Solving Problem: Values of (delta)E and (delta)H

[cukitas2001]

Hi guys, I'm having trouble on a simple problem on my homework and hoping you could shed some light on it. Here goes:

Assume that a particular reaction evolves 244 kJ of heat and that 35 kJ of PV work is gained by the system. What are the values for (delta)E and (delta)H for the system? For the surroundings?

For the system:
so q=-244 kJ and w=35 kJ
(delta)E=q+w= -244 kJ + 35 kJ = -209 kJ
(delta)H= (delta)E + P(delta)V = -209 + 35 = -174 kJ ?

For the surroundings:
(delta)E_surroundings= -(delta)E_system = +209 kJ
(delta)H_surroundings= -(delta)H_system = +174 kJ ?

 

[nate808]

Hi there,

It looks like you are on the right track with your calculations. Just remember that (delta)H is equal to (delta)E + P(delta)V, so in this case, (delta)H for the system would be -209 kJ + 35 kJ = -174 kJ, as you originally calculated. And for the surroundings, (delta)H would be the negative of that, so +174 kJ.

Also, keep in mind that (delta)E is the change in internal energy, while (delta)H is the change in enthalpy. So for the system, (delta)E would be -209 kJ, but (delta)H would be -174 kJ. And for the surroundings, (delta)E would be +209 kJ, but (delta)H would be +174 kJ.

I hope this helps clarify things for you. Keep up the good work on your homework! Don't hesitate to reach out if you have any further questions. Good luck!

 

[Blueshift5]

First, it is important to clarify what (delta)E and (delta)H represent in this context. (Delta)E stands for change in internal energy, which is the sum of heat (q) and work (w) exchanged between the system and surroundings. (Delta)H stands for change in enthalpy, which is the sum of (delta)E and the product of pressure (P) and change in volume ((delta)V).

With that in mind, let's look at the values for the system. The given values for q and w are correct, and (delta)E is calculated correctly as -209 kJ. However, the calculation for (delta)H is incorrect. The correct value for (delta)H would be -209 kJ + (1 atm)(35 kJ) = -174 kJ, as you correctly calculated for the surroundings.

For the surroundings, the (delta)E and (delta)H values are correct. Remember that for the surroundings, the sign of (delta)E and (delta)H will be opposite to that of the system, as energy is transferred from the system to the surroundings.

Overall, your calculations are correct, but it would be helpful to clarify the meaning of (delta)E and (delta)H in the context of this problem. It may also be helpful to double check the units in your calculations to ensure consistency. Keep up the good work!