1. If we reverse an equation

If A + B ⇔ C + D **K****or**_{P}_{ }**K****is reversed.**_{C}

_{ }**K**

**= 10**

_{C}then for C + D ⇔ A + B

_{ }K**= 10**

_{C }^{-1}

- ½ N

_{2}+ ½O

_{2 }⇔ NO

_{ }**K**

**= 5**

_{C}_{2}+ O

_{2 }⇔ 2NO

**K**

**= 5**

_{C}^{2}= 25

_{ }

3. If we divide each of the coefficient in a balanced equation by a factor n,then new equilibrium constant is nth root of the previous value.

2SO

SO_{2}+ O_{2}⇔ 2SO_{3 }_{ }**K****=25**_{C}_{2}+ 1/2 O

_{2}⇔ SO

_{3}B

_{ }**K**

**= (25)**

_{C}^{1/2}

4. When we combine individual equation,we have to multiply their equilibrium constant for net reaction.

If K

_{1 }, K

_{2}and K

_{3 }are stepwise equilibrium constant for A⇔B , B⇔C , C⇔D .

Then for A⇔D , equilibrium constant is K= K

_{1}K

_{2}K

_{3 }

#### Significance of the magnitude of equilibrium constant:

- If a large value of
**K**or_{P}_{ }**K**signifies that the forward reaction goes to completion or very nearly so._{C} - A large value of
**K**or_{P}**K**signifies that the forward reaction doesnot occur to any significiant extent._{C} - A reaction is most likely to reach a state of equilibrium in which both reactant and product are present if the numerical value of
**K**or_{P}_{ }**K**is neither very large nor very small_{C}

#### Units of equilibrium constant:

For the shake of simplicity in the units of Kc or Kp the relative molarity of pressure of reactants and products are used with respect to standard condition.(For solution standard state 1 mol/litre and for gas standard pressure=1 atm).Now resulting equilibrium constant becomes unitless by using relative molarity and pressure.But the numerical value may change in the value of standard states molarity and pressure.

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