Consider the following gas phase dissociation, PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g) with equilibrium constant K$_p$ at a particular temperature and at pressure P. The degree of dissociation ($\alpha$) for PCl$_5$(g) is
PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g)
$\alpha = \left(\frac{K_p}{K_p+P}\right)^{1/3}$
Assume the initial moles of PCl5 are 1 and the degree of dissociation is α. An ICE table based on moles can be set up:
| Reaction: | PCl5(g) | ⇌ | PCl3(g) | + | Cl2(g) | ||
|---|---|---|---|---|---|---|---|
| Initial (mol): | 1 | 0 | 0 | ||||
| Change (mol): | -α | +α | +α | ||||
| Equil. (mol): | 1-α | α | α |
The total moles at equilibrium is (1-α) + α + α = 1+α. The total pressure at equilibrium is P. Calculate partial pressures using pi = Xi × Ptotal, where Xi is the mole fraction. Therefore, pPCl5 = ( (1-α) / (1+α) ) P, pPCl3 = ( α / (1+α) ) P, and pCl2 = ( α / (1+α) ) P. The equilibrium constant Kp is expressed as: Kp = (pPCl3 × pCl2) / pPCl5. Substitute the partial pressures: Kp = (( αP / (1+α) ) ( αP / (1+α) )) / (( (1-α)P ) / (1+α) ). Simplify this to: Kp = (α2P2 / (1+α)2) / (( (1-α)P ) / (1+α) ). Further simplification gives Kp = (α2P2 / (1+α)2) × ((1+α) / ((1-α)P)), which reduces to Kp = (α2P) / ((1+α)(1-α)) and finally to Kp = (α2P) / (1-α2). Now, solve for α: Kp (1-α2) = α2 P, which simplifies to Kp - Kp α2 = α2 P. Rearranging gives Kp = α2 P + Kp α2, which factors to Kp = α2 (P + Kp). Then, α2 = Kp / (P + Kp), and α = √( Kp / (Kp + P) ) = ( Kp / (Kp + P) )1/2. This matches option (C).
For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 atm $mol^{–1} K^{–1}$]
$K_p$ for the reaction at 1000 K is: