Answer: A, C, D
Here −d[A]/dt=2k[A]2, so 1/[A]=2+0.020t. Then t1/2=1/(2k[A]0)=100 s, [A]300=1/(2+6)=0.125 mol L−1 and the initial rate of disappearance of A is 2k[A]02=5.0×10−3. Since one P forms per two A, d[P]/dt=r=k[A]02=2.5×10−3, half of the quoted value.