When 1.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 33.9 kJ are absorbed and PΔV for the vaporization process is equal to 2.90 kJ

Question

What is delta E?

note: does delta H = 33.9kJ ..?

Answer 1

∆E = q + w

q = ∆H @ constant pressure = +33.9 kJ (endothermic, + sign, heat absorbed)

w = P∆V = 2.90 kJ

∆E = 33.9 kJ + 2.90 kJ = 36.8 kJ

NOTE: delta H does equal 33.9 kJ