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JEE Main levelSingle correct optionDifficulty 2 of 3ck068

A second order reaction $2 A P$ obeys $- d [A] / d t = k [A]^{2}$ with $k = 0.025 L mo l^{- 1} s^{- 1}$ . Starting from $[A]_{0} = 0.80 mol L^{- 1}$ , the concentration of $A$ after $60 s$ is

A $0.179 mol L^{- 1}$
B $0.364 mol L^{- 1}$
C $0.435 mol L^{- 1}$
D $2.750 mol L^{- 1}$

Answer: B

Use

1/ [A] = 1/ [A]_{0} + k t = 1/0.80 + 0.025 \times 60 = 2.750 L mo l^{- 1}

. Therefore

[A] = 1/2.750 = 0.364 mol L^{- 1}

. Forgetting the final inversion gives 2.750, and using the first order equation gives 0.179.

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