Answer
$K_{p}$=$9.64\times10^{-3}$
$K_{c}$=$3.94\times10^{-4}$
Work Step by Step
Step 1: Let’s calculate the partial pressures of all three components at equilibrium to find $K_{p}$. When we divided by V on either side of the ideal gas equation PV=nRT, we can get P=$C_B$RT. From this we can tell concentration of ideal gas ($C_B$) is directly proportional to the pressure(P) at constant temperature. So in this problem, if the percent dissociation of NOBr is 34% (intrinsically meaning the concentration of NOBr reduces by 34%), its partial pressure also reduces by 34%.
Now, let’s assume the starting partial pressure of NOBr is x. According to the stoichiometry in this reaction 2NOBr(g) 2NO(g) + $Br_{2}$(g):
The final partial pressure of NOBr ($P_{NOBr}$)= x-34%x=66%x=0.66x
The final partial pressure of NO ($P_{NO}$)= 0+34%x=0.34x
The final partial pressure of $Br_{2}$ ($P_{Br_{2}}$)= 0+$\frac{34%x}{2}$=0.17x
Then, the total pressure = 0.66x+0.34x+0.17x=1.17x=0.25 atm; so x$\approx$0.21 atm.
When we blog the solved x value back to the expressions of final partial pressures for the three components, we can get:
$P_{NOBr}$=0.66(0.21 atm)$\approx$0.14 atm; $P_{NO}$=0.34(0.21 atm)$\approx$0.073 atm; $P_{Br_{2}}$=0.17(0.21 atm)$\approx$0.036 atm
$K_{p}$=$\frac{(P_{NO})^{2}(P_{Br_{2}})}{(P_{NOBr})^{2}}$=$\frac{(0.073)^{2}(0.036)}{(0.14)^{2}}$$\approx$$9.64\times10^{-3}$
Step 2: From the final partial pressures for the three components we got from step 1, we calculate their final concentrations to find $K_{c}$. For ideal gas, P=$C_B$RT; so, $C_B$=$\frac{P}{RT}$
[NOBr]=$\frac{P_{NOBr}}{RT}$=$\frac{0.14 atm}{(0.082 atm/(M*K))(298.15K)}$$\approx$$5.77\times10^{-3}$M
[NO]=$\frac{P_{NO}}{RT}$=$\frac{0.073 atm}{(0.082 atm/(M*K))(298.15K)}$$\approx$$2.97\times10^{-3}$M
$[Br_{2}]$=$\frac{P_{Br_{2}}}{RT}$=$\frac{0.036 atm}{(0.082 atm/(M*K))(298.15K)}$$\approx$$1.49\times10^{-3}$M
$K_{c}$=$\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}}$=$\frac{(2.97\times10^{-3})^{2}(1.49\times10^{-3})}{(5.77\times10^{-3})^{2}}$$\approx$$3.94\times10^{-4}$
