Question

For the reaction:
PCl3(g) + Cl2(g) PCl5(g) at 70.5°C, Kp = 1.05.
If one starts with 1.80 atm pressure of PCl3(g), 1.72 atm pressure of Cl2(g), and no PCl5(g), what is the partial pressure of PCl5(g) at equilibrium?

Answer 1

set up the equilibrium equation, write an ICE table, substitute the values from E (from ICE table) to the expression you initially wrote and solve.