Can someone explain kinetics to me?
814214
In physical chemistry, chemical kinetics or reaction kinetics study reaction rates in a chemical reaction. Analysing the influence of different reaction conditions on the reaction rate gives information about the reaction mechanism and the transition state of a chemical reaction. In 1864, Peter Waage pioneered the development of chemical kinetics by formulating the law of mass action, which states that the speed of a chemical reaction is proportional to the quantity of the reacting substances.\n\nKinetics deal with the experimental determination of reaction rates from which a rate law and reaction rate constant are derived. Essential rate laws exist for zero order reactions (for which reaction rates are independent of initial concentration), first order reactions, and second order reactions, and can be derived for others through calculus. In consecutive reactions the rate-determining step often determines the kinetics. In consecutive first order reactions, a steady state approximation can simplify the rate law. The activation energy for a reaction is experimentally determined through the Arrhenius equation and the Eyring equation. The main factors that influence the reaction rate include: the physical state of the reactants, the concentrations of the reactants, the temperature at which the reaction occurs, and whether or not any catalysts are present in the reaction.\n\n\n[edit] Factors that influence the rate of a reaction\n \nFig 1.1 - Collisions\nTop - Collisions before increased concentration.\nBottom - Increased collisions after the concentration is increased.\n[edit] Physical State\nThe physical state (solid, liquid or gas) of a reactant is also an important factor of the rate of change. When reactants are in the same phase, as in aqueous solution, thermal motion brings them into contact. However, when they are in different phases, the reaction is limited to the interface between the reactants. Reaction can only occur at their area of contact, in the case of a liquid and a gas, at the surface of the liquid. Vigorous shaking and stirring may be needed to bring the reaction to completion. This means that the more finely divided a solid or liquid reactant, the greater its surface area per unit volume, and the more contact it makes with the other reactant, thus the faster the reaction. To make an analogy, for example when you start a fire, first you put wood chips and small branches, you don't start with big logs right away. In organic chemistry On water reactions are the exception to the rule that homogeneous reactions take place faster than heterogeneous reactions.\n\n\n[edit] Concentration\nConcentration plays a very important role in reactions. According to the collision theory of chemical reactions, this is due to the fact that molecules must collide in order to react together. As the concentration of the reactants increases, the frequency of the molecules colliding increases, striking each other faster by being in closer contact at any given point in time. Imagine two reactants being in a closed container. All the molecules contained within are colliding constantly. By increasing the amount of one or more of the reactants you cause these collisions to happen more often, increasing the reaction rate (Figure 1.1).\n\n\n[edit] Temperature\nTemperature usually has a major effect on the speed of a reaction. Molecule at a higher temperature have more thermal energy. When reactants (reactant + reactant → product) in a chemical reaction are heated, the more energetic atoms or molecules have a greater probabilty to collide with one another. Thus, more collisions occur at a higher temperature, making a product in a chemical reaction. More importantly however, is the fact that at higher temperatures molecules have more vibrational energy, that is, atoms are vibrating much more violently, so raising the temperature not only increases the number of collisions but also collisions that can result in rearrangement of atoms withi
817765
kinetic just means movement. So it is like how molecules vibrate. The more they vibrate the further apart they get and more energy they have. So a solid can have molecules vibrating slower but add more heat it can turn to a liquid.
92261
1.  Heat speeds up a reaction by increasing vibrations and energy of the reactants.\n\n2.  Concentration speeds us a reaction by making more reactants available in a unit volume.\n\n3.  Catylists speed up a reaction (without being used up) by allowing reactants to come into closer proximity.\n\n4.  Increased surface area (by grinding, etc.) speeds up reactions by making more reactants exposed to reaction.
730275
It's called 'impulse' and any good, lower division, physics book will have an explanation of what it is and how it works.\n\n\nDoug
149491
This is a "rule of thumb" that is the average of a lot of kinetic data.  To blindly apply it is unwise, but as a zero order estimate, it should work OK.  Detailed measurements must be done on your system of interest to find out true kinetic parameters and their sensitivity.
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Well, the kinetics are actually pretty complicated. \n\nAll of the catalysts I'm aware of for this type of reaction are heterogeneous. This complicates things for several reasons. First, the rate is going to be much more dependent on reactor design since how well the catalyst is suspended and dispersed plays a big role in the rate of reaction. Second, the rate will depend on the active surface area of the catalyst, which can vary from batch to batch of catalyst and must be determined empirically for each sample.  \n\nAnother tricky issue is that in the course of the reaction you are going from a liquid (oil) to solid (fat). This will change the physical properties of your mixture, which changes the way the catalysts is distributed and will certainly change the kinetics. Solution viscosity could potentially become an issue, especially on the industrial scale. It really just depends on how you run the reaction.\n\nIn terms of amount of H2 needed, it's going to depend on precisely what kind of oil you are starting with. Most vegetable oils are mixtures of several oils made of the triacylgylcerides of unsaturated fatty acids such as oleic acid and linoleic acid. Since the degree of unsaturation varies from acid to acid (and therefore oil to oil), it depends entirely on which specific oil you have.\n\nIn terms of catalyst, in principle it's always regenerated and never lost.\n\nI wish I could give you more specifics, but the kinetics here are really messy and not easily amenable to a traditional analysis.
