Why do you think the enthalpy of activation, delta H, for a syn-periplanar elimination is higher than that for an anti-periplanar elimination?
In an elimination reaction, a pi-bond is formed by sidewise overlap of two orthogonal p-orbitals, one on each carbon undergoing elimination reaction. These p-orbitals evolve as the C-atoms re-hybridize from sp^3 to sp^2. Effectively, the sp^3 orbitals that were involved in C – H and C – X bond formation lose their s-character and becomes pure p-orbitals: correct alignment of the orbitals is essential for pi-bond formation. This is possible only if these two bonds are in periplanar (syn or anti) arrangement (see Fig.1).
Of these two conformers, the anti-periplanar conformation has a transition state in which bonds in the two carbons are in staggered conformation, thus possessing lower energy compared to the syn-periplanar conformation in which all the bonds are in ecclipsed conformation. There would be some difference in energy between the two starting materials too, but the difference in energy between the two transition states would be more conspicuous, as these are highly unstable species and any stabilizing factor there would have a more pronounced effect upon its stability, when compared to the stabler starting materials (see Fig.2). Consequently, the enthalpy of activation, `E_a` for the syn-periplanar elimination would be more than that of the anti-periplanar elimination reaction.