction to proceed the minimum energy must be equal to or higher than the activation energy.\n\nNow where your question can come in is, i think, is if the reaction is reversible. In one direction the reaction will be exothermic and in the opposite direction the reaction will be endothermic. For both the forward and backward reactions the activation energy is the same and there is nothing that can be done to change that.
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It's:\n\nSum of bond enthalpies of product - Sum of bond enthalpies of reactants
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Maybe this is of any help?\n\nhttp://www.chemistry.nmsu.edu/studntres/chem435/Lab7/intro.html
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Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
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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.