![]() Gasses are even higher entropy because the molecules no longer have any restrictions on their translational or rotational DOF from other molecules. Since atoms/molecules in a liquid are interacting with one another and tightly packed, there are restrictions on their translational DOF and rotational DOF (both entire molecule rotations and bond rotations). Liquids are higher entropy than solids because they gain back some or all (material dependent) rotational DOF and some (not all) translational DOF. This is why a crystalline solid of a pure material has zero entropy. Now why is gas phase higher entropy than liquid phase and why is liquid phase higher entropy than solid phase? For solid phase, you lose (nearly) all translational DOF and depending on the material, you likely lose some or all of your rotational DOF too. Thus, more smaller molecules will ALWAYS be more entropically favorable. However, that extra DOF comes at the cost of 6 DOF because each of the isolated atoms had 6 DOF (3 trans & 3 rot) so you have a net loss of 5 DOF. An isolated diatomic molecule still only gets 3 DOF for translation movement and 3 DOF for rotating the entire molecule, but it also gets one extra DOF for rotating the bond between the atoms. gas) atom gets 3 DOF for translation (moving around in X, Y, & Z axis) and 3 DOF for rotation (spinning about X, Y, & Z axis) plus others that don't matter here. The sum of the number of accessible DOF is a measure of the entropy of a system.Īn isolated (e.g. For a more quantitative answer, we need to consider degrees of freedom (DOF) which are the number of independent variables necessary to full describe a system at equilibrium. So since D is only a slight increase in complexity (a single new bond) and it was in the gaseous phase, a decrease in the moles of gas is more damaging to the entropy than a minor increase in flexibility.Įveryone else's answers were really qualitative. Since a gas is already highly entropic, you'd need a huge increase in complexity for it to be worth losing its ability to bounce around feely for the sake of gaining some flexibility. Where as in a solid or liquid form an increase in complexity might give additional degrees of freedom to a simple lattice of small atoms, since they're probably not moving around as much to begin with, adding in new ways for them to be flexible will help them move more. The bigger and chunkier they get the harder it is to do that as a gas. ![]() They like to tumble and bounce all over the place. Gases are much faster and much further apart. The longer answer involves the fact that this increase in complexity is usually more involved with solid->solid or liquid->liquid transitions. So in the grand scheme of things they are not that more "complex." There's a greater restriction on the degrees of freedom by forming this bond than if the bond doesn't form, so this is is actually a decrease in entropy. In that case you're going from 2 moles of solid and one mole of gas to 2 moles of gas, can you see how this wouldn't be a decrease? Then by process of elimination D must be the answer.īut for a further explanation, "complexity" is more complicated, the short answer is that if you were to draw out the structures of NO2 and N2O4, you'll see that they only differ by a single bond. You've outlined well how all of the others are an increase in entropy except for B. ![]() A screenshot is preferable to a picture of your laptop screen. Please do not ask for help acquiring, preparing, or handling illicit substances or for help with any activity that does not fall within the confines of whatever laws apply to your particular location.īonus points: If submitting a picture please make sure that it is clear. Any infractions will be met with a temporary ban at the first instance and a permanent ban if there is another. It is also important that you describe the specific part of the problem you are struggling with. It is OK if you are a little (or a lot!) stuck, we just want to see that you have made an effort. ![]() Please complete any questions as much as you can before posting. We will not do your homework for you, so don't ask. Please flair yourself and read over the rules below before posting. ![]()
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