Why do grammars in Chomsky Normal Form have derivations of length 2n-1?
Let $n$ be the length of a string. We start with the (non-terminal) symbol $S$ which has length $n=1$. 

Using $n - 1$ rules of form $(non-terminal) \rightarrow (non-terminal)(non-terminal)$ we can construct  a string containing $n$ non-terminal symbols.

Then on each non-terminal symbol of said string of length $n$ we apply a rule of form $(non-terminal) \rightarrow (terminal)$. i.e. we apply $n$ rules.

In total we will have applied $n - 1 + n = 2n - 1$ rules. 

example

Observe following grammar in Chomsky-normal form. 

$
\begin{align}
S &amp; \to AB \\
A &amp; \to BC | AC\\
A &amp; \to h|b\\
B &amp; \to a \\
C &amp; \to z \\
\end{align}
$

Consider following derivation

$
\begin{align}
\text{Current string} &amp; &amp; \text{rule applied} &amp; &amp; \text{#rules applied} &amp; &amp; \text{#length of string} \\
S &amp; &amp; \text{\\} &amp; &amp; 0 &amp; &amp; 1 \\
AB &amp; &amp; S \to AB &amp; &amp; 1 &amp; &amp; 2 \\
BCB &amp; &amp; A \to BC &amp; &amp; 2 &amp; &amp; 3 \\
\vdots &amp; &amp; \vdots &amp; &amp; \vdots &amp; &amp; \vdots \\
A\cdots CB &amp; &amp; \text{[multiple rules]} &amp; &amp; n-1 &amp; &amp; n
\end{align}
$

This last line represents a string containing only non-terminals. You can see that a string containing $n$ non-terminals is derived using $n-1$ rules. Let's continue. Applying $n$ rules of form $A \to a$ to each non-terminal in the string above gives you a string containing only terminals and thus a string from the language decided by the grammar. The length of the string has not changed (it's still $n$) but we applied an additional $n$ rules so in total we have applied $n-1 + n = 2n - 1$ rules.

While this explanation hopefully gives you an intuitive understanding, I think it would be an useful excercise to construct a formal proof using induction.