let's reason differently:
let's assume a reaction at equilibrium.
A+B <--> C
is it hard to reverse this? In other words, is it hard to make the reaction go backwards?
Well, the position of the equilibrium depends (amongst other things) on the amount of reactants available (for example, if C is a gas that evolves, the reaction will be indeed hard to reverse as one of the reactants, C, effectively gets removed from the solution/reactor)
But what if we forcibly supply more and more C, then the reaction will go backwards!
Read up on Le Chatelier's principle for equilibrium changes.
And yes, the equilibrium constant is related to the change in gibbs free energy (dG = -RT ln Keq), which is itself dependent on changes in enthalpy and entropy.