Electronic – How long will a DS1307 RTC run on a CR2032 coin cell battery

Like David says supercaps leak their charge to a certain extent, which is mainly a problem over longer periods of time. Let's make the required calculations ignoring the leakage.
The voltage drop across a capacitor at a constant current is given by

\$ \Delta V = \dfrac{I \cdot \Delta T}{C} \$

or rearranging for time:

\$ \Delta T = \dfrac{C \cdot \Delta V}{I} \$

\$V_{BAT}\$ is typically 3V, but for the given supercap is maximum 2.6V. Minimum for the RTC is 2V, so the allowable voltage drop is 0.6V. Filling in the other numbers this gives

\$ \Delta T = \dfrac{0.015F \cdot 0.6V}{500 nA} = 18000 s = \mbox {5 hours} \$

which isn't very long, but then you also chose a rather small supercap. A 1F/3V cap would increase your time to 23 days, but there we would have to take the cap's leakage into account, so in practice this may be around a week to a fortnight.

edit
Just picking the right RTC and supercap will dramatically improve longevity. The PCF2123 RTC can operate down to 1.1 V, and a PAS311HR supercap not only has a higher capacitance of 30 mF, but can also operate at 3.3 V. Then the equation becomes

\$ \Delta T = \dfrac{0.030F \cdot 2.2V}{110 nA} = 18000 s = \mbox {167 hours} \$

or just short of one week. A 1F/3.3V cap would be good for 7 months, or probably 2 to 3 months taking self-discharge into account.

The calculation is straightforward. The capacitor size is simply a question of how much voltage drop you can tolerate over the duration of the pulse. The average current from the battery is a function of the duty cycle.

ΔV = I × Δt / C

Solving for C gives:

C = I × Δt / ΔV

Let's assume you can allow ΔV = 0.1V. For your first example, this works out to:

C = 25 mA × 25 ms / 0.1 V = 6.25 mF

The average current draw is 25 mA * 25 ms / 2.5 s = 0.25 mA.

For the second example, the numbers work out to:

C = 50 mA × 100 ms / 0.1 V = 50 mF

Average current = 50 mA * 100 ms / 1.0 s = 5 mA.