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Ch.6 - Thermochemistry

Chapter 6, Problem 55

When 1 mol of a fuel burns at constant pressure, it produces 3452 kJ of heat and does 11 kJ of work. What are ΔE and ΔH for the combustion of the fuel?

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Hey everyone. And welcome back when one mole of fuel combusts at constant pressure. It produces 4462 kilojoules of heat and thus 25 kilojoules of work. What are Delta E and delta H for the combustion of the fuel? And we are given for answer choices. ABC and D, they essentially correspond to delta E and delta E rallies all given in killed Jools. So essentially what we are going to do in this problem is remember that the internal energy change delta E is equal to Q, which is heat plus W which is work before that we can also tell that the entropy change delta H is equal to the amount of heat Q at constant pressure, right? So that's what we know from the problem. Now, we can begin from here. So what is the value of Q? Well, essentially the problem states that we are producing 4462 kilojoules of heat. And essentially it states that fuel produces that heat. Fuel is our system. So if we, our system produces heat, that means heat is released into the surroundings and therefore, it must be negative. So we are adding that negative sign and we get our Q value of negative 4462 kilojoules, which by definition at a constant pressure is also equal to the entropy change of the reaction. So we are halfway done. That's our answer number one. And now we want to find the internal energy change. And for that purpose, we want to identify the work done. The problem states that our system which is our fuel does work. So whenever a system does work on the surroundings, it's once again negative. So instead of 25 kilojoules, we're adding a negative sign and we get negative 25 kilojoules of work done, right? Let's remember that the work done by by the surroundings on the system would be positive. So whenever the work is done on the system, it's positive when the work is done by the system, it's negative, right? So we can essentially state this is the work done by the system, which is our keyword. If it's on the system, it's positive. And now that we have both of our values, we can say that the internal energy change delta E is equal to Q which is negative 4462 kilojoules plus, what is our work? Well, it's negative 25 kilojoules. So in this case, our final answer will be negative 4487 kilojoules. We have our and that be change, right? As we previously said, that's our delta H and we have our change in internal energy. Now, we essentially want to determine the correct answer choice. And in this case, let's see which one is the correct one. It's basically option A which states that the in internal energy change is negative 4487 kilojoules. And the change in en entropy is negative 4462 kilojoules. Thank you for watching.
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