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NEET levelSingle correct optionDifficulty 1 of 3ck060

For $2 A P$ with $- d [A] / d t = k [A]^{2}$ , the quantity $1/ [A]$ rises from $2.0 L mo l^{- 1}$ to $2.5 L mo l^{- 1}$ in $100 s$ . The value of $k$ is

A $2.2 \times 1 0^{- 3} L mo l^{- 1} s^{- 1}$
B $2.0 \times 1 0^{- 2} L mo l^{- 1} s^{- 1}$
C $5.0 \times 1 0^{- 3} L mo l^{- 1} s^{- 1}$
D $2.5 \times 1 0^{- 2} L mo l^{- 1} s^{- 1}$

Answer: C

For second order,

1/ [A]_{t} - 1/ [A]_{0} = k t

. Hence

k = (2.5 - 2.0) /100 = 0.0050 L mo l^{- 1} s^{- 1}

. The value 0.0022 (option with

2.2 \times 1 0^{- 3}

) comes from wrongly using

ln (2.5/2.0) /100

, the first order equation.

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