CO (g) + Cl2 (g) ↔ COCl2 (g
In one experiment, 0.93 bar of CO was combined with 1.01 bars of Cl¬2 in a reaction vessel at 700K. After the system reached equilibrium, the total pressure was 1.24 bars. Calculate Kp0 for the system.
To calculate Kp0 for the system, we need to know the equilibrium partial pressures of CO, Cl2, and COCl2.
Given:
Initial pressure of CO, PCO = 0.93 bar
Initial pressure of Cl2, PCl2 = 1.01 bar
Total pressure at equilibrium, Ptotal = 1.24 bar
At equilibrium, let's assume the partial pressure of COCl2 is PCOCl2. The reaction stoichiometry tells us that the number of moles of COCl2 formed is the same as the number of moles of CO or Cl2 consumed.
Since the reaction is at equilibrium, we can set up an ICE table (Initial, Change, Equilibrium) to determine the values we need:
CO (g) + Cl2 (g) ↔ COCl2 (g)
P | P | P
-------------------------------------------
CO | 0.93 | -x | 0.93 - x
Cl2 | 1.01 | -x | 1.01 - x
COCl2 | 0 | +x | x
Based on this table, we can write the expression for the equilibrium constant Kp0 using the partial pressures:
Kp0 = (PCOCl2) / (PCO × PCl2)
To find the value of x, we can use the equation for Dalton's Law of Partial Pressures, which states that the total pressure is the sum of the partial pressures:
Ptotal = PCO + PCl2 + PCOCl2
Substituting the given values:
1.24 bar = 0.93 bar - x + 1.01 bar - x + x
Simplifying,
1.24 bar = 1.94 bar - x
x = 1.94 bar - 1.24 bar
x = 0.7 bar
Now, let's substitute the values back into the expression for Kp0:
Kp0 = x / [(0.93 bar - x) × (1.01 bar - x)]
= 0.7 bar / [(0.93 bar - 0.7 bar) × (1.01 bar - 0.7 bar)]
Kp0 = 0.7 / (0.23 × 0.31)
= 10.04
Therefore, the value of Kp0 for the system is approximately 10.04.