For the reaction:
PCl3(g) + Cl2(g) --> PCl5(g) at 85 degreeC, Kp = 1.19
If one starts with 2.00 atm pressure of PCl3, 1.00 atm pressure of Cl2 and no PCl5, what is the partial pressure of PCl5(g) at equilibrium?
vchn
To calculate the partial pressure of PCl5(g) at equilibrium, we need to use the equilibrium constant expression (Kp) and the initial pressures of the reactants.
The equilibrium constant expression for the given reaction is:
Kp = (PCl5)^x / (PCl3)^y * (Cl2)^z
Where (PCl3), (Cl2), and (PCl5) are the partial pressures of PCl3, Cl2, and PCl5 at equilibrium, respectively. And x, y, and z are the coefficients of PCl5, PCl3, and Cl2, respectively.
Given values:
Initial pressure of PCl3 (PCl3) = 2.00 atm
Initial pressure of Cl2 (Cl2) = 1.00 atm
No initial pressure of PCl5 (PCl5) = 0 atm
We can assume that the equilibrium partial pressure of PCl5 (PCl5) is 'x' atm.
Substituting the given values into the equilibrium constant expression, we have:
1.19 = x^1 / (2.00)^1 * (1.00)^1
Simplifying the expression, we get:
1.19 = x / 2.00
To solve for x (the partial pressure of PCl5), we need to isolate x by multiplying both sides of the equation by 2.00:
2.00 * 1.19 = x
2.38 = x
Therefore, the partial pressure of PCl5(g) at equilibrium is 2.38 atm.