Q1(a)(ii). Residual entropy of CO and N₂O at 0 K

Answer in brief

A CO molecule can lie as CO or OC in the crystal (N₂O as NNO or ONN) with almost the same energy. This random orientation is frozen in on cooling, so W=2^N at 0 K and Sₘ(0)=R ln 2=5.76 J K⁻¹ mol⁻¹, not zero.

Key points

• The third law gives S=0 at 0 K only for a perfect crystal: one arrangement, W=1, S=k ln 1=0.
• CO and N₂O are linear, nearly symmetrical in shape and have a very small dipole moment. A molecule lying CO or OC (NNO or ONN) has almost the same energy in the lattice.
• On freezing each molecule takes either orientation at random. On cooling the energy difference is too small to drive reordering, and molecules cannot turn over in the rigid solid: the disorder is frozen in.
• For N molecules with two orientations each, W=2^N: S=k ln 2^N=Nk ln 2; for one mole Sₘ(0)=R ln 2=8.314×0.6931, Sₘ(0)=5.76 J K⁻¹ mol⁻¹