Question

The equilibrium constant, K. for the following reaction is 0.636 at 600 K COC13(Q)= CO(g) + Cl() Calculate the equilibrium partial pressures of all species when COCK() is introduced into an evacuated flask at a pressure of 1.62 atm at 600K atm PcoCl, " Рco Рci, atm 8 more group attempts remaining Retry Entire Group Submit Answer

Answer 1

cocl₂(g) ⇌ co(g) + cl₂(g)

initially partial pressures (atm) 1.62 0 0

equilibrium partial pressures (atm) (1.62-x) x x

The equilibrium constant, Kp for the reaction is 0.636 at 600 K

By definition, Kp = (P_products)/(P_reactant)

So we get for this reaction, Kp = {(P_CO)(P_Cl2)}/P_COCl2)

P_CO = P_Cl2 for this reaction. The partial pressures P_CO and P_Cl2 are  zero at the start. At equilibrium, let P_CO = P_Cl2 = x atm

So, P_COCl2 = (1.62 - x) atm [ as total pressure is said to be 1.62 atm]

Putting the values in the expression for Kp, we get

0.636 = x²/(1.62 - x)

or, x² + 0.636x -1.03 = 0

or, x= 0.746

So, at equilibrium, P_CO = P_Cl2 = 0.746 atm , P_COCl2 = (1.62-.746) atm = 0.874 atm