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For integrated rate laws, I have attempted to find the shelf-life and half-life of second-order reaction:
However, the answers I have obtained (which are labelled with a red cross) are very different from the correct answers (bottom-right corner). I am wondering which parts were done incorrectly.
Your mathematics is correct until you calculated for specific cases. Let's go back to your $n$th order version:
$$\frac{[\ce{A}]^{1-n}}{1-n} = -k_nt + \frac{[\ce{A}]_\circ^{1-n}}{1-n} \ \text{where } n \ne 1 \tag1$$
At this point, since you are working on second order kinetic, it is easy if you substitute $n = 2$ in equation $(1)$:
$$\frac{[\ce{A}]^{-1}}{-1} = -k_2t + \frac{[\ce{A}]_\circ^{-1}}{-1} $$
Once simplify, it becomes:
$$\frac{1}{[\ce{A}]} = k_2t + \frac{1}{[\ce{A}]_\circ} \tag2$$
For self life, substitute $t = t_{10\%}$ and ${[\ce{A}]} = 0.9{[\ce{A}]_\circ}$ in equation $(2)$:
$$\frac{1}{0.9[\ce{A}]_\circ} = k_2t_{10\%} + \frac{1}{[\ce{A}]_\circ} $$
$$\therefore \; t_{10\%} = \frac{1}{k_2}\left(\frac{1}{0.9[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.9} - 1 \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{9} \right) = \bbox[yellow]{\frac{0.11}{k_2[\ce{A}]_\circ}}$$
Similarly, for half life, substitute $t = t_{50\%}$ and ${[\ce{A}]} = 0.5{[\ce{A}]_\circ}$ in equation $(2)$:
$$\frac{1}{0.5[\ce{A}]_\circ} = k_2t_{50\%} + \frac{1}{[\ce{A}]_\circ} $$
$$\therefore \; t_{50\%} = \frac{1}{k_2}\left(\frac{1}{0.5[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.5} - 1 \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{0.5}{0.5} \right) = \bbox[yellow]{\frac{1}{k_2[\ce{A}]_\circ}}$$
