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, K_{c} and K_{p}.

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

Equilibrium constant, K

_{p}= (P_{C})^{c}(P_{D})^{d}/ (P_{A})^{a}(P_{B})^{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, P_{T} 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, P_{Y} = Partial Pressure of Z, P_{Z}

Since equilibrium partial pressure of X, P_{X} 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, P_{Y} = 3/7 P

Partial pressure of Z, P_{Z} = 3/7 P

Using the formula on Equilibrium Constant, we will have:

K_{p} = (P_{Y})(P_{Z}) / (P_{X}) = (3/7 P)(3/7 P) / (1/7 P) = 9/7 P

Answer: B

Are you learning something? 😀

Hope you do!

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

Anon-y-mouse says

Nice job explaining this out.

Keep it up 😉

maisie says

Hi sir,

the “heated at constant pressure” in the qn refers to the total equilibrium pressure and not the initial pressure?

Thanks