Consider the following reaction: a(g)⇌2b(g). find the equilibrium partial pressures of a and b for each of the following different values of kp. assume that the initial partial pressure of b in each case is 1.0 atm and that the initial partial pressure of a is 0.0 atm. make any appropriate simplifying assumptions. kp = 1.4?
kp = 2.0 * 10^-4?
kp = 2.0 * 10^5?

a. Kp=1.4

b.Kp=2.0 * 10^-4

c.Kp=2.0 * 10^5

Explanation:

For the reaction  

A(g)⇌2B(g)

Kp is defined as:

The conditions in the system are:

          A                    B

initial   0                1 atm

equilibriumx       1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.    

Replacing these values in the expression for Kp we get:

Working with this equation:

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

(equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

With x1 we get a partial pressure of:

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

Not sense.

With x2

X2 is the solution for this equation.  

Kp=2*10^5

X1=50001

With x1

Not sense.

With x2

X2 is the solution for this equation.  

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