Question

1.1. Write the equilibrium constant expression for the following reactions:
CaCOs (s) Ca0(s)+COz(g)
(2)
1.2. The reaction below was carried out at 503 K in a 5.00 L vessel. At equilibrium, the
reaction vessel contained 0.0185 mol of PCla, 0.0158 mol of PCs, and 0.0870 mol
of Cle. Calculate the value of Kc at this temperature.
Reaction: PCh (g) + Ch (g) PCs (g)
(6)

Answer 1

Solutions

1.1

 CaCO₃ (s)⇋CaO(s)+CO₂ (g) 

1.2

Δn=(0+1)−0=1

Kc​= Kp/{(RT)Δn}

​Kc

​= 167bar/{(0.0821Lbar/K)×(1073K)}

​=1.896mol/L