For the reaction below, Kp= 28.63 at 800 K. Calculate the equilibrium partial pressures of the reactants and products if the initial pressures are PpCl5=0.5600 atm and PpCl3= 0.4700 atm
PCl5(g) <--> PCl3(g) + Cl2(g)
PCl5=??
PCl3=??
Cl2=??
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To calculate the equilibrium partial pressures of the reactants and products, we can use the given equilibrium constant (Kp) and the initial pressures.
First, let's define the initial partial pressures of the reactants and products:
P[PCl5] = 0.5600 atm
P[PCl3] = 0.4700 atm
Using the equation:
PCl5(g) <--> PCl3(g) + Cl2(g)
Let's assume that at equilibrium, the partial pressures become:
P[PCl5] = x
P[PCl3] = y
P[Cl2] = z
Now, we can use the equation for the equilibrium constant (Kp) to set up an expression for the equilibrium partial pressures:
Kp = (P[PCl3] * P[Cl2]) / P[PCl5]
Substituting the given values and the assumed values, the expression becomes:
28.63 = (y * z) / x
To solve for x, we can rearrange the equation as:
x = (y * z) / 28.63
Now, let's substitute the initial pressures into the equation:
x = (0.4700 atm * 0.5600 atm) / 28.63
Simplifying the expression:
x = 0.009230 atm
Therefore, the equilibrium partial pressure of PCl5 is 0.009230 atm.
To calculate the equilibrium partial pressure of PCl3 and Cl2, we can use the value of x we obtained:
P[PCl3] = y = 0.009230 atm (since 1 mole of PCl3 is formed for every 1 mole of PCl5)
P[Cl2] = z = 0.009230 atm (since 1 mole of Cl2 is formed for every 1 mole of PCl5)
Therefore, the equilibrium partial pressures are as follows:
PCl5 = 0.009230 atm
PCl3 = 0.009230 atm
Cl2 = 0.009230 atm