Answer
$$K_p = 1.50 \times 10^{2}$$
The products are favored at equilibrium.
Work Step by Step
- As we can observe, all the coefficients of that reaction were divided by 2 (or multiplied by $\frac 12$). Thus, the new equilibrium constant is equal to the old one to the power of $\frac 1 2$:
$$K_p = (2.26 \times 10^{4})^{1/2} = 1.50 \times 10^{2}$$
- Since $K_p \gt 1$, the products will be favored at equilibrium.
