For the reaction, $$2N_2O_5 \to 4NO_2 + O_2$$, the rate equation can be expressed in two ways:
$$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$ and $$+\frac{d[NO_2]}{dt} = k'[N_2O_5]$$. k and k' are related as:

For the reaction $$2N_2O_5 \to 4NO_2 + O_2$$, the rate of reaction can be defined using the stoichiometric coefficients. The rate is given by:

$$ \text{rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} $$

This equation comes from the fact that for every 2 moles of $$N_2O_5$$ consumed, 4 moles of $$NO_2$$ are produced. Therefore, the rate of disappearance of $$N_2O_5$$ and the rate of appearance of $$NO_2$$ are related by the stoichiometry.

Now, the problem gives two expressions:

First, $$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$.

Second, $$+\frac{d[NO_2]}{dt} = k'[N_2O_5]$$.

From the stoichiometric relationship:

$$ -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} $$

Multiply both sides by 4 to simplify:

$$ 4 \times \left( -\frac{1}{2} \frac{d[N_2O_5]}{dt} \right) = 4 \times \left( \frac{1}{4} \frac{d[NO_2]}{dt} \right) $$

$$ -2 \frac{d[N_2O_5]}{dt} = \frac{d[NO_2]}{dt} $$

Using the first given expression, $$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$, multiply both sides by 2:

$$ 2 \times \left( -\frac{d[N_2O_5]}{dt} \right) = 2 \times k[N_2O_5] $$

$$ -2 \frac{d[N_2O_5]}{dt} = 2k[N_2O_5] $$

But from earlier, $$-2 \frac{d[N_2O_5]}{dt} = \frac{d[NO_2]}{dt}$$, so substitute:

$$ \frac{d[NO_2]}{dt} = 2k[N_2O_5] $$

The second given expression is $$\frac{d[NO_2]}{dt} = k'[N_2O_5]$$. Therefore:

$$ 2k[N_2O_5] = k'[N_2O_5] $$

Since the concentration of $$N_2O_5$$ is not zero, we can divide both sides by $$[N_2O_5]$$:

$$ 2k = k' $$

This means that $$k'$$ is twice $$k$$, or $$2k = k'$$.

Looking at the options:

A. k = k'

B. 2k = k'

C. k = 2k'

D. k = 4k'

Hence, the correct answer is Option B.