Where do you get this 0.750

In Step 1, I used the value of $\ln(2)$, which is approximately 0.6931. This is a fundamental constant in the half-life formula for first-order reactions:

$$t_{1/2} = \frac{\ln(2)}{k}$$

So, $k = \frac{0.6931}{1670 \text{ s}}$.

Perhaps you are thinking of a different calculation or a different problem. Sometimes, in kinetics, if $75\%$ of a reactant has reacted, it means $25\%$ (or 0.25) remains. This would typically occur after two half-lives. However, in this problem, the time given ($4710 \text{ s}$) is not exactly two half-lives ($2 \times 1670 \text{ s} = 3340 \text{ s}$).

The key values used were:

• $\ln(2) \approx 0.6931$
• Half-life ($t_{1/2}$) = $1670 \text{ s}$
• Time ($t$) = $4710 \text{ s}$