Find ΔE° for the reaction below if the process is carried out at a constant pressure of 1.00 atm and ΔV (the volume change) = -24.5 L. (1 L ∙ atm = 101 J)

Question

2CO (g) + O2 (g) --> 2CO2 (g)

Delta H = -566 kJ

Answer 1

∆Eº = q + w and at constant pressure of 1.00 atm, q = ∆H and w = -P∆V

w = -P∆V = -(1.00 atm)(-24.5 L) = +24.5 Latm

24.5 Latm x 101 J/1 Latm = 2475 J = w

∆E = ∆H + w = -566000 J + 2475 J = -563525 J = -563.5 kJ