What is the enthalpy change (ΔH) for a reaction at a constant pre...

Question

What is the enthalpy change (ΔH) for a reaction at a constant pressure of 1.00 atm fi the internal energy chagne (ΔE) is 44.0 kJ and the volume increase is 14.0 L? (1 L-atm = 101.325 J)

Accepted Answer

Hey everyone. So here it says the reaction of an ideal gas occurs at a constant pressure of atmospheres and absorbs 53.7 killer jewels of heat. The volume of the gas changes from 1.50 L to 6.00 L, calculate the change in internal energy which is delta e. For the reaction. Right? So first of all, sometimes you might see internal energy being described as delta U. It's the same thing. Both of them equal Q plus W. Now here they're saying we absorb this much heat. Remember if we're saying absorbing, that means that Q will be positive. That's a positive 53.7 kg plus. Now we have to figure out what work is what work equals negative pressure times change in volume. So we have negative 25 atmospheres which is our pressure given and changing volume, it's final minus initial. So that would be 6.00 L -1.5 zero leaders Together. That will give me -112.5 L, times atmospheres. Now realize that there is a conversion factor here and the conversion factor is that one leaders, times atmospheres equals 101.3 jewels. So here, that's going to give me negative 11, 3 96.25 jewels. And if we were to change that to kill a jules, one kg is 10 to the three jewels that equals negative 11. Killing Jewels. So take that and plug it here, so negative 11. kg jewels. So when we add those together, we're going to get 42.30 killer jewels here, I'm just writing it to four sig figs. But since all the numbers are 36 figs, let's just do it as 42.3 kg as our final answer. So that's how we go about in finding our internal energy for this given reaction a k. A. Our system.

What is the enthalpy change (ΔH) for a reaction at a constant pressure of 1.00 atm fi the internal energy chagne (ΔE) is 44.0 kJ and the volume increase is 14.0 L? (1 L-atm = 101.325 J)

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