The normal freezing point of n-octane 1C8H182 is -57 °C. (b) In w...

Question

The normal freezing point of n-octane 1C8H182 is -57 °C. (b) In what temperature range is the freezing of n-octane a spontaneous process?

Accepted Answer

Welcome back everyone. We have an organic compound and obtain which has a normal freezing point of negative 950.55 or sorry, negative 90.55°C. We need to identify the range of temperature such that the freezing of pain occurs spontaneously. So we want to recall our normal freezing point condition Where our pressure should equal 1 80 m for our normal freezing point. And so the next thing we want to recall is our expression where our free energy change is set equal to our entropy change subtracted from temperature multiplied by our entropy change. So we want to also recognize that at our freezing point Our change in free energy is equal to zero, meaning that according to the prompt, our normal freezing point being negative 90.55°C means our system is an equilibrium and so for and obtain to freeze spontaneously we want to recall that this occurs. We can say this will occur at temperatures below -90.55°C. And this statement would be our final answer. To complete this example. I hope that everything I went through is clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

The normal freezing point of n-octane 1C8H182 is -57 °C. (b) In what temperature range is the freezing of n-octane a spontaneous process?

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