This might be a simple concept for you guys but I'm bit confused. We are learning rates of reaction and why does a crushed tablet react quicker than a full tablet. Any diagrams to help me with this would also be great. And while we are at I got few questions on equilibirum constant. Let's say we have to write our own equation and let's say its a combustion reaction. We can choose if the product water is liquid or gas. But If choose liquid I won't be able to use it in the equillibirum constant K( normal one not the gas one). So which state do I use Why is temperature the only thing that changes th equilibirum constant? Any help in any of these questions would be greatly appreciated.Please Register or Log in to view the hidden image!Thank you!!

because in the case of a crushed tablet you have all the surface area of each of the bits where as stuck together you only have a sufface area of the outside of the tablet. Its like asking why a sponge has a greater surface area than a block of wood does

mathematical answer... one (perfect) balloon having one liter of water...(1000 cc's..) 1000 balloons having one cc of water... do the math and see which (combined)) surface area is the greatest.. (and nope..YOU do the math..I'm laxy and I already know..HAH..)

Others have dealt with the surface area issue. As I'm not sure what you mean with your question about phase of the water produced (maybe if you give an example it might be clearer?) I'll just for now answer the query about temperature dependence. Thermodynamically, ΔG, the change in Gibbs free energy of the process, = -RT ln K. But we also know ΔG = ΔH - TΔS. So you can view it as being that the balance point between the enthalpy change and the entropy change x T term alters with temperature, affecting K. Another way to think of it is via Le Chatelier's Principle. If the reaction is exothermic, i.e. gives off energy going from left to right (i.e. ΔH is -ve in this direction), then at higher temperatures it is more able to go "backwards" from right to left as this requires energy to be input. Think of it as: reactants <-> products + heat. If you boost the heat on the right it will tend to go more to the left, to oppose the applied change.

You guys do realise this thread is over 6 years old and the member only has 4 posts so I doubt he is still here