Calculating equilibrium constants: adding two or more equations If two or more reactions are added to give another, the equilibrium constant for the reaction is the product of the equilibrium constants of the equations added. 

How is this true? What is the reason?
This is a direct result from the law of mass action.

When you have an equilibrium

$$\ce{A &lt;=&gt; B}$$  

with the equilibrium constant 

$$K_1=\frac {[\ce{B}]}{[\ce{A}]}\tag1$$

and a second equilibrium

$$\ce{B &lt;=&gt; C}$$  

with the equilibrium constant 

$$K_2=\frac {[\ce{C}]}{[\ce{B}]}\tag2$$

you can combine both chemical equations:

$$\begin{align}
\ce{A \;&amp;&lt;=&gt; B}\\
\ce{B \;&amp;&lt;=&gt; C}\\
\hline
\ce{A \;&amp;&lt;=&gt; C}
\end{align}$$

Then you can rearrange Equation $\text(1)$

$$\begin{align}
K_1&amp;=\frac {[\ce{B}]}{[\ce{A}]}\tag1\\[6pt]
\Leftrightarrow [\ce{B}]&amp;=K_1\cdot[\ce{A}]
\end{align}$$

and substitute $[\ce{B}]$ in Equation $\text(2)$

$$\begin{align}
K_2&amp;=\frac {[\ce{C}]}{[\ce{B}]}\tag2\\[6pt]
K_2&amp;=\frac {[\ce{C}]}{K_1\cdot[\ce{A}]}\\[6pt]
\Leftrightarrow K_1\cdot K_2&amp;=\frac {[\ce{C}]}{[\ce{A}]}
\end{align}$$

Thus the equilibrium constant for the combined equilibrium is the product $K_2\cdot K_1$.