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Ch.11 - Liquids, Solids & Intermolecular Forces

Chapter 11, Problem 67

How much energy is released when 65.8 g of water freezes?

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Hey everyone, we're told that hydrogen peroxide freezes at a temperature just around that of water, calculate how much energy is released. 1 55 g of hydrogen peroxide freezes and they provided us our entropy of fusion which is 12.50 kg joules Permal. So as we've learned, positive entropy of fusion is going to be equal to the negative of our entropy of freezing. So in this case we have a positive entropy of fusion which is 12.50 kg jewels Permal. And for our entropy of freezing this will be negative 12. kg joules per mole. So let's go ahead and use this information into our entropy formula, which is going to be Q equals our mass times our entropy of freezing and Q is going to represent the energy released. So we have Q equals 55.0 g, which is our mass. But we first want to convert this into moles. So using our dimensional analysis, taking the molar mass of hydrogen peroxide, we know that we have 34. grams of hydrogen peroxide per one mole of hydrogen peroxide. Next we're going to multiply this by our entropy of freezing, which we said to be negative 12.50 kg joules per mole. So when we calculate this out and cancel out our units, We end up with a queue of negative 20.212266 killer jewels. So as we've learned a negative sign tells us that energy is released. And since the question asked us for the energy released when freezing hydrogen peroxide, this means that we need to omit the negative sign. And so our final answer is going to be 20.2 kg jewels. And since we also know that the freezing of hydrogen peroxide is an ex a thermic reaction, we know that energy is going to be released. So our final answer here is going to be 20.2 kg jewels. Now, if they asked us if the energy involved in the freezing of hydrogen peroxide, the answer would be negative 20.2 killer jewels. But since they only asked us for the energy release, we have to omit that negative sign. Now I hope this made sense and let us know if you have any questions.