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JEE Advanced levelSingle correct optionDifficulty 3 of 3ck206

A decomposes by two parallel first-order paths giving B and C in the molar ratio $3 : 1$ . The half-life of A is $20.0$ min. Using $ln 2 = 0.693$ , the rate constant of the path forming C, in $mi n^{- 1}$ , is

A $2.60 \times 1 0^{- 2}$
B $1.73 \times 1 0^{- 2}$
C $8.66 \times 1 0^{- 3}$
D $3.46 \times 1 0^{- 2}$

Answer: C

Overall

k = 0.693/20.0 = 3.465 \times 1 0^{- 2} mi n^{- 1}

. In parallel paths the products form in the ratio

k_{1} : k_{2}

, so C receives

1/4

of the total:

k_{2} = 3.465 \times 1 0^{- 2} /4 = 8.66 \times 1 0^{- 3} mi n^{- 1}

(the value

2.60 \times 1 0^{- 2}

is

k_{1}

).

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