Here's a (slightly) different write-up of David's example, combined with Emil's comment: We want to show that the partial order $1<2$, $1<6$, $3<2$, $3<4$, $5<4$, $5<6$ is not a convex combination of total orders. Now the only total orders that can contribute here are the extensions of the given partial order. We observe that there are only four extensions that make $4<1$, namely $$ 3<5<4<1<2<6 , $$ and here we may switch $3,5$ and/or $2,6$. Since initially $1, 4$ were not comparable (so $x_{14}=0$), these four total orders must get combined weight $1/2$ in the convex combination we are looking for. We can now similarly consider the (again four) total order extensions that make $6<3$, and again these must get combined weight $1/2$. We have used up all our coefficients, but all orders considered so far have $5<2$, so this will not work (since $x_{25}=0$).