Here we have to get the [tex]K_{p}[/tex] of the reaction at 520 K temperature.
The [tex]K_{p}[/tex] of the reaction is 1.705 atm
We know the relation between [tex]K_{p}[/tex] and [tex]K_{c}[/tex] is [tex]K_{p}=K_{c}(RT)^{N}[/tex], where [tex]K_{p}[/tex] = The equilibrium constant of the reaction in terms of partial pressure, [tex]K_{c}[/tex] = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.
Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅
Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole
The given value of [tex]K_{c}[/tex] is 4.0×10⁻²
The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.
On plugging the values in the equation we get,
[tex]K_{p} = 4.0 X 10^{-2}(0.082X520)^{1}[/tex]
Or, [tex]K_{p}[/tex] = 1.705 atm
Thus, the [tex]K_{p}[/tex] of the reaction is 1.705 atm
