![]() The microscopic level, we see that the system is So from a macroscopic point of view, nothing seems to change. A good way to think aboutĪ microstate would be like taking a picture of Microscopic arrangement of positions and energies So going back to our boxes,īox 1, box 2 and box 3, each box shows a different To the kinetic energies of the particles. With an ideal gas here, by energies, we're referring Microscopic arrangement of all of the positions andĮnergies of the gas particles. ![]() Of each particle is equal to 1/2 mv squared, where m is the mass of each Particles are meant to represent the velocities of the particles. And the magnitude and theĭirection give a velocity. However, when we put anĪrrow on each particle, that also gives us the direction. Of a particle tells us how fast the particle is traveling. Slightly different positions and the velocities might have changed. Particles in our system at one moment in time, in box 1, if we think about them atĪ different moment in time, in box 2, the particles might be in Slamming into each other and transferring energy from ![]() Slamming into the sides of the container and maybe Here in the first box, imagine these gas particles However, from a microscopic point of view, things are changing all of the time. So from a macroscopic point of view, nothing seems to be changing. Particles is at equilibrium, then the pressure, the volume, the number of moles, and the temperature all remain the same. ![]() Moles at a specific pressure, volume, and temperature. And to think about microstates, let's consider one mole of an ideal gas. Where this ‘per mole’ means ‘per the number of moles specified in the reaction equation’.Of entropy is related to the idea of microstates. The units we use for the entropy change of a reaction are joules per kelvin per mole. So if the entropy change for this equation was this, That means that when 2 moles of aluminium react with 1.5 moles of oxygen to produce 1 mole of aluminium oxide, the entropy change is -313.2 joules per kelvin per mole. It really means ‘per the number of moles specified in the reaction equation’ Well, confusingly, this doesn’t mean ‘the entropy change per every one mole’ So this raises the question – what on earth is this doing in the units of our answer? Well, the answer we worked out is actually the entropy change when exactly:Ĥ moles of ammonia react with 5 moles of oxygen Or the entropy change when one mole of products form Nor does it represent the entropy change when a total of 1 mole of reactants react Well, it doesn’t represent the entropy change when one mole of ammonia reactsĪnd it doesn’t represent the entropy change when one mole of oxygen reacts That’s great, but, what does this answer actually represent? We used our equation, did all the calculations and got this answer So, suppose we were asked to find the entropy change for this reaction. Last time we saw how to calculate the entropy change of a reactionĪnd now that we know how to calculate entropy changes, we’re a big step closer to figuring out why some reactions happen and others don’t.īut before we move onto that, there are a few loose ends we have to tie up… ![]() How Are Reversible Reactions Compatible With the Second Law of Thermodynamics? Calculating Gibbs Free Energy Change for Reverse ReactionsĢ4. The Limitations of Our Temperature-Finding EquationĢ1. Finding the Temperature Where Reactions Become FeasibleĢ0. Using Graphs to Find Enthalpy and Entropy Changesġ9. Exam Technique: Explaining Feasibilityġ7. Assessing Feasibility – Thermal Decomposition of Calcium Carbonateġ5. Why Do Some Feasible Reactions Not Happen?ġ0. Calculating the Entropy Change of the Surroundingsħ. ![]()
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