Question

If Kc = 4.0×10−2 for PCl3(g)+Cl2(g)⇌PCl5(g) at 520 K , what is the value of Kp for this reaction at this temperature?

Answer 1

Here we have to get the $K_{p}$ of the reaction at 520 K temperature.

The $K_{p}$ of the reaction is 1.705 atm

We know the relation between $K_{p}$ and $K_{c}$ is $K_{p}=K_{c}(RT)^{N}$, where  $K_{p}$ = The equilibrium constant of the reaction in terms of partial pressure, $K_{c}$  = 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  $K_{c}$ 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,

$K_{p} = 4.0 X 10^{-2}(0.082X520)^{1}$

Or, $K_{p}$ = 1.705 atm

Thus, the $K_{p}$ of the reaction is 1.705 atm