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Chemical Equilibrium

Chemical Equilibrium: Question on Equilibrium Constant, Kp

September 30, 2010 By Sean Chua 2 Comments

In the previous blogpost, we have discussed on the concepts of chemical equilibrium in a reversible reaction.

We have also discussed on the formula to determine Equilibrium Constant, Kc and Kp.

For Reversible Reactions involving Gases, we should use the following formula:

Equilibrium constant, Kp = (PC)c (PD)d / (PA)a (PB)b

Let’s apply the formula to a question that was given by my students from a local Junior College (JC).

Now, at equilibrium, reactant (X) and products (Y and Z) will be all present together.

Since the question stated that the pressure is constant (value is P) and the temperature is also constant, we can assume that the total pressure, PT of the gases at equIlibrium is P.

By stoichiometry ratio comparison of equation

X <=> Y + Z;

Since Mole of Y = Mole of Z at equilibrium, therefore

Partial Pressure of Y, PY = Partial Pressure of Z, PZ

Since equilibrium partial pressure of X, PX is 1/7 P, it means that the other 6/7 P of pressure must be shared equally among Y and Z.

Thus,

Partial pressure of Y, PY = 3/7 P

Partial pressure of Z, PZ = 3/7 P

Using the formula on Equilibrium Constant, we will have:

Kp = (PY)(PZ) / (PX)    =   (3/7 P)(3/7 P) / (1/7 P)   =   9/7 P

Answer: B

Are you learning something? 😀

Hope you do!

If you think the above information is useful to your friends, classmates or siblings, feel free to share it with them.

PS: Do leave me your thoughts in the comments section below. I would love to hear from you.

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