Concentration data for $A products$ (in $mol L^{- 1}$ ): $0.800$ at $t = 0$ , $0.400$ at $20 min$ , $0.200$ at $40 min$ , $0.100$ at $60 min$ . Take $ln 2 = 0.693$ . Select the correct statement(s).

AThe reaction is first order.
BThe concentration at $t = 50 min$ is the arithmetic mean of $0.200$ and $0.100$ , i.e. $0.150 mol L^{- 1}$ .
CThe rate constant is $0.0347 mi n^{- 1}$ .
DThe rate at $t = 40 min$ is $6.93 \times 1 0^{- 3} mol L^{- 1} mi n^{- 1}$ .

Answer: A, C, D

Constant successive half-lives (20 min each) show first order, with

k = 0.693/20 = 0.0347 mi n^{- 1}

. The rate at 40 min is

k [A] = 0.0347 \times 0.200 = 6.93 \times 1 0^{- 3}

. Concentration decays exponentially, so at 50 min it is

0.800 \times 2^{- 2.5} = 0.141 mol L^{- 1}

, not the arithmetic mean 0.150.