Using the equation to calculate the equilibrium constant K $(3\mathrm{.}12$ in your book$)$ and your knowledge of Le Chatelier's Principle predict what would happen to the reaction $($driven forward or reverse$)$ for the following circumstances

The concentration of the reactants are doubled.

The concentration of $D$ is reduced by half.

How do you know the answers in a and b are correct?

$G$ is the driving force behind your answers to a and $b$ above. At $K$ your reaction is at equilibrium, meaning $G$ is $0.$ Any perturbations to that equilibrium move $\frac{?}{{?}_{\frac{?}{?}}G}$ either negative or positive. To calculate this change we must compare the current perturbation of equilibrium to the reaction at the standard chemical state. What is the chemical standard state?

$K=(\frac{[C{]}^c[D{]}^d}{[(A){]}^a[(B){]}^b}{)}_{eq}$