Answer
(a)$Kp=4.42×10^{-5}$
The reactants are favored at equilibrium.
(b)$Kp=1.503×10^{2}$
The products are favored at equilibrium.
(c)$Kp=1.9578×10^{-9}$
The reactants will be favored at equilibrium
Work Step by Step
(a)Since the reaction was inverted, the new equilibrium constant will be equal to the multiplicative inverse of the original one.
$Kp$=$\frac{1}{(2.26\times10^{4})}$=$4.42×10^{-5}$
Since $Kp<1$, the reactants will be favored at equilibrium.
(b) It is observed that all the coefficients of that reaction were divided by 2 (or multiplied by $\frac{1}{2}$).Thus, the new equilibrium constant is equal to the old one to the power of $\frac{1}{2}$.
$Kp$=$(2.26×10^{4})^{\frac{1}{2}}$=$1.503×10^{2}$
Since $Kp>1$, the products will be favored at equilibrium.
(c)Since the reaction was inverted and also multiplied by 2,the new equilibrium constant will be equal to the multiplicative inverse of the square of original one.
$Kp$=$\frac{1}{(2.26\times10^{4})^{2}}$=$1.9578×10^{-9}$
Since $Kp<1$ ,the reactants will be favored at equilibrium.
