The equilibrium constnat depends on the way we write the chemical reaction equation. When we change the equation, we need to change the value of the equalibrium for the new equation.

The law of mass action gives an equilibrium constant expression for any chemical reaction equation. The law suggests that for a general chemical reaction equation between a moles of A and b moles of B producing c moles of C and d moles of D,

where

C, D - represent products

If the system is at equilibrium at a given temperature, then the following ratio is a constant.

[C]When a mixture of A, B, C, D is enclosed in a closed system, the law of mass action suggests that the system try to maitain a state such that K is a constant.^{c}[D]^{d}---------------- =K_{eq [A]a [B]b }

You can write the equilibrium constant (K) expression for any reaction equation according to the las of mass action. This has been done in the module on the Law of Mass Action.

The following rules are obvious for manipulating the K's when you change the equations.

**Example**

If the equilibrium constant K = 2 for

**Example**

If the equilibrium constant K = 2 for

**Example**

If the equilibrium constant K = 2 for

dH ln K = - ---- + B R Twhere B is a constant. For example if you double the number of moles of the reactants and products, the value for dH doubles, resulting in the new equilibrium constant having its value squared. The reasons behind the other rules follow similar arguments.

- Write the expression for the equilibrium constant K for a given chemical reaction equiation.
- Express the equilibrium constant for a chemical equation in terms of equilibrium constant(s) for one or more related equations.
- Calculate the equilibrium constant for a chemical equation from equilibrium constant(s) for one or more related equations.

**If the equilibrium constant for the reaction equation**I is 0.010, what is the equilibrium constant for the reaction equation_{2}= 2 I2 I = I _{2}?**Hint...**

Review rule 1 please**Answer***100*

**Consider...**

You have learned rule 1.**If the equilibrium constant is 1E9 for the reaction equation**H what is the equilibrum constant for the reaction equation_{2}+ (1/2)O_{2}= H_{2}O,2 H _{2}+ O_{2}= 2 H_{2}O?**Hint...**

Review rule 2**Answer***1e18*

**Consider...**

The value may be higher, but the computer can not handle a very high value.**If the equilibrium constant K1 = 1e11 for**H and the equilibrium constant K2 = 6e-8 for_{2}SO_{4}= H^{+}+ HSO_{4}^{-},HSO what is the equilibrium constant K_{4}^{-}= H^{+}+ SO_{4}^{2-},_{overall}forH _{2}SO_{4}= 2 H^{+}+ SO_{4}^{2-}?**Hint...**

The third equation is the sum of the first two. Apply rule 3 in the Instructions of this tutorial. Try K1*K2, K1 * K2, or K1K2.**Answer***1000*

**Consider...**

H_{2}SO_{4}is a strong acid; and K1 is very large. You have learned rule 3.**The equilibrium constant is a quantity depending on how the equation is written. Any change in the equation should be reflected in the expression for the equilibrium constant. True or false? (t/f)****Answer***f*

**Hint...**

Hope you are not serious!!**Answer***t*

**Consider...**

The statement is true. The three rules given here guide your manipulation of the K's.