J.R. S. answered 11/25/24
Ph.D. University Professor with 10+ years Tutoring Experience
The ∆H (qp or heat @ constant pressure) would simply by the +38.85 kJ. Now, if you meant to ask what is the ∆U (change in internal energy) then, we will also use the work done, as follows:
∆U = q + w
∆U = change in internal energy = ?
q = heat = + 38.95 kJ (positive since heat is being absorbed by the system)
w = work = -2.49 kJ (negative since the system is doing work on the surroundings)
∆U = 39.95 kJ - 2.49 kJ = 37.46 kJ