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JEE Main levelSingle correct optionDifficulty 2 of 3ck036

The plot of $lo g_{10} [A]$ against time for a first order reaction is a straight line of slope $- 0.0125 mi n^{- 1}$ . The rate constant is (use 2.303)

A $0.0125 mi n^{- 1}$
B $0.0288 mi n^{- 1}$
C $0.00543 mi n^{- 1}$
D $0.00288 mi n^{- 1}$

Answer: B

The integrated law

lo g_{10} [A] = lo g_{10} [A]_{0} - k t /2.303

has slope

- k /2.303

. So

k = 2.303 \times 0.0125 = 0.0288 mi n^{- 1}

. Using the slope directly as

k

is a common slip.

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