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mod 5 q (1 Viewer)

Masaken

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could someone pls explain what the prediction is and why, not sure what to do here
thanks in advance
 

carrotsss

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In a dynamic equilibrium system, even once equilibrium is reached, both the forward and reverse reactions still continue to occur.
PbCO3(s) <-> Pb2+(aq) + CO32-(aq)
As a result, some radioactive lead ions from the solid lead carbonate will react into Pb2+(aq), and some non-radioactive lead ions will react into PbCO3(s). Over the course of a few hours, these reactions will occur to a sufficient extent that the radioactive lead ions will be relatively evenly spread between the solution and the lump
 

Masaken

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In a dynamic equilibrium system, even once equilibrium is reached, both the forward and reverse reactions still continue to occur.
PbCO3(s) <-> Pb2+(aq) + CO32-(aq)
As a result, some radioactive lead ions from the solid lead carbonate will react into Pb2+(aq), and some non-radioactive lead ions will react into PbCO3(s). Over the course of a few hours, these reactions will occur to a sufficient extent that the radioactive lead ions will be relatively evenly spread between the solution and the lump
ur a legend, explained it better than the worked solutions for me 💀
thanks!
 

wizzkids

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I agree mostly with the account given by @carrotsss which is very well done. A good description of dynamic equilibrium.
However, there is one feature missing - the transport problem. The radioactive Pb2+ ions will leach out of the surface of the solid and be replaced with non-radioactive Pb2+ ions on the surface of the solid. On this part, we are in agreement. However the interior of the solid will be unchanged. There will be a slow build-up of radioactivity in the solution, but this build-up will slow down due to the transport problem in the solid. In order for radioactive Pb2+ ions to continue to reach the surface between the solid and the liquid, they have to diffuse through the solid phase to reach the surface, which is an extremely slow process compared to diffusion in the liquid phase. Hence we will get a negative exponential curve, and the distribution of radioactive Pb2+ ions will look something like this:
kinetics.png
You will encounter this phenomenon in Module 8 under Chemical Synthesis and Design - reaction conditions. You might like to think about what might be done by changing the reaction conditions to speed up the process of reaching equilibrium.
 

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