Initial rates for the reaction A+B→ products at a fixed temperature were measured: experiment 1 with [A]=0.10 M, [B]=0.10 M gave 2.0×10^-4 M s^-1…

Write $r=k[\mathrm{A}]^x[\mathrm{B}]^y$. Comparing experiments 1 and 2, doubling $[\mathrm{A}]$ quadruples the rate, so $x=2$. Comparing 1 and 3, doubling $[\mathrm{B}]$ doubles the rate, so $y=1$. Option A is correct. The overall order is $x+y=3$, so B is wrong. The rate constant is $k=\dfrac{2.0\times10^{-4}}{(0.10)^2(0.10)}=0.20\ \mathrm{M^{-2}\,s^{-1}}$, so C is correct. For D, with $[\mathrm{B}]$ constant at $1.0\ \mathrm{M}$ the rate law becomes pseudo-second order, $r=k'[\mathrm{A}]^2$ with $k'=k[\mathrm{B}]=0.20\ \mathrm{M^{-1}\,s^{-1}}$. For a second-order reaction $t_{1/2}=\dfrac{1}{k'[\mathrm{A}]_0}=\dfrac{1}{0.20\times0.10}=50\ \mathrm{s}$, not $5\ \mathrm{s}$, so D is wrong. The pseudo-order argument works only because the excess of $\mathrm{B}$ keeps $[\mathrm{B}]$ practically constant, so $k'=k[\mathrm{B}]$ absorbs it, and the integrated second-order law then fixes the half-life from the initial concentration of $\mathrm{A}$ alone.