Answer
Since $K_c \gt 1$, the concentration of the products is larger at equilibrium. Thus, trans-2-butene appears in a larger concentration at equilibrium.
Work Step by Step
$$K_c = \frac{k_{forward}}{k_{reverse}}$$
Therefore, if $K_c \gt 1$, $k_{forward} \gt k_{reverse}$, so more products (right side of the reaction) are produced.
