At 600 K and 5.0 bar, the degree of dissociation of PCl5(g) is 0.920. Phosphorus pentachloride thermally disassociates according to following reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g)a) What is the degree of dissociation at 600 K and 1.0 bar of pressure?b) If the degree of dissociation is 0.80 at 520 K and 1.0 bar, what are ΔG, ΔH and ΔS at this temperature?

(a) Let α be the degree of dissociation for the given reaction. Now, PCl5g⇌PCl3g+Cl2g1−α α αThe equilibrium constant (Kp) is defined as: Kp=PPCl3PCl2PPCl5 ...... (1) Further, PPCl3=αP1+αPCl2=αP1+αPPCl5=1−αP1+α Equation (1) will now become: Kp=PPCl3PCl2PPCl5Kp=αP1+ααP1+α1−αP1+αKp=α2P1−α1+αKp=α2P1−α2 ...... (2)Use the given value of the degree of dissociation at P = 5 bar and 600 K to get the value of Kp as: Kp=α2P1−α2Kp=0.920251−0.9202Kp=27.55 bar The value of Kp is dependent only on temperature and thus remains constant over different pressures. Calculate the degree of dissociation using the calculated value of Kp at 1 bar and 600 K as: Kp=α2P1−α227.55=α211−α2α=0.9823(b) At 520 K and 1 bar, α = 0.80. Use equation (2) to calculate the value of Kp at this condition as: Kp=α2P1−α2Kp=0.80211−0.802Kp=1.78 bar Now, calculate the change in the Gibbs free energy as: ΔG=−RTlnKpΔG=−8.314 Jmol⋅K520 Kln1.78ΔG=−2492.86 JmolUse the Van’t Hoff equation to calculate the change in enthalpy as: lnKp2Kp1=−ΔHR1T2−1T1 ....... (3) The values to be used are: R=8.314 Jmol⋅KT1=600 KT2=520 KKp1=27.55KP2=1.78 Equation (3) will now give the value of change in enthalpy as: lnKp2Kp1=−ΔHR1T2−1T1ln1.7827.55=−ΔH8.3141520−1600ΔH=88823.6 JmolChange in entropy will be calculated as: ΔS=ΔH−ΔGTΔS=88823.6−−2492.86520ΔS=175.61 Jmol⋅K