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JEE Advanced levelSingle correct optionDifficulty 3 of 3ck101

For a reaction of the type $A products$ , the concentration of $A$ is $0.80 mol L^{- 1}$ at $t = 0$ , $0.40$ at $t = 50 min$ and $0.20$ at $t = 150 min$ . The order and rate constant are

Asecond order, $k = 0.025 L mo l^{- 1} mi n^{- 1}$
Bsecond order, $k = 0.050 L mo l^{- 1} mi n^{- 1}$
Cfirst order, $k = 0.0139 mi n^{- 1}$
Dzero order, $k = 0.0080 mol L^{- 1} mi n^{- 1}$

Answer: A

Successive half-lives are 50 min (0.80 to 0.40) and 100 min (0.40 to 0.20), doubling as the concentration halves, so

t_{1/2} \propto 1/ a

and the order is 2. Then

k = 1/ (a t_{1/2}) = 1/ (0.80 \times 50) = 0.025 L mo l^{- 1} mi n^{- 1}

. Using

a = 0.40

would wrongly give 0.050.

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