Question

Introduced into a 1.50 L flask is 0.100 mol PCl_5(g). The flask is held at 227°C until equilibrium is established. What are the partial pressures and the total pressure of the gases in the flask at equilibrium?

PCl_5(g) ↔ PCl_3(g) + Cl_2(g)

1) Calculate ∆H^°_rxn

2) Calculate ∆S^°_rxn

3) Calculate K_eq

₄₎ Calculate P_initial (PCl₅)

5) Calculate equilibrium pressures for PCl_5(g), PCl_3(g), Cl_2(g)

6) calculate total pressure of this gas mixture at equilibrium

Answer 1

1)

HRxn = hproducts - hreactants

HRxn = (Pcl3 + Cl2) - (PCl5)

HRxn = (-287.0 +0) - (-374.9) = 87.9 kJ/mol

2)

SRxn = sproducts - sreactants

SRxn = (311.8+ 223.1) - (364.6)

Srxn= 170.3 J /K

3)

Keq = from dG

dG = dH - T*dS = 87.9 *1000 - (227+273)*170.3 = 2750 J/mol

K = exp(-dG/(RT)) = exp(-2750/(8.314*(227+273))

K = 0.51605

4)

Pinitial of PCl5 -->

PV = nRt

P = nRT/V = 0.1*0.082*(227+273)/1.5

P = 2.7333 atm

5)

in equilbiirum

K = 0.51605

(PPCl3)(PCl2)/(PCl5) = k

0.51605 = x*x/(2.7333 -x)

x^2 + 0.51605x - 0.51605 *2.7333 = 0

x = 0.95721

(PPCl3)= 0.95721

(PCl2) = 0.95721

(PCl5) = 2.7333 -0.95721 = 1.776

6)

Ptotal = 0.95721+0.95721+1.776 = 3.69042 atm