How much energy in kilojoules is required to convert 100 mL of wa...

Question

How much energy in kilojoules is required to convert 100 mL of water at its boiling point from liquid to vapor, and how does this compare with the result calculated in Problem 9.68 for diethyl ehter? [ΔHvap(H2O) = + 40.7 kJ/mol]

Accepted Answer

Hi everyone for this problem, we're told that the following reaction shows the formation of ammonia. So we're given a reaction and we're also given the change of entropy. Okay, so we're being asked to calculate the amount of heat required and kill a jules, 55.60 g of ammonia decomposes to end two and H two at constant pressure. So we want to calculate the amount of heat which is killer jewels. So we want to know Killer jewels given 55.6 g of ammonia. So this is our starting point. This 55.6 zero grams of ammonia is where we're going to start. Okay. And we're told that it's at constant pressure. Now, what at constant pressure means is that our heat is equal to R N therapy. That's what that means there. Okay, so we're going to go from grams of ammonia to heat required. And what we're going to want to do here is pay attention to this entropy that they give us here the -92 killer jewels. Because what this translates to is if you look at the amount of ammonia We have two moles of ammonia and we have for every two moles of ammonia negative 92 kg joules of heat. So this basically translates to this entropy translates to kill a jules per mole of ammonia, which is going to be important for us to use in our conversion. Alright, so the first thing that we're going to want to do is go from grams of ammonia two moles of ammonia and we do that using molar mass. So in one mole of ammonia we have 17.04 g of ammonia. And now we can use that conversion using that entropy that killer joules per mole of ammonia. So we can say that in two moles of ammonia, how many kilo joules of heat is produced? It's 92. So we have 92 kg joules of heat. Okay, so making sure our units cancel here are grams of ammonia, cancel, our moles of ammonia cancel. And we're left with kilo jewels of heat. Once we calculate this, we get a final answer of 150.1 kila jules. So this is going to be the amount of heat required in kila jewels. If 55.60 g of ammonia decomposes to end to an age to at constant pressure. That's the end of this problem. I hope this was helpful

Watch next