For a certain reaction at constant pressure, ∆H = -15kJ, and 22 kJ of expansion work is done on the system. What is ∆U for this process?
So I know
∆= ∆U + Work flow
If work is being done on the system, shouldn't work flow be positive? So I assumed it was:
-15kJ = ∆U + 22 kJ
-37kJ = ∆U
but the answer is 7kJ.
What am I doing that's wrong? Thanks.
Expansion is lost, so 22kJ should be -22kj. Plugging in your values, you will get 7kj.
-15kJ=∆U + (-22kJ)
-15kj+22kj=∆U
7kj=∆U
You are correct that work done on the system is positive. However, there seems to be a small mistake in your calculation.
The correct equation relating ∆H, ∆U, and work is:
∆H = ∆U + P∆V
Since the reaction is at constant pressure, we can substitute P∆V with work:
∆H = ∆U + work
In this case, since work is being done on the system, work is positive (+22 kJ). Therefore, the correct equation becomes:
-15 kJ = ∆U + 22 kJ
Now you can solve for ∆U:
∆U = -15 kJ - 22 kJ
∆U = -37 kJ
So the correct value for ∆U is -37 kJ, not +37 kJ.
Therefore, it seems that the answer you mentioned (+7 kJ) is incorrect. The correct value for ∆U in this case is -37 kJ.
To calculate ∆U (the change in internal energy) for a process, you need to consider the relationship between ∆U, ∆H (the change in enthalpy), and the work done on or by the system.
The general formula is:
∆H = ∆U + P∆V
Where:
∆H is the change in enthalpy
∆U is the change in internal energy
P is the pressure
∆V is the change in volume
In this specific case, the question states that the reaction occurs at constant pressure. So, the equation becomes:
∆H = ∆U + P∆V
-15 kJ = ∆U + 22 kJ
Since work is being done on the system (expansion work), the work term, P∆V, is considered positive. Therefore, we have a positive work contribution of 22 kJ. So, you need to rearrange the equation as follows:
-15 kJ = ∆U + 22 kJ
∆U = -15 kJ - 22 kJ
∆U = -37 kJ
It seems like you've done the calculation correctly. However, the given answer of 7 kJ seems incorrect, as the value for ∆U should actually be -37 kJ, not 7 kJ. If the answer you have is different from this, there may be an error in the answer key or a misunderstanding of the problem.
Remember to double-check the values given and make sure there are no mistakes in the calculations.