A second-order step needs two molecules in the same place at the same instant. As the reactant thins, those meetings get rare fast — so a second-order reaction does not just slow down, it slows down and then slows down again. Integrated, that is $\dfrac{1}{[\mathrm{A}]} = \dfrac{1}{[\mathrm{A}]_0} + kt$: $[\mathrm{A}]$ is the concentration left at time $t$, $[\mathrm{A}]_0$ the concentration when the clock read zero, $t$ the elapsed seconds, and $k$ the rate constant in $\mathrm{L/(mol \cdot s)}$ — different units from every other order, which is the giveaway when a question refuses to say which order it is.

Work it as a straight line and it is easy: $1/[\mathrm{A}]$ starts at $1/[\mathrm{A}]_0$ and climbs by $k$ each second. Then — and this is where the marks go — flip it back. The law hands you the RECIPROCAL of the concentration; an answer of 20 is not 20 mol/L, it is 0.05 mol/L wearing its reciprocal upside down.

Now the contrast worth memorising. Ask this law when half is gone and $[\mathrm{A}]_0$ refuses to cancel: $t_{1/2} = \dfrac{1}{k\,[\mathrm{A}]_0}$ — half-life in seconds, $k$ in $\mathrm{L/(mol \cdot s)}$, and $[\mathrm{A}]_0$ the STARTING concentration, genuinely in the answer. Start with less and the half-life is longer. Worse: after one half-life the concentration IS halved, so the next half-life is twice as long, and the one after that twice again. First order's clock ticks evenly forever; second order's clock stretches as you watch it.