A reversible reaction is one in which can proceed in both directions i.e. forward and reverse reactions.

The reactants are not completely converted to products. Instead, an intermediate position or **EQUILIBRIUM** is reached whereby both reactants and products are present.

This also means that reversible reactions are never complete, a mixture of reactants and products is obtained.

At equilibrium in a reversible reaction:

1. Rate of forward reaction = Rate of reverse reaction

2. Concentration of the reactants and the products are unchanged

It is important to know what is the **EQUILIBRIUM CONSTANT**, which is defined as the measure of the extent to which the reaction are coverted into products before equilibrium is reached.

Consider the equilibrium, aA + bB <=> cC + dD

Equilibrium constant, K

_{c}= [C]^{c}[D]^{d}/ [A]^{a}[B]^{b}and is a constant at a given temperature

For reactions involving **gases**, the equilibrium constant can be represented in terms of the partial pressure, P, of the gases and is give by the symbol, K_{p}.

Consider the equilibrium, aA (g) + bB (g) <=> cC (g) + dD (g)

Equilibrium constant, K

_{p}= (P_{C})^{c}(P_{D})^{d}/ (P_{A})^{a}(P_{B})^{b}and is a constant at a given temperature

Something to note for Partial Pressure:

Partial pressure, P, of a gas is equal to the product of its mole fraction and the total pressure, P_{T}.

Example, in a mixture of two gases A and B (where n_{x} = no. of moles of gas X)

Partial pressure of Gas A, P_{A} = [ n_{A} / (n_{A} + n_{B}) ] x P_{T}

Important Note:

1. Concentration of a SOLID is constant, and is omitted in expressions of K_{c} and K_{p}.

2. State symbols are not essential in writing the expressions for equilibrium constant.

So far so good? 😉

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.

30

Sep 10

## Chemical Equilibrium – Question on Chemical Constant, Kp

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;

Mole of Y = Mole of Z

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

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.

girl says

Nice website! But could you do more notes on transition elements? 😀

rida says

This is so helpful! It would be so awesome and helpful if u cover everything in the GCE A’level syllabus! Is that possible? 🙂