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A substance A decomposes simultaneously by two parallel first-order paths, A→ B with k1=2×10^-3 s^-1 and A→ C with k2=3×10^-3 s^-1. Starting with 1…

Single correct option

A substance A decomposes simultaneously by two parallel first-order paths,

\mathrm{A\to B}

with

k_1=2\times10^{-3}\ \mathrm{s^{-1}}

and

\mathrm{A\to C}

with

k_2=3\times10^{-3}\ \mathrm{s^{-1}}

. Starting with

1

mol of A, the moles of B present after

277\ \mathrm{s}

(take

\ln2=0.693

) are

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Answer: (A) $0.30$

Overall decay constant

k=k_1+k_2=5\times10^{-3}\ \mathrm{s^{-1}}

, so

t_{1/2}=0.693/k=138.6\ \mathrm{s}

. In

277\ \mathrm{s}

(two half-lives)

75\%

of A has reacted. The products form in the ratio of rate constants, so B receives

k_1/k=2/5=0.4

of the reacted A:

n_\mathrm{B}=0.75\times0.4=0.30

mol.

0.75

is the total reacted and

0.40

is the branching fraction alone.

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