#### Answer

a. $K_p$= $\frac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}$
b. $K_p$= $\frac{(P_{NO_2})^2}{(P_{N_2O_4})}$
c. $K_p$= $\frac{(P_{SiCl_4})(P_{H_2})^2}{(P_{SiH_4})(P_{Cl_2})^2}$
d. $K_p$= $\frac{(P_{PCl_3})^2(P_{Br_2})^3}{(P_{PBr_3})^2(P_{Cl_2})^3}$

#### Work Step by Step

To solve for $K_p$, you divide $\frac{products}{reactants}$, and the coefficient in front of each compound becomes the exponent.
a. $N_2 (g) + O_2 (g) 2NO (g)$
To solve for $K_p$ for problem a, you take the products, $2NO$ and divide the reactants $N_2 + O_2$. Don't forget to make the coefficients the exponents for each compound, like $2NO$ becomes $(NO)^2$, but $N_2$ and $ O_2$ don't need to have exponents written because they have coefficients of 1. Now you have the basic idea of writing equations for $K_p$, but there is one last step: you must write the compound as a subscript of $P$ because $K_p$ is the equilibrium constant for partial pressures or gases. Thus $2NO$ should look like $(P_{NO})^2$, with the exponent going on the outside. Your final answer should look like this: $K_p$= $\frac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}$
Follow these same steps for the next three problems.
b. $N_{2}O_4 (g) 2NO_2 (g)$
To solve for $K_p$ for problem b, you take the products, $2NO_2$ and divide the reactants $N_{2}O_4$. Don't forget to make the coefficients the exponents for each compound, like $2NO_2$ becomes $(P_{NO_2})^2$, but $N_{2}O_4$ doesn't need to have an exponent written because it has a coefficient of 1. Now you must write the compound as a subscript of $P$ because $K_p$ is the equilibrium constant for partial pressures or gases. Your final answer should look like this: $K_p$= $\frac{(P_{NO_2})^2}{(P_{N_2O_4})}$
c. $SiH_4 (g) + 2Cl_2 SiCl_4 (g) + 2H_2 (g)$
To solve for $K_p$ for problem c, you take the products, $SiCl_4 (g) + 2H_2 (g)$ and divide the reactants $SiH_4 (g) + 2Cl_2$. Don't forget to make the coefficients the exponents for each compound, like $2Cl_2$ becomes $(P_{Cl_2})^2$. Now you must write the compound as a subscript of $P$ because $K_p$ is the equilibrium constant for partial pressures or gases. Your final answer should look like this: $K_p$= $\frac{(P_{SiCl_4})(P_{H_2})^2}{(P_{SiH_4})(P_{Cl_2})^2}$
d. $2PBr_3 (g) + 3Cl_2 (g) 2PCl_3 (g) + 3Br_2 (g)$
To solve for $K_p$ for problem d, you take the products, $2PCl_3 (g) + 3Br_2 (g)$ and divide the reactants $2PBr_3 (g) + 3Cl_2 (g)$. Don't forget to make the coefficients the exponents for each compound, like $3Cl_2$ becomes $(P_{Cl_2})^3$. Now you must write the compound as a subscript of $P$ because $K_p$ is the equilibrium constant for partial pressures or gases. Your final answer should look like this: $K_p$= $\frac{(P_{PCl_3})^2(P_{Br_2})^3}{(P_{PBr_3})^2(P_{Cl_2})^3}$