677540
A lot of physics (especially beginning physics) is just finding the right equation and plugging in what numbers you are given.\nEquations usefull for your problem:\nKE = 1/2*m*v^2 (KE = kinetic energy, m = mass, v = velocity)\nAND\nWork = Force * distance     This is a very important equation\nit's easy to memorize so do it!
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
757057
Conservation of momentum\nP = mv before must equal mv after\nmass remains the same\nvelocity is a vecor quantity (magnitude and direction)
845875
Ah, time to dig out the physics from last year.  If I remember correctly:\n\nStatic friction is when the force of friction is greater than the force applied; thus basically what amount of force it takes to move the object from rest.\n\nKinetic friction is when you apply a force for an object to keep sliding.  In this, it "breaks free" of your push and continues sliding.\n\nAnd the force needed to initially break the static friction (to get it moving) is much more than what is needed to keep it going.\n\nHope this helped a bit.  Just remember that static friction is associated with objects at rest and kinetic is objects in motion, and don't get the two mixed up. :*P\n\nHave fun with physics!
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Kinetic energy Ek\n\nlet's consider that an object with a mass m moves on a distance d = x1 - x2, with an acceleration a and its velocity changes from v1 to v1, under the action of a force F\n\naccording to Galileo's law:\nv2^2 - v1^2 = 2.a.d\n\nbut according to Newton's second law:\nF = m.a\n\n-> a = F/m \n-> v2^2 - v1^2 = 2.F.d/m\n-> m.v2^2 - m.v1^2 = 2.F.d\n-> m.v2^2/2 - m.v1^2/2 = F.d\n\nbut W = F.d = the work done while the force F moves the object on the distance d, it caracterizes a process\n\nBut energy is the capacity of an object of doing work, it caracterizes a state\n\nthe kinetic energy is due to the velocity of the object:\n\ndeltaEk = E2 - E1 = W = F.d \n-> E2 = m.v2^2/2\n-> E1 = m.v1^2/2\n\nPotential energy Ep\n\nLet's consider an object with a mass m is falling from a height h1 to h2. \n\nThe work done by gravity is:\n\nW = G.(h1-h2) = m.g.(h1-h2) = m.g.h1 - m.g.h2\n\nThe potential energy is depending on the position of the the object as compared to the Earth and its equal with the work done by gravity taken with opposite sign.\n\ndelta Ep = Ep2 - Ep1 = - W\n\n-> Ep2 = m.g.h2\n-> Ep1 = m.g.h1
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Force times distance is not kinetic energy. Force times distance is work. Work can appear in different forms. If you expend 100 Newton-meters of work pushing a dog very slowly up a frictionless inclined plane, the dog will have 100 Newton-meters of potential energy when you are done, and no kinetic energy, for it will stop moving when you stop pushing.\n\nOn the other hand, if you expend 100 Newton-meters of work pushing the dog along a horizontal frictionless slippy-slide, the dog will continue sliding after you stop. It will have no potential energy and 100 Newton-meters of kinetic energy.\n\nDo you want to know how fast the dog will be going when you stop pushing? Use\n\nE = (1/2) * m *  v^2
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.
837367
Kinetic Energy\nKinetic energy is the energy of motion. An object which has motion - whether it be vertical or horizontal motion - has kinetic energy. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another). To keep matters simple, we will focus upon translational kinetic energy. The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) which an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.\n                     \nke=1/2*m*v^2\n\nwhere m = mass of object\n\nv = speed of object
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http://www.chem.purdue.edu/courses/chm333/enzyme_kinetics.swf\nnow be a good boy, study the link, and then explain it to  me....  :)
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http://en.wikipedia.org/wiki/Kinetic_energy
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Conserve both momentum and total energy (i.e., potential energy plus kinetic energy).\n\nMy coordinate system will have the pendulum lifted to the left and the block on the right moving left, so that the initial momentum is - M2 Vo2.\n\nIt will also have the block at height 0, so the initial potential is M1 g h where h is the initial height of the pendulum, which looks like L(1 - sin(10 degrees)).\n\nThe initial kinetic energy is of course (1/2) M2 (Vo2)^2.\n\nSo ...\n\ninitial momentum: - M2 Vo2\ninitial energy: M1 g h + (1/2) M2 (Vo2)^2\n\nAfter the collision the potential energy has decreased to zero, and if w is the combined velocity of the two masses then ...\n\nfinal momentum: (M1 + M2)w\nfinal energy: (1/2)(M1 + M2)w^2 + [the energy "loss" from the slight heating and deformation of the two masses sticking].