Question

Calculate the PV work that occurs in the decomposition of 0.500 mol NI₃ at constant pressure, and at the temp of 25^o C. (assume that the volume of the solid is negligible)

2 NI_{3 (s)} --> N_{2 (g)} + 3 I_{2 (g)}

Answer 1

2 NI3 gives total 4 gas moles ( 1 N2 and 3I2)

0.5 mol NI3 gives ( 4/2) x 0.5 = 1 mole of gas

work done is calculated by formula PV = nRT          where n = 1mole , R = gas conatant = 0.08206 literatm/molK

T =25C = 25+273 = 298 K

work done = nRT = 1 x 0.08206 x 298

           work done = 24.454 atmliter

1 atm liter = 101.325 Joules

Hence work done = 24.454 x 101.325 J = - 2478 Joules

( -ve sign indicates expansion work )