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Ch.7 - Thermochemistry

Chapter 7, Problem 56

The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is 5084.3 kJ. If the change in enthalpy is 5074.1 kJ, how much work is done during the combustion?

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Hey everyone in this example we're told that at standard pressure the formation of nitrogen monoxide gas has a change in internal energy of 89.5 kg jewels. And we have a change in entropy of 91.3 kg jules. We need to find the value for work corresponding to this process. So we should recall that our formula for work is going to be found by taking our change in internal energy and subtracting that from our value for heat. Q. We should also recall that our value for heat is going to be equal to our change in entropy at constant pressure. And because we do have constant pressure, there's no mention of our pressure changing. We can assume that our value for Q is going to be equal to our given change in entropy of 91.3 kg jewels. So let's go ahead and use are given information to solve for work. What we would have is that work is equal to our change in our internal energy. Which is given as 89.5 kg jewels subtracted from our value for Q. Which again is equivalent to our change in entropy given as 91.3 kg joules. And so this difference is going to give us a value for work equal to negative 1.8 kg joules. So our final answer to complete this example is negative. 1.8 kg joules is equal to r value for work corresponding to the formation of nitrogen monoxide. So I hope that everything I reviewed was clear. If you have any questions please leave them down below, and I will see everyone in the next practice video.